Expansion Of Gas 6133b

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Expansion of Gas Basic facts: • Thermal expansion or contraction occurs in a gas similar to that in a liquid. As in the case of a liquid, the expansion of volume is only important. The discussion of the expansion of length or surface of gas is useless as a gas has no definite shape. • The amount of volume expansion in a gas is much larger compared to that in a solid or a liquid. Thus the thermal expansion of the solid vessel or container, which contains the gas, is usually neglected in comparison to the expansion of the gas. • Experiments show that all gases at very low densities exhibit the same expansion behaviour. A thermometer that uses a gas therefore shows the same readings irrespective of which gas is used. • The change in volume of a gas also occurs due to pressure applied on it. The role of pressure on the volume of liquid or solid is not important, in general. The volume of a gas increases when the pressure over it is decreased and it decreases when the pressure is increased. Ideal Gas and Gas Laws: The volume expansion in a gas occurs due to both temperature and pressure. To know the behaviour or the state of a certain quantity (mass) of a gas, we have to know its pressure ( P ), volume ( V ), and temperature ( T ). These three variables are related to one other. If one of the variables is kept constant and the other two are allowed to vary, we find three different relations. These are called gas laws. Within a certain range of temperatures and pressures, many gases have been found to follow three simple laws, namely Boyle’s law, Charles’ law and Gay-Lussac’s law. A gas that follows these laws completely is an idealization called an ideal gas. Note:

1. The temperature referred to in the gas laws is always the absolute temperature. 2. The pressure that is referred to is the pressure exerted due to the gas molecules on the inner walls of the vessel. In equilibrium, the pressure due to gas is equal to the external pressure on the gas. Boyle’s law:

When the temperature is held constant, the volume of a certain mass of a gas is inversely proportional to its pressure. The relation was discovered in 1660 by the British Chemist Robert Boyle (1627-1691).

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If V is the volume of a certain mass of a gas and P is the pressure, we can write 1 at constant temperature. according to Boyle’s law: V ∝ P Thus we have, PV = K (constant); the value of the proportionality constant K depends on the nature, mass and temperature of the gas. If the volume of the gas is V1 at pressure P1 , V2 at pressure P2 and so on, we can write, P1V1 = P2V2 =….= K .

Therefore, Boyle’s law can also be stated as follows: When temperature is held constant, the product of pressure and volume of a certain amount of gas is constant: PV = constant. The relation between density and pressure from Boyle’s law: Let us suppose, the mass of a certain amount of gas is m . The gas is held at constant temperature. The volume and density of the gas are V1 and ρ1 at pressure P1 and that are V2 and ρ 2 at pressure P2 . According to Boyle’s law, PV PV P1V1 = P2V2 Or, 1 1 = 2 2 [Dividing both sides by m ] m m P P m m Or, 1 = 2 [Q ρ1 = and ρ 2 = ] ρ1 ρ 2 V1 V2 P ∴ = const. Or, P ∝ ρ

ρ

Therefore, we can write: If the temperature is held constant, the density of a certain mass of a gas is proportional to its pressure. Graphical representation of Boyle’s law: Boyle’s law can be expressed by a number of ways through graphical plots. The plots involve pressure ( P ) and volume ( V ) at constant temperature. Thus they are called isotherms. • P − V graph: If we plot a graph considering the volume, V as abscissa and the pressure, P as ordinate keeping the temperature of a gas constant, we obtain a rectangular hyperbola. This is an isotherm. We get different isotherms (different rectangular hyperbolas) for different constant temperatures as shown in the following plot [fig.**].

Fig. to be included

•

P−

1 grap V

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1 as V ordinate keeping the temperature of the gas constant, we get a straight line ing through origin. Different straight lines (with different slopes) are obtained for different constant temperatures [see fig.**]. 1 The V − graphs will also be similar straight lines. P When the pressure P is plotted as abscissa and the inverse of volume

Fig. to be included

•

PV − P graph: When the pressure P is plotted as abscissa and the product PV as ordinate for a gas whose temperature is kept constant, we get a horizontal straight line (parallel to P -axis). Different parallel straight lines are obtained for different constant temperatures [fig.**]. The PV − V graphs will also be similar straight lines.

Fig. to be included Charles’ law:

When the pressure is held constant, the volume of a quantity (mass) of gas is directly proportional to the absolute temperature of it.

If the volume of a gas is V and its temperature in absolute scale isT , we can write according to Charles’ law: V ∝ T . V V V We can then write, = const. Or, 1 = 2 …and so on (at constant pressure). T T1 T2 If the volume of the gas is Vt at t 0 C and V0 at 0 0 C , then we can write, Vt V  t + 273  [Q t 0 C = (t + 273) K ] = 0 Or, Vt = V0   t + 273 273  273  t   Vt = V0 1 + ∴  ………………..(1)  273  Similarly, if the volume of the gas is Vt at − t 0 C , then we can write, t   Vt = V0 1 − .  273 

Therefore, Charles’s law can also be written in the following form:

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When the pressure is held constant, the volume of a certain mass of gas increases or 1 fraction of its volume at 0 0 C for one degree Celsius increase or decreases by 273 decrease of temperature.

In 1787, French physicist Jacques Charles (1747-1823), who made early hot air balloon, observed that the volume of a gas under constant pressure increases or decreases with temperature. This behaviour was quantified around 1808 by another French scientist, Joseph Gay-Lussac, who measured the thermal expansion of a gas as 1/267 of its original volume per degree Celsius. In 1847, Henri Regnault refined this value to 1/273, and also discovered that many gases violate this rule, which in principle holds only for so-called ideal gases. Gay-Lussac’s law:

When the volume is held constant, the pressure exerted by a gas is directly proportional to the absolute temperature of the gas.

This law is named after Joseph Louis Gay-Lussac as he made the observation in 1802. The above law is sometimes called pressure law. If P is the pressure exerted by the gas at temperature T measured in absolute scale, then according to Gay-Lussac’s law: P ∝ T . P P P We can then write, = const. Or, 1 = 2 …and so on (at constant volume). T T1 T2 If the pressure of the gas at t 0 C is Pt and at 0 0 C it is P0 , then we can write as before, Pt P t   = 0 Or, Pt = P0 1 +  ……………………(2) t + 273 273  273 

:

Boyle’s law: PV = const., when T is constant

V = const., when P is constant T P Gay-Lussac’s law: = const., when V is constant. T Charles’ law:

Examples with Solutions

Example#1 The volume of a gas of certain mass is 500 cc at standard temperature and pressure. What will be the volume at a pressure of 700 mm if the temperature is kept constant?

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Solution: Since the temperature is held constant, we get from Boyle’s law, P1V1 = P2V2 . Here, P1 = 76 cm Hg. (pressure of 76 cm long mercury column), V1 = 500 cc, P2 = 700 mm Hg = 70 cm Hg, V2 = ? 76 × 500 ∴ 76 × 500 = 70 × V2 Or, V2 = = 542.85 cc. 70 Example#2 During recording of pressure against volume of a certain mass of gas at a fixed temperature, a student forgets to write some reading. Fill in the blanks in the following recording table.

Pressure (mm) Volume (cc)

100 80

125 ----

200 40

---32

Solution: As the temperature is fixed, we can apply Boyle’s law: PV = constant. Here, in the 1st observation, P1V1 = 100 × 80 = 8000 and in the 3rd observation, P3V3 = 200 × 40 = 8000 . 8000 = 64 cc. Thus in the 2nd observation, P2V2 = 8000 Or, 125 × V2 = 8000 Or, V2 = 125 8000 = 250 cc. In the 4th observation, P4V4 = 8000 Or, P4 × 32 = 8000 Or, P4 = 32 1 Example#3 The volume of a gas is reduced to th of its initial volume by applying 6 pressure at constant temperature. If the initial pressure is equal to atmospheric pressure, what will be the final pressure? Solution: As the temperature remains constant, we get from Boyle’s law: P1V1 = P2V2 . 1 Here, P1 = 1 atm. (atmospheric pressure), V1 = x cc (say,), P2 = ?, V2 = x cc. 6 1 ∴ 1 × x = P2 × x Or, P2 = 6 atm. 6 Example#4 A bubble of 1 mm diameter is formed at the depth of 238 ft in a lake. What will be the diameter of the bubble when it reaches the surface of the lake? The temperature is same everywhere and the height of the water barometer is = 34 ft. [J.E.E. ‘00] Solution: 4 The volume of the bubble at the bottom of the lake, V1 = π .(0.05) 3 cc. 3 The pressure on the bubble at the bottom, P1 = 238 + 34 = 272 ft water column’s pressure. If the diameter of the bubble is d cm at the surface of the lake, the volume of the bubble 3

4 d  at that place, V2 = π   cc. 3 2

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The pressure on the bubble at that place is P2 = 34 ft water column’s pressure. The pressure inside the bubble is outside pressure on the bubble at equilibrium. Keeping in mind the temperature to be same everywhere, we can apply Boyle’s law: 3

4 d  4 π .(0.05) 3 = 34 × π   3 2 3 3 272 × (0.05) × 8 = 0.008 Or, d = 0.2 cm. Or, d 3 = 34 Example#5 An air bubble of volume 20 cc is formed at the 40 m depth of a lake. This bubble rises up the water surface. What will be its volume just before it comes up the water surface? (The standard atmospheric pressure = 76 cm Hg.) [H.S. ‘01] Solution. The volume of air bubble, V1 = 20 cc at the bottom of the lake. The pressure on the bubble at the bottom of the lake, P1 = (76 × 13.6 + 4000 × 1) × 980 dyne/cm 2 Let the volume of the air bubble just before it comes out of water surface is V2 . The pressure on the air bubble at this time, P2 = 76 × 13.6 × 980 dyne/cm 2 . As the temperature of the bubble does not change, we apply Boyle’s law, P1V1 = P2V2 Or, (76 × 13.6 + 4000) × 980 × 20 = 76 × 13.6 × 980 × V2 (76 × 13.6 + 4000) × 20 = 97.4 cc. Or, V2 = 76 × 13.6 P1V1 = P2V2 Or, 272 ×

Example#6 A movable piston is fitted at the upper end of a 100 cm long vertical cylinder. The cylinder contains an ideal gas. At the initial stage, when the system is at equilibrium with the piston between the confined gas and the atmosphere, the length of the column of gas in the cylinder is 90 cm. Now mercury is being poured on the piston from top. When the mercury is about to overflow, the piston goes down by 32 cm. Find the atmospheric pressure. Assume that the temperature of the gas remains constant. The thickness and the mass of the piston are negligible. [I.I.T.] Solution:

Fig. to be included Let us say, the atmospheric pressure = P cm Hg. At the initial stage at equilibrium, the pressure of the confined gas = the atmospheric pressure, P cm Hg. The initial volume of the gas, V = 90α cc [ α = area of cross-section of the cylinder, in sq. cm] When the mercury is about to overflow, the length of mercury column on the piston = 42 cm (see fig.**). The pressure of gas, P1 = ( P + 42) cm Hg, the volume of gas, V1 = 58α cc. As the temperature remains constant, we have from Boyle’s law: PV = P1V1 Or, P.90α = ( P + 42).58α Or, 90.P = 58.P + 58 × 42

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Or, 32 P = 58 × 42 Or, P =

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58 × 42 = 76.125 cm Hg. 32

Example#7 The volume of a gas is made double at constant pressure by increasing temperature. What is the final temperature of the gas if the initial temperature is 13 0 C ? [H.S.(T)] Ans. V V As the pressure is kept constant, we apply Charles’ law: 1 = 2 . T1 T2 According to question the final volume, V2 = 2.V1 . We have, T1 = 273 + 13 = 286 K , T2 = ? V1 2.V1 ∴ = Or, T2 = 572 K 286 T2

