Kinetic Theory of Gases

Questions for Short Answer

1. When we place a gas cylinder on a van, and the van moves, does the kinetic energy of the molecules increase?

Answer: If we consider the mechanical kinetic energy, then the kinetic energy of the gas increases because the gas has a mass. But the kinetic energy of molecules, which is the measure of the internal energy, does not increase.

2. While the gas from a cooking gas cylinder is used, the pressure does not fall appreciably until the last few minutes. Why?

Answer: The cooking gas is filled at very high pressure in the cylinder. At this pressure, the gas liquefies. On the liquefied surface of the gas, there is a saturated vapor of the gas. As soon as some gas is used in cooking, the saturation balance is disturbed, and some of the liquid converts to gas to maintain the balance. Thus, the pressure of the gas does not appreciably drop until there is liquid gas in the cylinder.

3. Do you expect the gas in a cooking gas cylinder to obey the ideal gas equation?

Answer: No. The density of the gas is very high here, so it will not obey the ideal gas equation.

4. Can we define the temperature of a vacuum? The temperature of a single molecule?

Answer: No. The temperature of a vacuum can not be defined, because the temperature of a gas is proportional to the square of the rms speed of its molecules and hence to the total translational kinetic energy of its molecules. Since there are no molecules in the vacuum, no temperature is defined.

The temperature of a single molecule can not be defined, but for a bunch of molecules.

5. Comment on the following statement. The temperature of all molecules in a sample of a gas is the same.

Answer: The temperature is defined for a given mass of the substance, and it is proportional to the total translational kinetic energy of its molecules. The temperature of each molecule is not defined.

6. Consider a gas of neutrons. Do you expect it to behave much better as an ideal gas as compared to hydrogen gas at the same temperature and pressure?

Answer: Though it is hardly possible to have a gas of neutrons and very difficult to find a container that can hold it, if we assume it to be possible, then the answer will be yes. Because in the assumptions of the ideal gas, the size of a molecule is much smaller than the average separation between them, neutrons have a much smaller size than a hydrogen molecule. Also, neutrons do not exert electrostatic force on each other.

7. A gas is kept in a rigid cubical container. If a load of 10 kg is put on the top of the container, does the pressure increase?

Answer: No. Since the cubical container is rigid, the load will not be transferred to the gas, and the pressure will not increase.

8. If it were possible for a gas in a container to reach the temperature of 0 K, its pressure would be zero. Would the molecules not collide with the walls? Would they not transfer momentum to the walls?

Answer: At the temperature of 0 K, the kinetic energy of the molecules will be zero, and hence they will not have velocities and momentum. So the molecules will not collide with the walls and not transfer momentum to the walls.

9. It is said that the assumptions of the kinetic theory are good for gases having low densities. Suppose a container is so evacuated that only one molecule is left in it. Which of the assumptions of kinetic theory will not be valid for such a situation? Can we assign a temperature to this gas?

Answer: The assumption "when gas is left for sufficient time, it comes to a steady state. The density and distribution of molecules with different velocities are independent of position, direction and time" may be justified if the number of molecules is very large. In the given condition, this assumption will not be valid. This gas can not be assigned a temperature.

10. A gas is kept in an enclosure. The pressure of the gas is reduced by pumping out some gas. Will the temperature of the gas decrease by Charles' law?

Answer: The condition in Charles' law is that the volume and the mass of the gas should be constant. In the given condition, the mass of the gas in the enclosure is not constant. Hence, Charles' law can not be applied.

11. Explain why cooking is faster in a pressure cooker.

Answer: In open cooking, the water-salt mixture boils just above 100°C, but in a pressure cooker the pressure is about 4 times higher than atmospheric pressure and due to this the water boils around 140°C. Since the cooking in the pressure cooker is at a higher temperature, the cooking is faster.

12. If the molecules were not allowed to collide among themselves, would you expect more evaporation or less evaporation?

Answer: The molecules move randomly, and they can only be prevented from colliding among themselves if they have zero velocities. In this case, the evaporation will be less.

13. Is it possible to boil water at room temperature, say 30°C? If we touch a flask containing water boiling at this temperature, will it be hot?

Answer: A liquid boils when the saturated vapor pressure above its surface equals the external pressure. By reducing the external pressure over the water surface, it can be made to boil even at 30°C.

The hotness is related to the temperature, not to the boiling of water. So it will not be hot when touched at 30°C.

14. When you come out of a river after a dip, you feel cold. Explain.

Answer: After a dip in the river, the whole body is wet, and the evaporation from the wet body surface starts. In evaporation, the molecules with higher kinetic energy escape. This leaves the average kinetic energy of the left-behind molecules reduced, and hence the temperature of the wet body surface is reduced. That is why the feeling of coldness comes.

OBJECTIVE-I

1. Which of the following parameters is the same for molecules of all gases at a given temperature?

(a) mass

(b) speed

(c) momentum

(d) kinetic energy

Answer: (d)

Explanation: The absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules.

Mass, speed, and momentum of a gas do not depend on the temperature.

2. A gas behaves more closely as an ideal gas at

(a) low pressure and low temperature

(b) low pressure and high temperature

(c) high pressure and low temperature

(d) high pressure and high temperature

Answer: (b)

Explanation: A gas behaves more closely as an ideal gas if its density is very low. It can be achieved if the pressure is low and the temperature is high.

3. The pressure of an ideal gas is written as p = 2E/3V. Here E refers to

(a) translational kinetic energy

(b) rotational kinetic energy

(c) vibrational kinetic energy

(d) total kinetic energy.

Answer: (a)

Explanation: If U = Total translational kinetic energy of all the molecules then, U = (3/2)nRT

→nRT = 2U/3

Now the ideal gas equation is

pV = nRT = 2U/3

→p = 2U/3V

Since in the problem the pressure is written as p = 2E/3V, hence E = U. So E refers to total translational kinetic energy of all the molecules.

4. The energy of a given sample of an ideal gas depends only on its

(a) volume

(b) pressure

(c) density

(d) temperature

Answer: (d)

Explanation: The absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules. So the energy of a given sample of a gas depends only on the temperature, not on the pressure, volume, or density.

5. Which of the following gases has maximum rms speed at a given temperature?

(a) hydrogen

(b) nitrogen

(c) oxygen

(d) carbon dioxide

Answer: (a)

Explanation: Since the kinetic energy of a gas

U = (3/2)nRT.

So U depends only on T. For a given temperature, the average kinetic energy of a molecule is the same. The kinetic energy depends upon the product of mass and the square of the speed. So a lighter molecule will have the maximum speed. Out of the given options, hydrogen is the lightest, hence has the highest rms speed.

6. Figure 24-Q1 shows graphs of pressure vs density for an ideal gas at two temperatures T₁ and T₂.

(a) T₁ > T₂

(b) T₁ = T₂

(c) T₁ < T₂

(d) any of the three is possible.
The figure for Q-6

Answer: (a)

Explanation: The ideal gas equation can also be written as

p=ρRT

So for a given ρ, p is higher for higher T. In the given figure, for ρ=A, p is higher in T₁.

Hence T₁ > T₂.

7. The mean square speed of the molecules of a gas at absolute temperature T is proportional to

(a) 1/T

(b) √T

(c) T

(d) T².

Answer: (c)

Explanation: vᵣₘₛ² = 3kT/m =(3k/m)T

→vᵣₘₛ² ∝ T

8. Suppose a container is evacuated to leave just one molecule of a gas in it. Let vₐ and vᵣₘₛ represent the average speed and the rms speed of the gas.

(a) vₐ > vᵣₘₛ

(b) vₐ < vᵣₘₛ

(c) vₐ = vᵣₘₛ

(d) vᵣₘₛ is undefined.

Answer: (c)

Explanation: Since there is only one molecule, the speed will be constant, say v. The average speed, vₐ will be this constant speed. vₐ = v.

rms speed, vᵣₘₛ =√(Σv²/N) =√v² =v

(Since N=1)

Hence, vₐ = vᵣₘₛ.

9. The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at the same temperature is about

(a) 125 m/s

(b) 2000 m/s

(c) 8000 m/s

(d) 31 m/s.

Answer: (b)

Explanation: Molecular weight of Oxygen = 32 g/mole, and that of hydrogen = 2 g/mole. Let m₁, m₂ be the masses and v₁, v₂ be the rms speeds of an oxygen and a hydrogen molecule, respectively. Then m₁ = 16m₂.