∴ In Celsius scale, the final temperature is = 572 – 273 = 299 0 C . Example#8 The volume of a certain mass of gas at 47 0 C is 640 cc and the pressure is 75 cm Hg. When will the pressure be double if the gas is heated keeping the volume unchanged? [H.S.] Ans. Here, P1 = 75 cm Hg, T1 = 273 + 47 = 320 K , and P2 = 2 × 75 = 150 cm Hg, T2 = ? Applying Gay-Lussac’s law (pressure law), 75 150 = Or, T2 = 640 K . In Celsius scale, t 2 = 640 − 273 = 367 0 C . 320 T2 Example#9 The temperature of a gas having volume 5 litre is changed from 0 0 C to 35 0 C and its volume is increased by 640 cc. Find the absolute zero temperature for the gas in Celsius scale. [H.S.] Ans. Let the absolute temperature for the gas is = − T 0 C . ∴ The initial temperature, T1 = 0 0 C = T K and the final temperature is T2 = 35 0 C = (T + 35) K . The initial and final volumes are V1 = 5000 cc and V2 = 5000 + 640 = 5640 cc. Following Charles’ law, V1 V2 5000 5640 = Or, Or, 500.(T + 35) = 564.T = T1 T2 T T + 35 17500 Or, 64.T = 17500 Or, T = = 273.43 64 ∴ The absolute temperature at Celsius scale is − 273.430 C . Example#10 Someone measures the air pressure of his car tyre to be 30 lb/sq. inches. At this time, the temperature of air is 27 0 C and the air pressure is 15 lb/sq. inches. Later in

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the afternoon, the person goes to a city where the atmospheric temperature is 12 0 C and the atmospheric pressure is 10 lb/sq. inches. How much reading will the pressure gauge show for the pressure of a car tyre at that time? Consider the volume of a car tyre to be the same in two cases. [J.E.E. ‘89] Ans. The pressure gauge measures a pressure that is something above the atmospheric pressure. Therefore, the actual air pressure inside the tyre in the first case is P1 = 30 + 15 = 45 lb/sq. inches. The temperature in the first case, T1 = 273 + 27 = 300 K . Let the reading in the pressure gauge in the second case = x lb/sq. inches. ∴ The actual pressure of air in the tyre is P2 = ( x + 10) lb/sq. inches. The temperature in this case is T2 = 273 + 12 = 285 K . As the volume remains constant, P1 P2 x + 10 45 × 285 45 = Or, Or, x + 10 = = 42.75 Or, x = 32.75 lb/sq. inches. = 300 T1 T2 300 285 Example#11 A hydrogen cylinder can withstand an internal pressure of 1000 lb/sq. inches. When the temperature is 15 0 C , the pressure of hydrogen gas in the cylinder is 240 lb/sq. inches. At what temperature the cylinder will about to explode? [J.E.E.] Ans. The initial pressure, P1 = 240 lb/sq. inches and temperature, T1 = 273 + 15 = 288 K . When the cylinder is about to explode, the pressure inside will be P2 = 1000 lb/sq. inches; T2 = ? As the volume of the gas remains constant in the cylinder, P1 P2 240 1000 288 × 1000 = = Or, Or, T2 = = 1200 K . T1 T2 288 240 T2 ∴ The required temperature in Celsius scale = 1200 – 273 = 927 0 C . Volume and Pressure Coefficients of Gas:

A certain mass of gas can be heated in two ways: • When the pressure is kept constant, the volume changes according to Charles’ law 1   .t  …………….(3) as given by equation (1): Vt = V0 1 +  273  • When the volume is kept constant, the pressure changes according to Gay1   Lussac’s law as given by equation (2): Pt = P0 1 + .t  …………(4)  273  Thus the equations (3) and (4) can be viewed as the expressions for thermal expansion of volume and pressure of the gas:

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Vt = V0 (1 + γ p .t ) ……………….(5)

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and

Pt = P0 (1 + γ v .t ) ………………...(6) Here, γ p is called the volume coefficient at constant pressure as the expansion of volume is considered when the pressure is held constant. Similarly, γ v is the pressure coefficient at constant volume as the increase of pressure is considered when the volume is held constant. Comparing equations (5) and (6) with equations (3) and (4), 1 γ p = γv = = 0.00366 / 0 C . 273 Note that t is the increase in temperature measured from 0 0 C . The definitions of the volume and pressure coefficients can be obtained from equations (5) and (6): V − V 0 ∆V γp = t ……….(7) and = V0 .t V0 .t P − P0 ∆P γv = t . ………..(8) = P0 .t P0 .t If we put V0 = 1 and t = 1 in (7), we have, γ p = ∆V . For a value of t = 1 , we have to consider the increase in temperature from 0 0 C to 1 0 C . Thus we have the following definition of volume coefficient. Definition of Volume coefficient ( γ p ): When the temperature of a certain mass of gas is increased

from 0 0 C to 1 0 C at constant pressure, the increase in volume of the gas for each unit volume is called the volume coefficient. Similarly, if we put P0 = 1 and t = 1 in (8), we have, γ v = ∆P . Thus we have the following definition of pressure coefficient. Definition of Pressure coefficient ( γ v ): When the temperature of a certain mass of gas is increased

from 0 0 C to 1 0 C at constant volume, the increase in pressure of the gas for each unit pressure is called the pressure coefficient. Examples with Solutions

Example#1 The pressure of a gas becomes 900 mm Hg when the temperature is raised to 50 0 C at constant volume. Find the pressure coefficient of the gas. Solution.

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The pressure of a gas at 0 0 C is 760 mm Hg (the standard temperature and pressure). We know, Pt = P0 (1 + γ v t ) , γ v = the pressure coefficient at constant volume. Here, Pt = 900 mm Hg, P0 = 760 mm Hg, t = 50 − 0 = 50 0 C . P − P0 900 − 760 140 ∴γv = t = = = 0.00368 / 0 C . P0 .t 760 × 50 760 × 50 Example#2 The volume of a certain amount of gas is 500 cc at 80 0 C at 600 cc at 150 0 C . What is the volume expansion coefficient of the gas? Solution. We know, Vt = V0 (1 + γ p t ) , where γ p is the volume expansion coefficient at constant

pressure and V0 is the volume at 0 0 C . ∴ 500 = V0 × (1 + γ p × 80) ……….(1) and

600 = V0 × (1 + γ p × 150) ………..(2) 6 1 + 150.γ p = Or, 6 + 480.γ p = 5 + 750.γ p 5 1 + 80.γ p 1 0 Or, 270.γ p = 1 Or, γ p = / C. 270

Dividing (2) by (1),

Example#3 A glass vessel contains air at 30 0 C . At what temperature has the vessel to 1 0 1 be heated, keeping pressure constant, so that fraction of air comes out? γ p = / C. 273 3 [H.S. ‘02] Ans. Method-1 Suppose, the volume of air at 0 0 C is V0 and the volume is V30 at 30 0 C where the pressure is kept constant. So, we can write, V30 = V0 (1 + γ p × 30) ………………….(1)

Let the required temperature is t 0 C . If the volume is Vt at this temperature, we can write, Vt = V0 (1 + γ p × t ) ……………..(2) Dividing (2) by (1) we get, Vt 1 + 30.γ p = ………………..(3) V30 1 + t.γ p 1 According to question, fraction of the volume of gas comes out when heated up to a 3 0 temperature, t C . 1 Hence, fraction of gas remains and this volume is equal to the volume of the vessel 3 which in turn is equal to the volume of gas at 30 0 C .

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V 2 3 Vt = V30 Or, t = ……….(4) V30 2 3 Comparing (3) and (4) we get, 3 1 + 30.γ p = Or, 3 + 90.γ p = 2 + 2.t.γ p 2 1 + t.γ p

Therefore, we can write,

Or, t =

1 + 90 × γ p 2×γ p

=

273 + 90 = 181.5 0 C . 2

Method-2 (In this method, we do not need to use the pressure coefficient, γ p .)

Suppose, the initial volume of air is V1 cc.; the temperature, T1 = 273 + 30 = 303 K . If the volume of air is V2 cc at a temperature of T2 K , the volume of emergent air =

1 V2 3

cc; the volume of air that remains in the vessel should be V1 cc. 1 3 ∴ V2 = V1 + V2 Or, V2 = V1 cc. 3 2 As the pressure remains constant, 3V / 2 V1 V2 V 909 Or, 1 = 1 Or, T2 = = 454.5 K . = 2 303 T1 T2 T2 ∴The required temperature in Celsius scale is = 454.5 – 273 = 181.5 0 C . Equation of State of an Ideal Gas (Ideal Gas Law):

Let us consider Boyle’s law: PV = constant, when the temperature T is constant and V Charles’ law: = constant, when the pressure P is constant. T When P , V and T all vary at the same time, we can combine Boyle’s law and Charles’ law to get, PV = const. T ∴We can write, PV = k .T , where the value of the constant k depends on the mass of the gas and on the units of P , V and T . If the volume of a gas is V1 at a pressure P1 and at temperature T1 and the volume is V2 at pressure P2 and at temperature T2 , we can write, P1V1 P2V2 . = T1 T2 Examples with Solutions

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Example#1 Some amount of gas is taken at normal temperature and pressure and then it is heated to make its pressure and volume double. Find the final temperature of the gas in Celsius scale. [H.S. ‘93] Solution. From the equation of state of an ideal gas we can write, P1V1 P2V2 = T1 T2 Here, P1 = 76 cm, V1 = x cc (say,) T1 = 273 K , P2 = 2 × 76 cm, V2 = 2 x cc, T2 = ? PV T 76 × 2 × 2 x × 273 ∴ T2 = 2 2 1 = = 1092 K . 76 × x P1V1 The final temperature in Celsius scale = 1092 – 273 = 819 0 C . Example#2 At a temperature of 27 0 C and a pressure of 76 cm Hg, 100 cc gas is collected over water. The volume occupied by the gas is saturated by water vapour. What is the volume of dry gas at Normal temperature and pressure (N.T.P.)? The highest pressure of water vapour at 27 0 C is = 17.4 mm Hg. Solution. Initially, the volume of the gas is V1 = 100 cc, the temperature, T1 = 273 + 27 = 300 K and the pressure of dry gas is P1 = 76 − 1.74 = 74.26 cm Hg. The temperature and pressure at N.T.P. are P2 = 76 cm Hg, and T2 = 0 0 C =273 K . Suppose, the volume of dry gas at N.T.P. is = V2 cc. We apply gas law, P1V1 P2V2 74.26 × 100 76 × V2 = = Or, T1 T2 300 273 74.26 × 273 Or, V2 = = 88.91 cc. 3 × 76

Avogadro’s hypothesis:

Avogadro’s hypothesis, formulated in 1811, states that equal volumes of gas at the same pressure and temperature contain equal numbers of molecules. One mole or one gram-mole:

One mole of a gas is the quantity that contains N A = 6.023 × 10 23 molecules (Avogadro’s number) and one gram-mole is the molecular weight of one mole gas expressed in grammes. Example: The mass of one mole of oxygen is 32.0 gm.

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Universal Gas Constant:

When one gram-mole or one mole gas is taken, the constant k is denoted by R . According to Avogadro’s hypothesis, one gram-mole of all gas at same temperature and pressure has same volume. Thus the constant R for any one gram-mole ideal gas will be the same. For this reason, R is called universal gas constant (or molar gas constant). Therefore, we can write for one gram-mole of any gas: PV = RT . If we take n -gram mole a gas, the ideal gas equation becomes: PV = nRT . If we take a gas of m -gram whose molecular weight is M , we have n = ∴ PV =

m . M

m RT . M

If we now compare the above with the equation of state PV = kT , we find k = m.