At the same temperature, kinetic energy will be the same. So,

½m₁v₁² = ½m₂v₂²

→v₂² = (m₁/m₂)v₁²

→v₂ = √(m₁/m₂)*v₁

=√(16)*500 m/s

=2000 m/s.

10. The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be

(a) 100 kPa

(b) 200 kPa

(c) 400 kPa

(d) 800 kPa.

Answer: (a)

Explanation: Since pV = nRT, here V and T are constant. R is the universal gas constant. So, p ∝ n. Since half the gas is removed, n becomes n/2. Hence, p will now become p/2, i.e., 200/2 kPa =100 kPa.

11. The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become

(a) v

(b) v√2

(c) 2v

(d) 4v.

Answer: (c)

Explanation: vᵣₘₛ = √(3kT/m) =v

Here, temperature T' = 2T, and after dissociation of oxygen molecules, m' = m/2. Putting in the above equation,

v'ᵣₘₛ = √{3k*2T/(m/2)} =2√(3kT/m) =2v.

12. The quantity pV/kT represents

(a) mass of the gas

(b) the kinetic energy of the gas

(c) number of moles of the gas

(d) the number of molecules in the gas.

Answer: (d)

Explanation: The ideal gas equation can also be written as pV = NkT because of R=Nₐk and N=nNₐ. From this relation, N = pV/kT. So it represents the number of molecules in the gas.

13. The process on an ideal gas, shown in figure (24-Q2), is

(a) isothermal

(b) isobaric

(c) isochoric

(d) none of these.
The figure for Q-13

Answer: (c)

Explanation: The figure shows that p is directly proportional to T. So from the equation, pV=nRT

→p=(nR/V) T

V should also be constant. The constant volume process is isochoric.

14. There is some liquid in a closed bottle. The amount of liquid is continuously decreasing. The vapor in the remaining part

(a) must be saturated

(b) must be unsaturated

(c) maybe saturated

(d) there will be no vapor.

Answer: (b)

Explanation: In a liquid-vapor combination, the molecules from the liquid go into the vapor, and the molecules from the vapor go into the liquid. When the vapor is unsaturated, the number of molecules going from liquid to vapor is more than coming back. So the amount of liquid decreases. Since the amount of liquid is continuously decreasing here, the vapor is unsaturated.

15. There is some liquid in a closed bottle. The amount of liquid remains constant as time passes. The vapor in the remaining part

(a) must be saturated

(b) must be unsaturated

(c) maybe saturated

(d) there will be no vapor.

Answer: (a)

Explanation: Since the amount of liquid remains constant, the number of molecules going out and coming in is the same. It is because the remaining part can not hold more molecules, so it must be saturated.

16. Vapor is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel

(a) increases continuously

(b) decreases continuously

(c) first increases and then decreases

(d) first increases and then becomes constant.

Answer: (d)

Explanation: The pressure will first increase because the number of molecules increases in the constant volume. When the pressure equals the saturated vapor pressure, the injected vapor will not remain as vapor but condense, and the vapor pressure will be constant. Hence, the option (d).

17. A vessel A has volume V and a vessel B has volume 2V. Both contain some water, which has a constant volume. The pressure in the space above water is pₐ for vessel A and pᵦ for vessel B.

(a) pₐ = pᵦ

(b) pₐ =2pᵦ

(d) pᵦ = 4pₐ

Answer: (a)

Explanation: Since the volume of water in both the vessels is constant, the vapor above the liquid in both the vessels is saturated. Hence, both vapors are at saturated vapor pressure. So the option is (a).

OBJECTIVE-II

1. Consider a collision between an oxygen molecule and a hydrogen molecule in a mixture of oxygen and hydrogen kept at room temperature. Which of the following are possible?

(a) The kinetic energies of both the molecules increase.

(b) The kinetic energies of both the molecules decrease.

(c) The kinetic energy of the oxygen molecule increases, and that of the hydrogen molecule decreases.

(d) The kinetic energy of the hydrogen molecule increases, and that of the oxygen molecule decreases.

Answer: (c), (d)

Explanation: Since the collision is fully elastic, the kinetic energies of both the molecules can not increase or decrease simultaneously, because the total momentum and energy will not be conserved (Kinetic energy increases or decreases if the speed increases or decreases).

Hence, the options (c) and (d) depend upon the directions of velocity before the strike.

2. Consider a mixture of oxygen and hydrogen kept at room temperature. As compared to a hydrogen molecule, an oxygen molecule hits the wall

(a) with a greater average speed

(b) with a lower average speed

(c) with greater average kinetic energy

(d) with smaller average kinetic energy.

Answer: (b)

Explanation: The average kinetic energy of both the molecules will be the same because the temperature is the same. Since the oxygen molecule is heavier than hydrogen, it will have a lower average speed for the same kinetic energy. Thus, the option (b).

3. Which of the following quantities is zero on average for the molecules of an ideal gas in equilibrium?

(a) kinetic energy

(b) momentum

(c) density

(d) speed.

Answer: (b)

Explanation: Since the gas has mass and volume, the density can not be zero. Option (c) is wrong. The kinetic energy and the speed are scalars, i.e., they have only magnitudes; hence they will be simply added and can not be zero. The momentum is a vector quantity and has the same direction as the velocity. In the equilibrium condition, the gas has no net velocity; hence the average momentum for the molecules is zero.

4. Keeping the number of moles, volume, and temperature the same, which of the following are the same for all ideal gases?

(a) rms speed of a molecule

(b) density

(c) pressure

(d) average magnitude of momentum.

Answer: (c)

Explanation: The rms speed, density, and the average magnitude of the momentum will depend upon the mass of the molecules of the ideal gas under consideration. But the pressure will remain the same because p=nRT/V and here all on the right side are constants.

5. The average momentum of a molecule in a sample of an ideal gas depends on

(a) temperature

(b) number of moles

(c) volume

(d) None of these.

Answer: (d)

Explanation: As discussed in Q-3, the average momentum of a molecule in an ideal gas sample is zero. Hence, the option (d).

6. Which of the following quantities is the same for all ideal gases at the same temperature?

(a) the kinetic energy of 1 mole

(b) the kinetic energy of 1 g

(c) the number of molecules in 1 mole

(d) the number of molecules in 1 g.

Answer: (a), (c)

Explanation: The average kinetic energy of a molecule is the same for all gases at a fixed temperature. Since 1 mole of any gas contains the same number of molecules, the kinetic energy of 1 mole of all the gases at the same temperature will be the same. Hence, the options (a) and (c). One gram of different gases will have different numbers of molecules; hence, options (b) and (d) are not true.

7. Consider the quantity MkT/pV of an ideal gas where M is the mass of the gas. It depends on the

(a) the temperature of the gas

(b) the volume of the gas

(c) the pressure of the gas

(d) nature of the gas.

Answer: (d)

Explanation: MkT/pV = MkT/nRT

=Mk/nR = M*K,

(where K =k/nR = constant).

So the quantity in consideration does not depend on pressure, volume, or temperature of the gas. It only depends on the mass of the gas, M. But M = mN, where m is the molecular weight, and N is the number of molecules. The molecular weight depends on the nature of the gas. Hence, the option (d).

EXERCISES

1. Calculate the volume of 1 mole of an ideal gas at STP.

Answer: STP means T = 273 K and

p = 1 atm =1.01325x10⁵ N/m². Here n = 1 mole, and we know that R = 8.3 J/mol-K. Volume V =?

From the ideal gas equation,

pV =nRT

→V = nRT/p

= 1*8.3*273/1.01325x10⁵

= 2.24x10⁻² m³

= 22.4 liters

2. Find the number of molecules of an ideal gas in a volume of 1.000 cm³ at STP.

Answer:

Since the volume of 1 mole of an ideal gas at STP =22.4 liters =2.24x10⁻² m³, 1.000 cm³ of an ideal gas will contain 1x10⁻⁶/2.24x10⁻² =4.46x10⁻⁵ moles. Hence, the number of molecules in 1.000 cm³ of an ideal gas at STP