R ; the M

R is called specific gas constant. This is different for different gases as the M molecular weights of different gases are different.

constant, r =

Value of Universal Gas Constant: PV P0V0 For one gram-mole gas, R = = , where P0 = Normal pressure, T0 = Normal T T0 temperature and V0 = the volume of one gram-mole gas at normal temperature and pressure (N.T.P.). We know, P0 = 76 cm Hg = 76 × 13.6 × 980 dyne/cm 2 , T0 = 0 0 C = 273 Kelvin and

following Avogadro’s hypothesis, V0 = 22.4 litre/mole = 22400 cm 3 /mole. 76 × 13.6 × 980 × 22400 ∴R= = 8.31 × 10 7 dyne-cm/(mole Kelvin) 273 7 = 8.31 × 10 erg/(mole K) = 8.31 Joule/(mole K). The value of specific gas constant for different gases: Hydrogen: Molecular mass = 2

R 8.31 = = 4.16 J/(mole K) 2 2 Nitrogen: Molecular mass = 28 R 8.31 ∴r = = = 0.297 J/(mole K) 28 28 Oxygen: Molecular mass = 32 ∴r =

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R 8.31 = = 0.26 J/(mole K) 32 32 Carbon di-oxide: Molecular mass = 44 R 8.31 ∴r = = = 0.189 J/(mole K) 44 44 ∴r =

Boltzmann Constant:

The total number of molecules in a gas, N , is equal to the product of the number of moles of the gas and the number of molecules per mole (i.e., Avogadro’s number, N A ): N = nN A . Therefore, the ideal gas law for n -gram mole a gas can also be written in the following form: N PV = nRT = R.T = Nk B T , NA R = 1.38 × 10 − 23 J/K is known as Boltzmann constant. where k B = NA Examples with Solutions

Example#1 If the mass of 1 litre hydrogen at N.T.P. is 0.0896 gm, find the value of the universal gas constant R. Solution. At N.T.P., the volume of 0.0896 gm hydrogen is = 1000 cc. ∴At N.T.P., the volume of 1 gram-mole or 2 gm hydrogen , 1000 V= × 2 = 22321.4 cc. 0.0896 Normal (standard) pressure, P = 76 cm Hg = 76 × 13.6 × 980 dyne/cm 2 and normal temperature, T = 0 0 C = 273 K . PV 76 × 13.6 × 980 × 22321.4 ∴R= = = 8.28 × 10 7 erg/(mole Kelvin). T 273 Example#2 If the volume of 10 gm oxygen at 20 0 C and at 2 atmospheric pressure is 3.76 litre, find the universal gas constant R . Solution. We know, PV = nRT . Here, P = 2 atm. = 2 × 76 × 13.6 × 980 dyne/cm 2 , V = 3.76 litre = 3760 cc, 10 n= gm-mole, and T = 273 + 20 = 293 K . 32 PV 2 × 76 × 13.6 × 980 × 3760 ∴R= = 8.3 × 10 7 erg/(mole K). = 10 nT × 293 32

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Example#3 Find the gas constant k of 1 gm air. The density of air at NTP is = 1.293 gm/litre and the density of mercury = 13.6 gm/cc. [J.E.E. ‘96] Solution. PV If the gas constant for 1 gm air is k , we can write, k = . T Here, the volume of 1.293 gm air = 1000 cc. 1000 cc., the pressure P = 76 × 13.6 × 980 dyne/cm 2 , ∴ The volume of 1 gm air = 1.293 T = 273 K . 76 × 13.6 × 980 × 1000 ∴k = = 0.287 × 10 7 erg/(gm K). 273 × 1.293 Variation of Density with Pressure and Temperature:

Let us take an ideal gas of mass, m . If the volume of the gas is V1 and density, ρ1 at a temperature, T1 and the volume and density are V2 and ρ 2 respectively, at a temperature, m m T2 , we can write, V1 = and V2 = .

ρ1

ρ2

If P1 be the pressure of the gas at temperature, T1 and P2 be the pressure at temperature, T2 , we have, P1V1 P2V2 Pm Pm P P Or, 1 = 2 Or, 1 = 2 = T1 T2 ρ1T1 ρ 2T2 ρ1T1 ρ 2T2

∴

P = const. ρT

(i) When the temperature of the gas is changed, keeping the pressure constant, 1 ρT = const. Or, ρ ∝ T (ii) When the pressure of the gas is changed, keeping the temperature constant, P = const. Or, ρ ∝ P .

ρ

Examples with Solutions

Example#1 The temperature and atmospheric pressure on a hill are 7 0 C and 70 cm Hg, respectively; the temperature and pressure at the foot of the hill are 27 0 C and 76 cm Hg respectively. Compare the densities of air at the top and bottom of the hill.

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16 [H.S. ‘92]

Solution. P1 P = 2 ρ1T1 ρ 2T2 Here, P1 = 70 cm Hg, T1 = 273 + 7 = 280 K ; P2 = 76 cm Hg, T1 = 273 + 27 = 300 K . ρ 70 76 70 × 300 75 = ∴ Or, 1 = = ρ1 × 280 ρ 2 × 300 ρ 2 76 × 280 76 ∴ ρ1 : ρ 2 = 75 : 76.

We know,

Example#2 The density of air at N.T.P. is 1.29 gm/litre. Find the mass of 10 litre air at pressure of 5 times the normal pressure and at a temperature of 127 0 C . Solution. P P We know, 1 = 2 ρ1T1 ρ 2T2 Here, P1 = standard pressure = 76 cm Hg, T1 = 273 K , ρ1 = 1.29 gm/litre; P2 = 5 × 76 cm Hg, T2 = 273 + 127 = 400 K ρ T P 1.29 × 273 × 5 × 76 ∴ ρ2 = 1 1 2 = = 4.4 gm/litre P1T2 76 × 400 ∴ The mass of 10 litre air = 4.4 × 10 = 44 gm. Example#3 The density of Argon is 1.6 gm/litre at a temperature of 27 0 C and a pressure of 76 cm Hg. A 200 cc glass bulb is filled with Argon. If the pressure of Argon gas in the bulb is 75 cm Hg and the average temperature is 127 0 C , find the mass of Argon. Solution. Suppose, the density of argon at 75 cm Hg pressure and at 127 0 C temperature is ρ 2 gm/litre. P P We know, 1 = 2 . ρ1T1 ρ 2T2 Here, P1 = 76 cm Hg, ρ1 = 1.6 gm/litre, T1 = 273 + 27 = 300 K ; P2 = 75 cm Hg, T2 = 273 + 127 = 400 K . 1.6 × 300 × 75 76 75 ∴ Or, ρ 2 = = 1.184 gm/litre = 76 × 400 1.6 × 300 ρ 2 × 400 1.184 ∴ The mass of 200 cc Argon is = × 200 = 0.2368 gm. 1000 Example#4 The atmospheric pressure is 75 cm and the temperature is 27 0 C at some place. The atmospheric pressure is 70 cm and the temperature is 17 0 C at some other place. Compare the density of air at this two places. [J.E.E. ‘99] Solution.

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P1 P = 2 . ρ1T1 ρ 2T2 Here, P1 = 75 cm Hg, T1 = 273 + 27 = 300 K ; P2 = 70 cm Hg, T2 = 273 + 17 = 290 K . ρ 75 70 75 × 290 29 ∴ Or, 1 = = = ρ1 × 300 ρ 2 × 290 ρ 2 70 × 300 28 ∴ ρ1 : ρ 2 = 29 : 28

We know,

More Examples with Solutions

Example#1 A bubble of diameter 1 mm is created at the bottom of a lake. The diameter of the bubble becomes 2 mm when it comes to the water surface. If the air pressure is 76 cm Hg, what is the depth of the lake? The density of mercury is 13.6 gm/cc. [ H.S. ‘90] Solution. Let the depth of the lake = h cm. At the bottom of the lake, the pressure, P1 = atmospheric pressure + water pressure = 76 × 13.6 × 981 + h × 1 × 981 = (76 × 13.6 + h) × 981 dyne/cm 2 and 3

4  1  the volume, V1 = π   cc. 3  20  At the water surface, the pressure, P2 = the atmospheric pressure = 76 × 13.6 × 981 dyne/cm 2 and 3

4 1 the volume, V2 = π   cc. 3  10  According to Boyle’s law, P1V1 = P2V2 3

Or, (76 × 13.6 + h) × 981 ×

4  1  4 1 π   = 76 × 13.6 × 981 × π   3  20  3  10 

3

1 = 76 × 13.6 Or, h = 8 × 76 × 13.6 − 76 × 13.6 8 Or, h = 76 × 13.6 × (8 − 1) = 76 × 13.6 × 7 = 7235.2 cm = 72.352 m.

Or, (76 × 13.6 + h) ×

Example#2 During making an electric vacuum tube, it is sealed at 27 0 C and at a pressure of 12 × 10 −6 cm. The volume of the tube is 100 cc. Find the number of gas molecules that are left in the tube. The Avogadro number, i.e., the number of molecules [J.E.E.] of any gas in a volume, 22.4 litre at N.T.P. is 6.02 × 10 23 . Solution. Let us assume, there are n - molecules that are left in the sealed tube. 6.02 × 10 23 molecules/cc. The density of gas at N.T.P. = 22.4 × 10 3

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We know,

18

P1 P = 2 ρ1T1 ρ 2T2

Here, P1 = 12.6 × 10 −6 cm Hg, ρ1 =

n molecules/cc, T1 = 273 + 27 = 300 K ; 100

6.02 × 10 23 molecules/cc, T1 = 273 K . 22.4 × 10 3 1.2 × 10 −6 × 100 76 × 22.4 × 10 3 ∴ = n × 300 6.02 × 10 23 × 273 1.2 × 10 −6 × 100 × 6.02 × 10 23 × 273 Or, n = = 3.86 × 1013 molecules. 3 300 × 76 × 22.4 × 10 P2 = 76 cm Hg, ρ 2 =

Example#3 During making an electric bulb of volume, 250 cc has been sealed at a temperature of 27 0 C and at a pressure of 10 −3 mm. Find the number of molecules in this bulb. (Avogadro number = 6.0 × 10 23 ). [H.S. ’99; I.I.T.] Solution. The volume of air in the bulb, V1 = 250 cc, the air pressure inside the bulb, P1 = 10 −3 mm Hg = 10 −4 cm Hg, the temperature, T1 = 27 + 273 = 300 K . Let there are n -molecules inside the bulb. n ∴The density of air in the bulb is ρ1 = molecules/cc. 250 At normal temperature and pressure (N.T.P.), that is at a pressure, P2 = 76 cm Hg, and at

a temperature, T2 = 273 K the volume, V2 = 22.4 litre = 22400 cc. contains 6.0 × 10 23 number of molecules. 6.0 × 10 23 ∴ The density of air at NTP is ρ 2 = molecules/cc. 22400 P P We know, 1 = 2 ρ1T1 ρ 2T2 10 −4 × 250 76 × 22400 = ∴ n × 300 6.0 × 10 23 × 273 10 −4 × 250 × 6.0 × 10 23 × 273 25 × 6 × 273 = × 1016 = 8.018 × 1015 . Or, n = 300 × 76 × 22400 3 × 76 × 224 Example#4 If the temperature of a gas is increased from 15 0 C to 25 0 C at fixed pressure, the volume increases in the ratio 1:1.035. Determine the absolute zero temperature in Celsius scale. Solution. Let the absolute zero temperature is x 0 C below the freezing point (0 0 C ). Then the temperature of 15 0 C in absolute scale will be T1 = (15 + x) K and the temperature of 25 0 C in absolute scale will be T2 = (25 + x) K . According to Charles’ law,

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V1 V2 V T 15 + x Or, 1 = 1 = = T1 T2 V2 T2 25 + x 1 15 + x Or, 25 + x = 1.035 × (15 + x) Or, x = 270.7 = Or, 1.035 25 + x ∴The absolute temperature is = -270.7 0 C .

Example#5 A 5 litre and a 3 litre container contain air at 3 times and 7 times atmospheric pressure, respectively. What will be the common air pressure in two containers when they are ed by a narrow pipe? [I.I.T.] Solution. Let us assume, there are n1 and n 2 gm-moles of air in the two containers. Therefore, using PV = nRT formula we can write, 5 × 3 = n1 RT ; 3 × 7 = n1 RT . Adding the above two expressions, 15 + 21 = (n1 + n 2 ) RT ∴ 36 = (n1 + n2 ) RT …………….(1) When the two containers are connected by pipe, there will be total (n1 + n 2 ) gm-moles of air in the system of total volume, (5+3) = 8 litre, at the same temperature. Let the common pressure be P atm ( P times the atmospheric pressure). Then we can write, P × 8 = (n1 + n 2 ) RT …………….(2) Dividing (1) and (2), 36 P × 8 = 36 Or, P = = 4.5 atm. 6 Example#6 Two bulbs of equal volume are filled with a gas at NTP after they are ed by a narrow pipe. Now, one bulb is kept in melting ice and the other is immersed in water of 62 0 C . What will be the pressure of gas? Ignore the volume of t he pipe? [I.I.T ‘85] Solution. Let the volume of each bulb is V cc and there are n gram-mole gas in each of them. Since, the bulbs are filled with the gas at NTP, we can write for each of the bulb, 76 × V = nR × 273 [Q PV = nRT ] 76 × V ∴Total number of gm-moles in the two bulbs is 2n = ×2. R × 273 If the common pressure in the system in second case is P , PV the number of gram-moles in one bulb, n1 = and R × 273 PV . the number of gram-moles in the other bulb, n2 = R × (62 + 273) We can write, n1 + n2 = 2n . 76 × V PV PV ×2 ∴ + = R × 273 R × (62 + 273) R × 273

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1  76 × 2  1 + Or, P = 273  273 335 

Or, P =

20

76 × 2 × 273 × 335 = 84.36 cm Hg. 273 × (273 + 335)

Example#7 An air bubble is rising from the bottom of a lake. The diameter of the bubble at the bottom is 3.6 mm and at the top it is 4 mm; the depth of the lake is 2.5 m and the temperature above is 40 0 C . What is the temperature at the bottom of the lake? Neglect the variation of density of water with depth. Atmospheric pressure = 76 cm Hg [I.I.T.] and g = 980 cm/s 2 . Solution. The pressure at the bottom of the lake, P1 = 76 × 13.6 × 980 + 350 × 1 × 980 = (76 × 13.6 + 350) × 980 dyne/cm 2 . The volume of the bubble at the bottom of the lake, 4 V1 = π (0.18) 3 cc; the temperature, T1 = ? 3 4 At above, the pressure, P2 = 76 × 13.6 × 980 dyne/cm 2 , the volume, V2 = π (0.2) 3 cc; 3 the temperature, T2 = 273 + 40 = 313 K . PV PV We know, 1 1 = 2 2 T1 T2 4 4 (76 × 13.6 + 250) × 980 × π (0.18) 3 76 × 13.6 × 980 × π (0.2) 3 3 3 Or, = T1 313

1283.6 × (0.18) 3 × 313 1283.6 × (0.18) 3 1033.6 × (0.2) 3 Or, Or, T1 = = T1 313 1033.6 × (0.2) 3 ∴ T1 = 283.37 K ; the temperature in Celsius scale is = 283.37 – 273 = 10.37 0 C . Example#8 A balloon can lift a total 175 kg of weight at NTP. During the rise, when the barometer reads 50 cm and the temperature is -10 0 C , how much weight the balloon may lift? Assume the volume of the balloon to be constant. Solution. Suppose the volume of the balloon is = V .cc According to Archimedes’ principle, the volume of the displaced air is V cc. If the density of air is ρ1 kg/cc at NTP, we can write, 175 Vρ1 = 175 Or, ρ1 = V Let us assume that at the given height, the balloon can carry a mass of M kg. If the M density of air at that height be ρ 2 kg/cc, we can write, Vρ 2 = M Or, ρ 2 = V P P P1 P2 We know, 1 = 2 Or, = ρ1T1 ρ 2T2 175 × T1 M × T2

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76 50 = [Q T1 = 0 0 C = 273 K ; T2 = −10 0 C = 263 K ] 175 × 273 M × 263 175 × 273 × 50 Q M = = 119.5 kg. 76 × 263 Or,

Example#9 If a tyre is filled with air at 27 0 C and then the temperature is raised to 57 0 C , what will be the percentage increase of pressure inside the tyre? Solution. In the first case, the pressure = P1 , the temperature, T1 = 273 + 27 = 300 K . In the second case, the pressure = P2 , the temperature, T2 = 273 + 57 = 330 K . As the volume remains constant, we can write, P1 P2 P − P1 P − P1 T2 − T1 = 2 Or, 2 . = = T1 T2 T2 − T1 P1 T1 Here the quantity, ( P2 − P1 ) is the increase of pressure. P − P1 ∴The percentage increase in pressure = 2 × 100% P1 T −T 330 − 300 × 100% = 10% . = 2 1 × 100% = T1 300 Example#10 A compartment contains a gas of mass m1 at pressure, P1 and another compartment contains the gas of mass m2 at pressure, P2 . If a age is created between the two compartments, find the pressure of the gas mixture. Solution. Let the volumes of the 1st and 2nd compartments are V1 and V2 , respectively. In the 1st compartment, there is gas of mass m1 at a pressure, P1 and in the 2nd compartment, there is gas of mass m 2 at a pressure, P2 . Therefore, we can write, m m P1V1 = 1 RT and P2V2 = 2 RT at a constant temperature, T ; where R = universal M M gas constant and M = molecular weight of the gas. (m + m2 ) ∴ P1V1 + P2V2 = 1 RT ……………….(1) M After mixture, the total volume = (V1 + V2 ) and the total mass = (m1 + m2 ) . If the pressure of the gas mixture is now P , we can write, (m + m2 ) P (V1 + V2 ) = 1 RT …………………(2) M Comparing (1) and (2), P V + P2V2 ………………….(3) P= 1 1 V1 + V2 m m Again we have, V1 = 1 RT and V2 = 2 T . MP1 MP2

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∴The expression (3) becomes m1 m RT + 2 RT (m1 + m2 ) (m1 + m2 ) P1 P2 M = = . P= M m1 m2 m1 m2 m1 P2 + m2 P1 + RT + RT MP1 MP2 P1 P2 Example#11 The length of a capillary glass tube, having two sides sealed, is 100 cm. In horizontal position, a mercury column of 10 cm is in the middle of it. On the two sides of this mercury column, there are air columns (of same length) at a pressure of 76 cm Hg and at a temperature of 27 0 C . If now the tube is kept horizontal in such a way that the temperature of the air column on one side becomes 0 0 C and the temperature of the air column on the other side becomes 127 0 C . Find the length and pressure of the air column at 0 0 C . Neglect the thermal expansion of glass and mercury. [I.I.T.] Solution. Let the area of cross-section of the tube = α cm. ∴The volume of mercury column = 10α cc, the volume of air column, in the first case, on either side of mercury column, V1 = 45α cc. [see fig**]

Fig. to be included In the first case, the temperature, T1 = 273 + 27 = 300 K ; the air pressure, P1 = 76 cm Hg. In the second case, the air temperature at the left end is T2 = 0 0 C = 273 K . If the length of the air column is l cm, the volume of air on the left, V2 = lα cc. Suppose the air pressure in the second case is P (Since the system is in equilibrium, the air pressures on two sides of mercury column are same.). Applying the following formula, P1V1 P2V2 , we get, = T1 T2 76 × 45α P × lα ………………..(1) = 300 273 Now the air temperature at the right end is 127 0 C = 273 + 127 = 400 K . The length of this air column is (90 − l ) cm. Thus we can write, 76 × 45α P × (90 − l )α = ………………(2) 300 400 Comparing (1) and (2) we get, P × lα P × (90 − l )α 90 × 273 = Or, l = = 36.5 cm. 273 400 673 From equation (1),

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76 × 45α P × 36.5α = Or, P = 85.3 cm Hg. 300 273 ∴The length of air column at 0 0 C is = 36.5 cm and the pressure is = 85.3 cm Hg. Example#12 Two thermally insulated containers of volume 1 litre and 3 litre are connected by a pipe. The first container is filled with Nitrogen at a temperature, 0 0 C and at a pressure, 0.5 atm and the second container is filled with Argon at a temperature, 100 0 C and a pressure, 1.5 atm while keeping the tap on the connected pipe closed. If now the tap is made open, the temperature of the gas mixture becomes 79 0 C . What will be the pressure of the gas mixture? Solution. In the first container, the pressure of gas, P1 = 0.5 atm, the volume, V1 = 1 litre, the temperature, T1 = 273 K . In the second container, the pressure, P2 = 1.5 atm, the volume, V2 = 3 litre, the temperature, T2 = 100 + 273 = 373 K . The total volume of the gas mixture, V = 1 + 3 = 4 litre and the temperature, T = 273 + 79 = 352 K . Suppose, the pressure of the gas mixture after the tap is open is = P . Applying ideal gas law we can write, P1V1 P2V2 PV 0.5 × 1 1.5 × 3 P × 4 Or, + = + = T1 T2 T 273 373 352 P = 0.0018 + 0.0120 = 0.0138 Or, 88 ∴ P = 0.0138 × 88 = 1.2144 atm. Example#13 A 3 litre and a 1 litre glass bulb are ed by a narrow pipe. This system is filled with air at a temperature of 30 0 C and a pressure of 76 cm. The 3 litre bulb is immersed in vapour of 100 0 C and the other one is kept at 30 0 C . What will be the air pressure in the two bulbs? Neglect the expansion of 3 litre bulb. [J.E.E.] Solution. In the first case, the total volume of the two bulbs, V1 = 3 + 1 = 4 litre, the air pressure, P1 = 76 cm Hg, air temperature, T1 = 273 + 30 = 303 K . In the second case, the air pressure = P2 cm Hg (say,). The temperature of air in the 3 litre bulb = 100 + 273 = 373 K , the temperature of air in the 1 litre bulb = 30 + 273 = 303 K . PV We know the ideal gas law is = nR , where n is number of moles in the gas and R is T the universal gas constant. As the total number of moles in the gas remains constant, we have the total number of moles of gas in the system in the first case should be equal to the sum of moles of gas in the two bulbs kept at two different temperatures: n = n1 + n 2 We have here,

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P ×3 P ×1 P1V1 , n1 R = 2 and n 2 R = 2 T1 373 303 PV P × 3 P2 × 1 76 × 4 1   3 ∴ 1 1 = 2 = P2  + Or, +  T1 303 373 303  373 303  76 × 4 × 373 Or, P2 = = 88.4 cm Hg. 3 × 303 + 373 nR =

Example#14 One end of a narrow and uniform tube is sealed. Some air is trapped inside the tube with the help of a mercury column of length h cm. If the pipe is kept vertical with the closed end up, the length of air column is l1 cm. When the tube is held vertical keeping the open end up, the length of air column is l 2 cm. What is the air pressure? Solution. Fig. to be included

Suppose, the air pressure = P cm Hg; the area of cross-section = α sq. cm. In the case when the closed end of the tube is on the upper side (see fig.**): the volume of enclosed air, V1 = l1α cc; the pressure of this enclosed air pressure, P1 = ( P − h) cm Hg. In the other case, when the open end is up, the volume of enclosed air, V2 = l 2α cc; the pressure of this enclosed air pressure, P2 = ( P + h) cm Hg. According to Boyle’s law, P1V1 = P2V2 Or, ( P − h)l1α = ( P + h)l 2α l +l ∴ P = 1 2 .h cm Hg. l1 − l 2 Example#15 Some air is trapped inside a one end open tube by a 20 cm long mercury column. The length of the air column becomes 3 cm when the tube is held vertical, keeping the open end up. When the tube is held upside down, the air column becomes 6 cm. Find the atmospheric pressure. Solution. Let the atmospheric pressure = P cm Hg; the area of cross-section = α sq. cm. In the first case, the pressure of enclosed air = ( P + 20) cm Hg; the volume = 3α cc. In the second case, the pressure of enclosed air = ( P − 20) cm Hg; the volume = 6α cc. ∴According to Boyle’s law: ( P + 20) × 3α = ( P − 20) × 6α , solving we get, P = 60 cm Hg. Example#16 A one side open glass tube of uniform cross-section contains some air at 27 0 C which is confined by 4 cm long mercury column. If the tube is held vertical, the length of air column becomes 9 cm when the open side is up and the air column is 10 cm

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when it is held upside down; find (i) the atmospheric pressure, (ii) the temperature when the air column is 9 cm in upside down position. [I.I.T.] Solution. Fig. to be included (i) Let the atmospheric pressure is P cm Hg, the area of cross-section is α sq. cm. When the open side of the tube is up in a vertical position (see fig.**), the volume of enclosed air, V1 = 9α cc, the pressure of enclosed air, P1 = ( P + 4) cm Hg. When the open side is below, the volume of enclosed air, V2 = 10α cc, the pressure of enclosed air, P2 = ( P − 4) cm Hg. According to Boyle’s law, P1V1 = P2V2 Or, ( P + 4) × 9α = ( P − 4) × 10α Or, 9 P + 36 = 10 P − 40 Or, P = 76 cm Hg. (ii) Suppose, the temperature in this case is = T2 K . We have, the pressure, P2 = 76 − 4 = 72 cm Hg, the volume, V2 = 9α cc. PV PV (76 + 4) × 9α 72 × 9α 80 72 ∴ 1 1 = 2 2 Or, = Or, = Or, T2 = 270 K . T1 T2 273 + 27 T2 300 T2 ∴ The required temperature in Celsius scale is = 270 – 273 = − 3 0 C . Example#17 A glass tube of uniform cross-section has its both sides sealed. In the horizontal position, there is a 5 cm long mercury column in the middle of the tube and the air columns are on two sides having same length. The pressure of this air is P . The tube is now made inclined by 60 0 with respect to the vertical. In this position, the lengths of air column above the mercury column and that below the mercury column are 46 cm and 44.5 cm, respectively. Find the value of P . The temperature of the system remains fixed at 30 0 C . [I.I.T. ‘86] Solution. Fig. to be included

Let in the horizontal position, the length of each air column on two sides is = l cm. This implies, l1 = l 2 = l according to the picture (see fig.**). The length of the tube = 44.5 + 46 + 5 = 95.5 cm. 95.5 − 5 ∴l = = 45.25 cm. 2 1 In the inclined position, the air pressure, P2 = P + 5 cos 60 0 = P + 5. = ( P + 2.5) cm Hg. 2 If the volume of air above is V1 and the volume of air below is V2 , we can write according to Boyle’s law, P1V1 = P2V2 . ∴ P1 .46α = ( P1 + 2.5) × 44.5α [ α = area of cross-section of the tube] Solving, P1 = 74.17 cm Again we can write, PV = P1V1 Or, P × 45.25α = 74.17 × 46α

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Or, P =

26

74.17 × 46 = 75.4 cm Hg. 45.25

Example#18 When the barometer reading is 75 cm, an amount of 10 cc air is inserted in the empty space of the barometer in atmospheric pressure. The mercury column of the barometer goes down by 25 cm due to this. How much space is occupied by the air inside the barometer tube? [H.S.] Solution. Initially, the pressure of the confined air, P1 = 75 cm Hg; the initial volume of air, V1 = 10 cc. In the second case, the volume of air = V2 (say,) and the pressure, P1 = 75 − 25 = 50 cm Hg. PV 75 × 10 = 15 cc. According to Boyle’s law, P1V1 = P2V2 Or, V2 = 1 1 = P2 50 ∴ The air will occupy a volume of 15 cc in the barometer tube. Example#19 The reading of a faulty barometer comes down from 75 cm to 65 cm due to air gets into the empty space of it. The length of the empty column, up in the barometer tube, was 6 cm at the initial stage. If the cross-section of the tube is 1 sq. cm, how much space will be occupied by enclosed air at normal atmospheric pressure? [H.S.] Solution. Normal atmosphere, P1 = 76 cm Hg Let us say, the volume of enclosed air at that pressure is = V1 cc Pressure of enclosed air, P2 = 75 − 65 = 10 cm Hg; Volume of enclosed air, V2 = {6 + (75 − 65)} × 1 = 16 cc. PV 10 × 16 = 2.105 cc. Applying Boyle’s law: P1V1 = P2V2 Or, V1 = 2 2 = P1 76 Thus the enclosed air occupies 2.105 cc volume at normal pressure. Example#20 When the readings of an errorless barometer are 28.5 in and 31 in, the corresponding readings of the faulty barometer are then 28 in and 30 in. What will be the reading of the errorless barometer when the reading of the faulty barometer is 29 in? [H.S. ’03; J.E.E.] Solution. Suppose, the length of the tube of barometer = l in and the area of cross-section = α in. In the first case, we can write, the volume of enclosed air in the faulty barometer tube, V1 = (l − 28) × α cubic in; the pressure of this enclosed air, P1 = (28.5 − 28) = 0.5 in Hg. In the second case, the volume of enclosed air in the faulty barometer tube, V2 = (l − 30) × α cubic in; the pressure of this enclosed air, P2 = (31 − 30) = 1 in Hg. Applying Boyle’s law: P1V1 = P2V2 , we get, (l − 28)α × 0.5 = (l − 30)α × 1 Or, l = 32 in. Let the reading in errorless barometer be h in when the faulty barometer reading is 29 in. ∴The volume of enclosed air = (l − 29) × α cubic in; the pressure = (h − 29) in Hg.

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Again applying Boyle’s law, (l − 30) × α × 1 = (l − 29) × α × (h − 29) Or, (32 − 30) × α × 1 = (32 − 29) × α × (h − 29) Or, 2 = 3 × (h − 29) 2 Or, h = 29 + = 29.67 in. 3 ∴The required reading in the errorless barometer is 29.67 in.

Note that in a faulty barometer, air introduced in the empty space, above the

mercury column, forces the mercury column to go down. So, the reading in a faulty barometer will be somewhat less than the correct reading. The difference in readings is equal to the pressure (in of length of mercury column) exerted by air column.

Example#21 A small thread of mercury separates some air from surroundings in a tube. The tube can be rotated in the vertical plane. The length of air column in tube is l1 when the tube is horizontal and it is l 2 in the vertical position (the air column is up) of the tube. If the tube makes an angle α with the vertical, find the length of air column in it. [J.E.E. ‘88] Solution. Suppose, the atmospheric pressure = P , the length of the mercury thread = h , the area of cross-section of the tube = A . In the horizontal position of the tube, the air pressure in the tube = P ; volume = l1 A . In the vertical position, the air pressure in the tube = P − h ; volume = l 2 A . Let the length of air column is l3 when the tube makes an angle α with the vertical. According to Boyle’s law, P × l1 A = ( P − h) × l 2 A ………..(1) P × l1 A = ( P − h cos α ) × l3 A ………..(2) l1 l h h Or, = 1 − 1 ……(3) = 1− l2 P P l2 l P 1 …….(4) = From (2), Pl1 = ( P − h cos α )l3 Or, 3 = h l1 P − h cos α 1 − cos α P Now putting (3) in (4) we get, l1l 2 l1 . l3 = = l 2 − (l 2 − l1 ) cos α  l1  1 − 1 −  cos α  l2 

From (1), Pl1 = ( P − h)l 2 Or,

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Example#22 When a barometer reads 75 cm, the length of Torricelli’s vacuum is 20 cm. When 10 cc air is introduced into the barometer tube at atmospheric pressure, the barometer reading is 45 cm. Find the diameter of the barometer tube. [J.E.E. ‘95] Solution. Let the area of cross-section of the barometer tube = A sq. cm, the diameter of the barometer tube = d cm. We have the initial volume of air, V1 = 10 cc and the pressure of air, P1 = 75 cm Hg; the final volume, V2 = [20 + (75 − 45)]× A = 50 A cc and the final pressure, P2 = 75 − 45 = 30 cm Hg. According to Boyle’s law, P1V1 = P2V2 Or, 75 × 10 = 30 × 50 A 75 × 10 1 Or, A = = sq. cm. 30 × 50 2 2 πd 2 1 ∴ = Or, d = = 0.798 cm. 4 2 π Discussions of a few Questions Q.1 For expansions of solid and liquid, only temperature is referred whereas in the case of volume expansion of a gas, temperature and pressure both are mentioned. Why? Ans. When temperature is changed, volume of all kinds of matter -solid, liquid, gas changes. Thus the temperature is mentioned in case of volume expansion. For solid and liquid, the influence of pressure in the expansion of volume is negligible. However, for a gas, volume is dependent on pressure. The volume of a gas decreases as the pressure is increased and the volume increases as the pressure is decreased. Hence, the volume of a gas is dependent on both pressure and temperature. For this reason, temperature and pressure both are mentioned in the case of expansion of volume of a gas. Q.2 Liquid has a coefficient of apparent expansion whereas for a gas such a coefficient is not mentioned. Why? Ans. Liquid is heated in a container. Along with the expansion of liquid, there is an expansion of the solid container too. The coefficient of expansion of solid is less but not negligible in comparison to the coefficient of expansion of liquid. Thus the expansion of the container is not ignored. If we do not consider the expansion of the container, the coefficient of expansion of the liquid is called the coefficient of apparent expansion. When the coefficient of expansion of the container is added with the coefficient of expansion of liquid, we get the coefficient of real expansion of liquid. Like a liquid, a gas is also heated keeping it in a container. But the expansion coefficient of a gas is approximately 100 times the expansion coefficient of a solid container. Thus the expansion of container is ignored with respect to the expansion of gas. Hence, it is assumed that the coefficients of real and apparent expansions of gas are essentially the same unless a very accurate measurement is required. For a gas, only one coefficient of expansion is mentioned instead of using ‘real’ and ‘apparent’ words.

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Q.3 Two similar spherical glass bulbs , filled with air, are ed by a horizontal glass tube which contains a small thread of mercury. The air temperatures in the two bulbs are 0 0 C and 20 0 C . If now the temperatures of both the bulbs are increased by 10 0 C , will there be any change of position of the mercury thread? If so, then in which direction will it shift? [J.E.E. ‘88] Ans. The mercury thread will shift towards the bulb which has higher temperature. We can assume that the mercury thread was in equilibrium in the horizontal tube at the initial stage. Let the common pressure at this stage is P . When the temperatures of the bulbs are increased by 10 0 C , the temperatures of the bulbs are now 0 + 10 = 10 0 C and 20 + 10 = 30 0 C . Considering the volume of the bulbs to remain constant, we can write for the first bulb, P T 10 + 273 283 P1 P Or, 1 = 1 = = = 1.037 = P T 0 + 273 273 T1 T Similarly, for the second bulb, P2 T2 30 + 273 303 = = = = 1.034 P T 20 + 273 293 ∴ We can write, P1 > P2 . Hence, the mercury thread shifts towards the bulb of higher temperature. Q.4 In the definition of the coefficient of expansion of gas, the initial volume or pressure is always referred to at 0 0 C . But for a solid or liquid, this is not done – why? Ans. The value of the coefficient of expansion for a solid or liquid is small. Thus the value of pressure or volume at any temperature can be considered as the initial value. The error involved due to this is negligible. But the value of the coefficient of expansion of a gas is large. Thus the calculations done by assuming the initial value of volume or pressure of gas at different temperatures give significantly different results. Hence, for a gas, the initial value of volume or pressure is always referred to be that at 0 0 C . Q.5 What is universal gas constant? Is it same for all gases? What is the value of this constant? Ans. PV For one gram mole of a gas, the value of obtained from the ideal gas law is a T PV   constant. This constant  R =  is called the universal gas constant. T   The value of R is same for all ideal gases. R = 8.31 × 10 7 erg/(mole Kelvin). Q.6 Determine the value of the universal gas constant R and the gas constant k for 1 gm mole air. [At N.T.P., the density of air = 1.293 gm/litre and the density of mercury = 13.6 gm/cc.] [J.E.E. ‘96]

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Ans. Pressure, P = 76 × 13.6 × 980 dyne/cm 2 , Temperature, T = 0 0 C = 273 K . At NTP, the volume of 1 gm-mole of any gas, V = 22.4 litre = 22400 cc. PV 76 × 13.6 × 980 × 22400 ∴R = = = 8.31 × 10 7 erg/(mole K). T 273 1 1000 litre = cc. The volume of 1 gm-mole air, V ′ = 1.293 1.293 PV ′ 76 × 13.6 × 980 × 1000 ∴k = = 0.287 × 10 7 erg/(gram K). = 1.293 × 273 T Q.7 Draw P − t 0 C and P − T K graphs for a certain mass of gas at constant volume. Can one find the value of absolute zero from the first graph? [H.S. ’04, ‘90] Ans.

Fig. to be included If for a certain mass of gas at constant volume, the temperature, t 0 C is considered as abscissa and the pressure, P is considered as ordinate, we get a straight line as shown in the figure**. The P − t relationship is the following: t   P = P0 1 + .  273  The pressure, P0 is the pressure of gas at t = 0 0 C which is the intercept on the vertical P P -axis by the straight line; 0 is the slope of the line. For different constant volumes 273 of gas, we get different values of P0 . Thus we obtain a set of straight lines having different slopes and different intercepts. If the straight lines are extrapolated towards the negative temperature, they all cut the t − axis at a certain point ( P = 0) . The value of this point is − 2730 C . The pressure of gas is zero at this point as can be seen from the graph. Therefore, the temperature t = −2730 C is the absolute zero temperature. The straight lines can not be extrapolated further as that will indicate negative pressure which is unphysical. P The P − T relationship is = constant. T If we plot temperature T (in absolute scale) as abscissa and pressure P as ordinate of a certain mass of gas at constant volume, we get a straight line ing through the origin. For different constant volumes, we get different P − T straight lines and all of them through the origin (see fig.**). Q.8 A certain mass of gas is heated first in a small vessel and then in large vessel. Assume that the volumes of the vessels remain unchanged during heating. How will be the pressure-temperature (P − T ) graphs in two cases? Ans.

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Fig. to be included P = constant. T Thus the (P − T ) graphs will be straight lines ing through origin in both the cases. The equation for ideal gas is PV = kT , k is a constant. Suppose, the volumes of the two vessels are V1 and V2 , where V1 < V2 . If the pressure of the gas is P1 in the first vessel and P2 in the second vessel at a certain temperature T , we can write, P1V1 = P2V2 . ∴We have, P1 > P2 . This is true for any temperature. Thus the pressure of the gas is always higher in small volume than the pressure of gas in larger volume. Therefore, the (P − T ) straight line corresponding to smaller volume ( V1 ) is above the one corresponding to larger volume ( V2 ) at all temperatures. The slope in the first case will be higher than the slope in the second case. As the gas is heated, keeping the volume constant, we can write

Q.9 At first m gm and then 2m gm of a gas are heated in a container of fixed volume. Draw pressure-temperature ( P − T ) graph in each case. Ans.

Fig. to be included P = constant. T So, the ( P − T ) graphs for different masses of gas will be straight lines ing through origin. Now, the ideal gas law is PV = nRT , where n is the number of gram moles of the gas and R is the universal gas constant. If M is the molecular weight, we can write for m gm gas, m mR RT Or, P = PV = .T ……….(1) and for 2m gm gas, M MV 2m 2mR PV = RT Or, P = .T ……….(2) M MV mR 2mR Therefore, in the ( P − T ) graph, is the slope in the first case and is the slope MV MV in the second case. Hence, the slope in the second case is higher than the first one. Thus the ( P − T ) straight line graph corresponding to 2m gm gas will be above the graph corresponding to m gm gas. If some amount of gas is heated in a fixed volume, we can write

The above can also be analyzed in a different way. Suppose, P1 and P2 are the pressures of m gm gas and 2m gm gas, respectively at any temperature, T ′ . As the volume, V is constant, we can write, P1 RT ′ P RT ′ = constant; 1 = = constant. = m MV 2m MV

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P2 ; the value of pressure in the ordinate is higher in the first case than the second 2 case. So, the ( P − T ) straight line for m gm gas will be above the one for 2m gm gas.

∴ P1 =

Q.10 In the fig.**, V − T graphs for a certain amount of gas have been shown at two pressures P1 and P2 . It can be said from the graphs that the pressure P1 is greater than [I.I.T.] the pressure P2 . Is it correct? Ans.

Fig. to be included It is seen from the graphs, at some temperature T ′ (say), the volume of the gas V1 at pressure P1 is greater than the volume V2 at pressure P2 . P V According to Boyle’s law, P1V1 = P2V2 Or, 1 = 2 . P2 V1 Q V2 > V1 ∴ P1 > P2 . Hence, we can say from the given graphs, the pressure P1 is greater than the pressure P2 . Q.11 When a gas expands, it obeys the law: PV 2 = constant. Show that there will be cooling due to this kind of expansion. [J.E.E. ‘98] Ans. Let the initial volume, pressure and temperature of the gas are V1 , P1 and T1 , respectively. If the corresponding quantities are V2 , P2 and T2 after expansion, we can write from ideal gas law: P1V1 P2V2 PV T Or, 1 1 = 1 = T1 T2 P2V2 T2

Or,

P1V1

2

P2V2

2

=

V1T1 V2T2

[Multiplying by

V1 on both sides] V2 2

2

Since PV 2 = constant, we can write P1V1 = P2V2 . VT T V ∴ 1 1 = 1 Or, 1 = 2 . V2T2 T2 V1 As the gas expands, the final volume will be greater than the initial volume: V2 > V1 . ∴ T1 > T2 which implies that the final temperature is lower than the initial temperature. Hence, in this kind of expansion, the gas cools down. Q.12 A faulty barometer tube contains some air above mercury. How can the correct value of atmospheric pressure be determined by this? Ans. Let the correct barometer reading = H cm Hg and the area of cross-section of the barometer tube = α sq. cm.

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Suppose, in the faulty barometer, we find the height of mercury column = h1 cm and the length of air column above mercury = l1 cm. ∴The volume of the air column, V1 = l1α cc and the pressure, P1 = ( H − h1 ) cm Hg. Next, the mercury column is altered by slightly lifting the barometer tube while keeping its open end still immersed in mercury. Suppose this time, the height of mercury column = h2 cm and the length of air column above mercury = l 2 cm. ∴In this case, the volume of the air column, V2 = l 2α cc and the pressure, P2 = ( H − h2 ) cm Hg. Applying Boyle’s law: h l − h2 l 2 ( H − h1 )l1α = ( H − h2 )l 2α or, H = 1 1 . l1 − l 2 ∴ The correct barometer reading or the actual atmospheric pressure H can be determined if we know h1 , h2 , l1 , l 2 . Questionnaire Very Short Questions:

Mark: 1 (Answer in one or two words)

1 0 / C] 273 1 0 / C] 2. What is the pressure coefficient of gas? [ 273 3. What will be the temperature in degree Celsius at which the volume of gas becomes zero according to Charles’ law? [-273 0 C ] 4. What is the volume of one mole of any ideal gas at NTP? [22.4 litre] 1. What is the volume coefficient of gas?

[

(Fill in the blanks)

1. At constant temperature, the volume of some amount of gas is _________ its pressure. [Inversely proportional to] 2. At constant pressure, the volume of some amount of gas is ____________ its absolute temperature. [Proportional to] 3. At constant volume, the pressure of some amount of gas is ____________ its absolute temperature. [Proportional to] (Multiple choice type)

1. The coefficient of volume expansion of gas, when compared to solid and liquid, is (a) same (b) relatively large (c) relatively small. [(b)]

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2. The volume and pressure of some amount of gas are seen to increase. This is possible when the temperature of the gas (a) remains same (b) decreases (c) increases. [(c)] 3. If the volume expansion coefficients of solid, liquid and gas are γ S , γ l and γ g ,

respectively, (a) The values of γ S , γ l and γ g for different solids, liquids and gases are different (b) The values of γ S , γ l for different solids and liquids are different, but γ g is same for all gases (c) The value of γ S is different for different solids, but γ l is same for all liquids and γ g is same for all gases (d) The values of γ S , γ l and γ g for different solids, liquids and gases, respectively are same.

[(b)]

4. The volume expansion coefficients of solid, liquid and gas are γ S , γ l and γ g ,

respectively. In general, (a) γ S < γ l < γ g (b) γ S > γ l > γ g (c) γ l < γ S < γ g (d) γ l > γ S > γ g

[(a)]

5. The volume of a gas at N.T.P. is 150 cc. If the temperature at constant volume is 25 0 C , the pressure becomes 850 mm. What is the pressure coefficient of the gas? [(a)] (a) 4.73 × 10 −3 / 0 C (b) 5.73 × 10 −3 / 0 C (c) 6.73 × 10 −3 / 0 C (d) 1 / 0 C 6. The value of relative gas constant for hydrogen is (a) 4.16 × 10 7 erg/gm K (b) 0.26 × 10 7 erg/gm K (c) 4.80 × 10 7 erg/gm K (d) 5.16 × 10 7 erg/gm K

[(a)]

7. When the temperature is increased keeping the pressure constant, the density of a gas (a) remains same (b) decreases (c) increases (d) may increase or decrease, depends on gas. [(b)] 8. For determining the volume expansion coefficient of gas, the volume at some temperature is always taken as initial value. The temperature is [(b)] (a) 273 0 C (b) 0 0 C (c) 100 0 C (d) -273 0 C 9. The isotherm of a gas is (a) P − V graph (b) P − T graph (c) V − T graph (d) PV − T graph.

[(a)]

10. At constant pressure, the graph of volume of a certain amount of gas with temperature, ( V − T ) graph is (a) horizontal (b) straight line with a positive slope (c) straight line with a negative slope (d) rectangular [(b)] 11. The PV − P graph is

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(a) parallel to P -axis (b) parallel to PV -axis (c) straight line ing through the origin (d) rectangular hyperbola [(a)] 12. The volume coefficient and the pressure coefficient are same for (a) ideal gas (b) real gas (c) hydrogen gas (d) inert gas

[(a)]

13. There is 1 mole O2 gas (relative molecular weight 32) in a vessel at a temperatureT . The pressure of this gas is P . There is 1 mole He gas (relative molecular weight 4) at a temperature 2T in a similar vessel. The pressure of this gas will be P (b) P (c) 2 P (d) 8 P [I.I.T. ‘97] (a) 8 [(c)] 14. Two containers A and B having frictionless pistons attached, contain same ideal gas. In both the containers, the volume of gas ( V ) and temperature are same. The mass of gas in the containers A and B are m A and m B , respectively. Now the gas is made to expand up to a volume 2V at fixed temperature in both the containers. The change in pressure in A and B containers are ∆P and 1.5∆P , respectively. Which of the following statements is correct? (a) 4m A = 9m B (b) 2m A = 3m B (c) 3m A = 2m B (d) 9m A = 4m B [I.I.T. ‘98] [(c)] 15. A closed horizontal cylinder of length 50 cm, has a freely moving piston separating two chambers in it. The left chamber contains 25 mg and the right chamber contains 40 mg of He gas. When equilibrium is established, what will be the ratio of the lengths of left and right chambers? (a) 1:2 (b) 3:2 (c) 5:8 (d) 8:5 [(c)] Short Questions:

Marks: 2

1. How is the expansion of gas different with respect to solid and liquid? 2. State Boyle’s law and explain. [H.S. ’95, ‘92] 3. State Charles’ law and explain. [H.S. ’02, ’97, ’95, ’93, ‘92] 4. Name the law of variation of volume of a gas with temperature at constant pressure and explain. 5. How does the concept of absolute temperature come from Charles’ law? [H.S. ’05, ‘02] 6. Establish the equation for ideal gas from Boyle’s law and Charles’ law. [H.S. ’01, ’99, ’95, ‘92] 7. Determine the combined form of Boyle’s law and Charles’ law. [H.S. ‘93] 8. What is universal gas constant? [H.S. ‘02] 9. Establish the mathematical form of Charles’ law using absolute temperature. 10. Determine the relation among pressure, temperature and density of gas. 11. ‘To determine the state of a gas, we need to know three variables- volume, temperature and pressure’ – Explain the meaning of this statement. 12. What are the values of absolute zero in Celsius and Fahrenheit scale?

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13. Why the absolute zero temperature rather than 0 0 C in Celsius scale is considered fundamental? 14. What do you mean by ideal gas law? Is this law applicable for real gas? 15. What is the value of universal gas constant in C.G.S. unit? 16. Why the expansion coefficient of gas is of two types? 17. Why do we consider 0 0 C as the initial temperature when we calculate expansion coefficient of a gas? Why is this difference with respect to solid and liquid? 18. Is the value of expansion coefficient same for all gases? 1 0 19. What do you mean by the pressure coefficient of gas to be / C? 273 20. What is Normal temperature and pressure (N.T.P.)? 21. What is the relation between the volume coefficient and pressure coefficient of a gas? What are the values? 22. Is it possible to reach a temperature below the absolute temperature? Why? 23. A container is filled with oxygen gas and it is taken to Moon. What will happen if the container is a (i) steel cylinder and (ii) rubber balloon? 24. When is the density of a gas proportional to the pressure? 25. When is the density of a gas inversely proportional to its absolute temperature? 26. For the expansions of solid and liquid, only temperature is referred whereas in the case of volume expansion of a gas, temperature and pressure both are mentioned. Why? 27. There is a coefficient of apparent expansion for liquid, but no such coefficient is mentioned for a gas. Why? 28. What is the relative gas constant? Is this constant same for all gases. 29. Equal numbers of hydrogen and helium molecules are kept in two similar containers. What will be the ratio of their pressures? 30. What is an ideal gas? 31. What is a mole? What is Avogadro’s number? 32. Draw PV − P isotherm for an ideal gas of a certain mass and explain. [H.S. ‘03] 33. Draw V − t 0 C and V − T K graphs for a certain mass of a gas at constant pressure. Is it possible to get the value of absolute temperature from the first graph? 34. A certain amount of gas is first heated in a container of small volume and then it is heated in a container of larger volume. Assume that the volume of the containers remain unchanged during heating. What will be the pressure temperature ( P − T ) graphs in the two cases? 35. During blowing up, the volume of a balloon increases as well as the pressure also increases. Is Boyle’s law violated here? [H.S. ‘04] Medium level Questions:

Marks: 4

1. Define the volume coefficient and pressure coefficient of a gas. Show that for an ideal gas, the values of the two coefficients are same. [1+1+2] [H.S.(XI) ’06; H.S. ’05, ’03, ’00, ’98, ’96, ‘94] 2. What do you mean by absolute scale of temperature? How does the concept come from Charles’ law? Why is it called absolute scale? What will be the volume of an ideal gas at zero degree of the absolute scale? [1+1+1+1] [H.S. ’00, ’97, ‘93]

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3. What is universal gas constant? Why is it called universal? Calculate the value of it for one gram-mole of any gas. [1+1+2] [H.S. ‘01] 4. State the law of ideal gas when volume is constant (Pressure law). Write the expression when absolute temperature is used and draw a graph. [2+1+1] 1 5. Draw P − V , P − , PV − P , and PV − V graphs following Boyle’s law. V [1+1+1+1] 6. How does the melting point of ice depend on external pressure? [H.S. ‘05] Short Problems:

Marks: 2

1. At 27 0 C and 70 cm pressure, the volume of some oxygen gas is 400 cc. What will be its volume at normal temperature and pressure? What will be the change in product of the pressure and volume of the gas due to this variation? [Ans. 335.3 cc, 2517 C.G.S. unit] 2. The volume of some amount of gas is 2 litre at normal pressure and temperature. What is the volume of this gas at temperature 91 0 C and at pressure 570 mm? [Ans. 3.5 litre] 3. The volume of some amount of nitrogen is 50 cc at 50 0 C . What will be the volume of [Ans. 34.5 cc] this gas at -50 0 C if the pressure does not change? 0 4. The pressure of a gas is 60 cm Hg at a temperature -73 C . What will be the pressure [Ans. 90 cm Hg] at 27 0 C if volume remains constant? 5. When the temperature of 5 litre of a gas is raised from 0 0 C to 35 0 C at constant pressure, the volume is increased by 640 cc. Determine the value of absolute zero in Celsius scale from this data. [Ans. − 273.4 0 C ] 6. The atmospheric pressure is 74 cm Hg and temperature is 27 0 C at some place. The atmospheric pressure and temperature are 70 cm Hg and 23 0 C , respectively at some other place. Compare the densities of air at the two places. [Ans. 2738 : 2625] 7. One litre helium gas of temperature 27 0 C and of pressure twice the atmospheric pressure is heated such that the volume and pressure of the gas both become double. Find the final temperature. [Ans. 927 0 C ] 8. A glass vessel is filled with air at 50 0 C . Up to what temperature is the vessel to be 1 heated so that fraction of air comes out? Assume the pressure to remain constant in the 4 vessel. [Ans. 157.67 0 C ]

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9. What will be the volume of 1 mole oxygen at 27 0 C temperature and two atmospheric [Ans. 12.31 litre] pressure? R = 8.31 × 10 7 erg/(mole Kelvin). 10. The volume of a dry gas is 1520 cc at 27 0 C temperature and at 700 cc Hg. What is the volume of that gas at NTP? [Ans. 1274 cc] 11. The weight of 1 litre gas is 1562 gm at NTP. What will be the weight of 1 litre gas at 25 0 C temperature and 78 cm Hg pressure? [Ans. 1460 gm] 12. The volume of a bulb is 1 litre. What percentage of air will come out if the bulb is heated from 0 0 C to 27 0 C ? Assume that the pressure inside remains constant. [Ans. 9.89%] 13. The density of oxygen is 1.429 gm/litre at normal temperature and pressure. Find the mass of 2.5 ltre oxygen gas at 27 0 C temperature and 780 mm pressure contained in a cylinder. [Ans. 3.336 gm] 14. If the mercury pressure in a barometer is 70 cm, the volume of a gas is 650 cc. What is the volume of that gas at standard temperature and pressure? [Ans. 598.68 cc] 15. When the mercury pressure in a barometer is 75 cm, the volume of some amount of hydrogen gas is 150 cc. If the volume of this quantity of hydrogen gas becomes 160 cc in the next day, what will be the barometer reading? [Ans. 70.31 cm] Medium level Problems:

Marks: 4

1. An air bubble comes up from the bottom of 34 ft deep river. The temperature of water

at the depth of river us 7 0 C and the volume of the bubble is 14 cc. The temperature of water above is 27 0 C and the pressure, 75 cm Hg. If the density of mercury is 13.6 gm/cc, what will be the volume of the bubble just above the water surface? [Ans. 65 cc] 2. The volume of an air bubble becomes 4 times as it rises above from the depth of a sea. If the atmospheric pressure is 76 cm Hg and the temperature below and under and above the sea is same what is the depth of the sea? Density of mercury = 13.6 gm/cc. [Ans. 3101 cm] 3. The perimeter of the tyre of a car is 1 m and diameter 10 cm. How much air has to be introduced in the tyre at atmospheric pressure so that the pressure inside it will be 10 times the atmospheric pressure? [Ans. 78.5 litre] 4. The density of air at standard temperature and pressure is 0.00129 gm/cc. If the barometer height comes down from 76 cm to 74 cm, what will be the difference in the weight of 15 litre air? [Ans. 0.51 gm] 5. The volume of a both side sealed cylinder is 22.4 litre and it contains 4 gm hydrogen at 0 0 C . What will be the pressure if the temperature is 60 0 C ? If now 14 gm nitrogen is

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inserted in the cylinder at 0 0 C instead of hydrogen what will be the pressure of this gas [Ans. 2.44 atm; 0.683 atm] at 100 0 C ? 6. The volume of a vessel is 10 litre. This is filled with O2 gas at NTP. The vessel is heated to 27 0 C and opened at 75 cm Hg. How much gas will come out? At NTP, 1 litre O2 gas weighs 1.43 gm. [Ans. 1.452 gm] 7. There is a 10 cm long thread of mercury resting in the middle of a both side sealed narrow horizontal tube. Air is trapped at 76 cm Hg on the two sides of the mercury thread. If now the tube is kept vertical, what will be the displacement of the mercury column? The length of the tube = 100 cm. [Ans. 2.95 cm] 8. A car tyre is filled with 15 litre air at a temperature of 17 0 C and at a pressure of 2.5 times the atmospheric pressure. If the temperature is increased to 37 0 C and the volume is increased to 15.5 litre, find the pressure inside the tyre. The atmospheric pressure = 15 pound/in 2 . [Ans. 38.8 pound/in 2 ] 9. A container is filled with 4 gm gas at 12 0 C and then heated to 50 0 C . Some gas comes out and the inside pressure remains the same. What is the mass of outgoing gas? [Ans. 0.471 gm] 0 10. Some air, at 20 C and at atmospheric pressure, is confined in a flask with the help of a cork. Due to rise in temperature, the pressure increases by 1.7 times which pushes the cork out. What is the value of increased temperature? [Ans. 225.1 0 C ] 11. When a air bubble comes up to the surface of water from the bottom of the lake, its volume increases 10 times. Find the depth of water in the lake if the barometer height is 75 cm. [Ans. 91.8 m] Harder Problems: 1. A weightless and freely movable piston is fitted inside a cylindrical tube which contains some ideal gas. The tube is held vertical. The piston is in equilibrium at a temperature 27 0 C when it is at 10 cm above the bottom. The whole system is now immersed in a water bath. The temperature of the bath is then gradually increased to 100 0 C . Calculate the present height of the piston. [Ans. 12.43 cm]

2. The combined volume of some amount of gas and a piece of glass in it is 100 cc at 27 0 C . If the pressure and temperature are increased by two times, the volume becomes 60 cc. What is the volume of the glass piece? [Ans. 12.1 cc] 3. The diameter of a glass sphere is 20 cm. This is filled with 1 litre air at 30 in Hg pressure and at a temperature of 0 0 C . What is the pressure inside the sphere when it is [Ans. 7.86 in Hg] heated to 27 0 C ? Neglect the expansion of the sphere.

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4. The volume of an oxygen tank is 50 litre. The pressure of this is measured by a pressure gauge (Pressure gauge measures the amount in excess of atmospheric pressure). When oxygen is taken out from the tank, the reading of the pressure gauge drops from 300 lb/in 2 to 100 lb/in 2 , and the temperature of the gas that remains in the tank, drops from 30 0 C to 10 0 C . How much oxygen was there in the tank initially? How much oxygen have been withdrawn? [Ans. 1.376 kg; 0.838 kg] 5. The volume of some air with saturated water vapour is 80 cc at a pressure of 74 cm. If the pressure is made 146 cm keeping the temperature constant, the volume becomes half the initial volume. What will be the pressure due to water vapour at this time? [Ans. 2 cm Hg] 6. The area of cross-section of a barometer tube is 1 sq. cm and the height of this is 85 cm from the surface of mercury contained in the cup. The height of mercury column is 75 cm. What will be the height of mercury column when 1 cc air is introduced into the Torricelli’s empty space at atmospheric pressure? [Ans. 70 cm] 7. A helicopter is flying at the height of 400 m from the ground. If the average density of air is 1.2 × 10 −3 gm/cc below this and the air pressure is 1010 m bar on the ground, what is the air pressure inside the helicopter? [Ans. 962.96 m bar] 8. There is some air above the mercury column in an erroneous barometer. When the barometer readings are 760 mm and 742 mm, the correct readings are 770 mm and 750 mm. What is the length of the confined air column in the first case? What is the correct barometer reading if the erroneous barometer reading is 752 mm? Assume the temperature is unchanged. [Ans. 7.2 cm; 761 mm] 9. A mercury barometer gives erroneous reading because an air bubble enters in the empty space above the mercury column. When the air pressure is equal to the pressure of 760 mm mercury column, the barometer reads 740 mm. Again, the reading is 710 mm at a pressure of 727.5 mm Hg. Find the length of barometer tube (above the mercury surface in the cup). [Ans. 95 cm] 10. A bottle of volume 500 cc is being immersed in a water tank keeping the open face down. To what depth the bottle has to be taken so that 100 cc of water will enter in it? The height of mercury column in barometer is 75.6 cm and the density of mercury is 13.6 gm/cc. [Ans. 2.57 m] 11. The temperatures at the bottom and at the surface of a lake are 7 0 C and 27 0 C , respectively. At what ratio the diameter of the bubble will change when it reaches the surface from the bottom of the lake? The atmospheric pressure is 75 cm Hg and the density of mercury is 13.6 gm/cc. [Ans. 1:1.61] 12. The space between the lower sealed end of a vertical tube and a mercury column in the tube is filled with some ideal gas. The open end of the tube is exposed to the atmosphere (atmospheric pressure = 76 cm Hg). The lengths of the mercury column and that of the column of gas are 20 cm and 43 cm, respectively. When the tube is slowly

Fundamental Physics-I: by Dr. Abhijit Kar Gupta (email: [email protected])

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inclined with respect to the vertical by 60 0 , what will be the length of the column of gas? Assume the temperature of the gas to remain constant. [Ans. 48 cm] 13. A 80 cm long both side open narrow tube is immersed halfway in mercury. Now, as the tube is taken out from mercury by closing its upper end, a 23 cm long mercury column remains in the tube. What is the atmospheric pressure? [Ans. 77.12 cm Hg] 14. A one end sealed glass tube contains some air in it with the help of a 8 cm long mercury thread. If the tube is held vertical keeping the sealed end up, the length of air column becomes 40 cm at 27 0 C . What will be the length of the air column if the temperature is increased to 60 0 C and the tube is inclined by an angle of 60 0 with the vertical? The area of cross-section of the tube is 0.5 cm 2 and the atmospheric pressure is 76 cm Hg. What is the mass of air confined in the tube? The density of air at standard [Ans. 41.5 cm; 21.05 × 10 −3 gm] temperature and pressure = 1.293 × 10 −3 gm/cc. 15. One end of a uniform glass tube is closed; some air is trapped in it by a mercury column. The length between the closed end and the edge of the mercury column is 10 cm at 20 0 C . How far does the mercury column shift when the temperature increases to [Ans. 11.7 cm] 70 0 C ? γ P = 0.00366 / 0 C . 16. The volume of a cylindrical container is 10 4 cc and it is filled with oxygen gas at a pressure of 2.5 × 10 6 dyne/cm 2 and at a temperature of 27 0 C . When some gas comes out, the pressure of the gas inside becomes 1.3 × 10 3 dyne/cm 2 . What is the mass of gas that comes out? The molecular weight of oxygen = 32. [Ans. 15.36 gm] 17. At normal condition, 40 cc of oxygen gas is inserted in a one end sealed tube which is then held vertical keeping it immersed in mercury taken in a container. The height of mercury column in the tube is seen to be 15.6 cm above the surface of mercury. If the air pressure is 75.6 cm and the room temperature is 31 0 C , what length of the tube was filled with the gas? [Ans. 47.02 cm] 18. A both side open glass tube is immersed vertically in mercury such that a length of 13 cm remains above mercury surface. Now the tube is pulled upwards by another 35 cm while closing the upper end. Find the length of air column above mercury inside the closed tube. Atmospheric pressure = 76 cm Hg. [Ans. 20.41 cm] 19. A vessel of volume 800 cc is immersed in water holding the side of its opening downward. At what depth the vessel has to be taken so that 300 cc of water enters into it? The barometer height above water = 76 cm Hg and the density of mercury = 13.6 gm/cc. [Ans. 620.2 cm] 20. Some air is trapped in a one end open capillary tube with the help of a 15 cm long mercury thread. If the tube is held vertical with the open end up, the length of the air column becomes 10 cm at 27 0 C . On the other hand if the tube is held upside down, the

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length of air column becomes 15 cm. At what temperature the length of air column will become 20 cm in this upside down position? [Ans. 127 0 C ] 21. A glass tube of narrow bore has two sides sealed. A mercury thread is inside the tube such that it divides the tube in a ratio of 3:1. If the temperature of the system is increased from 0 0 C to 273 0 C , what will be the change in air pressure inside the tube? [Ans. increases twice in each half] 22. The volume and temperature of a vessel are V and T , respectively. Three gases are inserted in this. What is the resultant pressure of the gas mixture if the initial volume, temperature and pressure of the gases are ( P1 , V1 , T1 ); ( P2 , V2 , T2 ); ( P3 , V3 , T3 ), T  P1V1 P2V2 P3V3   ]  + + V  T1 T2 T3  23. The length between the sealed end of a uniform cylindrical tube of a barometer and 0 the mercury surface in the tube is l mm. An air bubble is inserted into the tube at t1 C and at a pressure of P mm Hg. The height of mercury column reduces to l1 mm due to

respectively?

[Ans.

this. If the barometer reading is l 2 mm at t 0 C , what will be the correct reading? Assume 1 + α .t l − l1 the expansion coefficient of air = α . [Ans. l 2 + . ( P − l1 ) mm] 1 + α .t1 l − l 2 24. A small mercury thread separates some air from surroundings in a tube. The tube can be rotated in the vertical plane. The length of air column in the tube at its horizontal position is l1 and in the vertical position (open face up), the length becomes l 2 . Find the length of the air column when the tube makes an angle θ with the vertical. l1l 2 ] [Ans. l 2 − (l 2 − l1 ) cos θ 25. The bottom end of a 1 m long vertical tube is closed and an air tight piston is fitted at the other end at top. The piston can move up and down without friction. Some ideal gas is trapped inside the tube. The gas occupies 94 cm in the tube when the piston is in equilibrium. Now mercury is being poured over the piston up to the edge of the tube. The piston now comes down to 30 cm. Find the atmospheric pressure assuming the temperature to be constant. The weight of the piston can be neglected. [Ans. 76.4 cm Hg] 26. A container contains m1 gm gas at P1 pressure. Another container has m 2 gm gas at P2 pressure. If the two containers are now ed by a tube, what will be the temperature P P (m + m2 ) of the gas mixture? [Ans. 1 2 1 ] P2 m1 + P1 m2 27. A closed vessel of volume 0.02 m 3 contains a mixture of Neon and Argon gas at a temperature of 27 0 C and at a pressure of 1 × 10 5 Newton/m 2 . The total mass of the gas mixture is 28 gm. The molecular weights of Neon and Argon are 20 and 40 gm/mole, respectively. Determine the mass of each gas. Assume the two gases are ideal (universal gas constant = 8.314 joule/ mole K). [I.I.T. ‘94] [Ans. Mass of Neon = 4.08 gm, Mass of Argon = 23.92 gm]

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28. An ideal gas is seen to obey an additional law: VP 2 = constant. The initial temperature of the gas is T and volume, V . If the volume of the gas becomes 2V as it expands, what is the temperature? [Ans. 2T ] 29. There is a 4 cm long mercury thread in a uniform glass tube which has one end open. The space between the mercury thread and the closed end contains some air at 27 0 C . The length of air column is 9 cm when the tube is held vertical with the open end facing up. The length of the air column becomes 10 cm when the tube is held up side down. (i) Find the atmospheric pressure. (ii) At what temperature the length of air column becomes 9 cm in the inverted position? [Ans. (i) 76 cm Hg, (ii) − 30 C ] 30. An annulus tube (as shown in the fig.**) contains two ideal gases of equal mass. The relative molar masses of the gases are M 1 = 32 and M 2 = 28 . The two gases are kept separated by a fixed partition and a movable cork. The cork can slide in a frictionless [Ans. 192 0 ] manner inside the tube. Determine the angle α as shown in the figure. I Fig. to be included

31. The lower end of a narrow and uniform vertical tube is sealed. There is some ideal gas in the space between the sealed end and a mercury column. The upper end of the tube is exposed to the atmosphere. The lengths of the mercury column and the confined air column are 20 cm and 43 cm, respectively. The tube is slowly inclined to make an angle of 60 0 with the vertical. What is the length of the column of gas in this situation? Given, atmospheric pressure = 76 cm Hg and the temperature is the gas is constant. [Ans. 48 cm] 32. The lower end of a narrow and uniform 100 cm long vertical tube is sealed while the upper end is open. A movable piston is attached in the tube at 10 cm from the open end such that there is 90 cm long air column under this piston. Now mercury is being poured over the piston until it starts overflowing. The piston comes down to 32 cm. What is the atmospheric pressure? [Ans. 76.125 cm Hg] 33. Show that if the rate of variation of pressure with density and height in Earth’s dP atmosphere is given by = − ρg , the air pressure at any height h measured from dz ground is given by P = P0 e − Mgh / RT , where M is the average molecular weight of air. Assume the atmospheric air is ideal gas and the temperature to be uniform throughout.