=4.46x10⁻⁵*6.02x10²³

=2.685x10¹⁹

3. Find the number of molecules in 1 cm³ of an ideal gas at 0°C and at a pressure of 10⁻⁵ mm of mercury.

Answer: V = 1 cm³ =1x10⁻⁶ m³

T = 0°C = 273 K

p = 10⁻⁵ mm of mercury

=136009.8(10⁻⁵/1000) N/m²

(Since p =ρgh, ρ of mercury =13600 kg/m³)

=1.33x10⁻³ N/m²

R = 8.3 J/mol-K

Hence n =pV/RT

→n =1.33x10⁻³1x10⁻⁶/(8.3273)

= 5.869x10⁻¹³ moles

Hence the number of molecules,

N =n*Nₐ

=5.869x10⁻¹³*6.02x10²³

= 3.53x10¹¹

4. Calculate the mass of 1 cm³ of oxygen kept at STP.

Answer: 1 cm³ = 1x10⁻⁶ m³

Since 22.4 liters i.e. 2.24x10⁻² m³ of an ideal gas contains 1 mole or 6.02x10²³ number of molecules, hence 1x10⁻⁶ m³ of gas will contain

1x10⁻⁶*1/2.24x10⁻² moles

=4.46x10⁻⁵ moles.

(The molecular weight of oxygen =2*16 =32 amu

Hence 1 mole of oxygen = 32 g)

=4.46x10⁻⁵*32 g

=1.43x10⁻³ g

=1.43 mg.

5. Equal masses of air are sealed in two vessels, one of volume V₀ and the other of volume 2V₀. If the first vessel is maintained at a temperature of 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

Answer: Here, for the first vessel, V = V₀, T = 300 K, and let the pressure = p. So,

pV₀ = nRT ------ (i)

Similarly, for the second vessel, V =2V₀, T'=600 K, and let the pressure =p'. So,

p'*2V₀ = nRT' ------(ii)

Dividing (i) by (ii),

pV₀/2p'V₀ =nRT/nRT' =T/T'

→p/2p' =300/600 =1/2

→p/p' =1

Hence, the ratio of p:p' = 1 : 1.

6. An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10⁻³ mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6x10²³ per mol, the density of mercury =13600 kg/m³, and g = 10 m/s².

Answer: V =250 cc =250/10⁶ m³ =2.5x10⁻⁴ m³.

p= ρgh =1360010(10⁻³/1000) N/m²

=0.136 N/m²

R =8.3 J/mol-K

T =27+273 K =300 K

Hence, from the ideal gas equation,

pV=nRT

→n = pV/RT

=0.136*2.5x10⁻⁴/(8.3*300) moles

= 1.365x10⁻⁸ moles

=1.36x10⁻⁸*6.02x10²³ molecules

=8.1x10¹⁵ molecules

7. A gas cylinder has walls that can bear a maximum pressure of 1.0x10⁶ Pa. It contains gas at 8.0x10⁵ Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

Answer: Since at a constant volume the pressure is directly proportional to the absolute temperature (Charles ' law of pressure), p/p' = T/T'.

Here, p =8x10⁵ Pa, p' =1x10⁶ Pa, T=300 K, T'=?

So, T' =p'T/p =1x10⁶*300/8x10⁵

→T' = 3000/8 = 375 K.

8. 2 g of hydrogen is sealed in a vessel of volume 0.02 m³ and is maintained at 300 K. Calculate the pressure in the vessel.

Answer: 2 g of hydrogen = 2/2 mole =1 mole of hydrogen. So, n = 1, V=0.02 m³, T = 300 K, R=8.3 J/mol-K. p=?

From the ideal gas equation,

pV = nRT

→p = nRT/V =18.3300/0.02 Pa

→p = 1.24x10⁵ Pa.

9. The density of an ideal gas is 1.25x10⁻³ g/cm³ at STP. Calculate the molecular weight of the gas.

Answer: Volume of 1 mole of an ideal gas at STP = 2.24x10⁻² m³. Hence the mass of 1 mole of the given gas at STP

= 2.24x10⁻²1.25x10⁻³10⁶ g

= 28 g

Hence, the molecular weight of the given gas = 28 g/mol.

10. The temperature and pressure at Shimla are 15.0°C and 72.0 cm of mercury, and at Kalka, these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Shimla.

Answer:

The ideal gas equation is

pV=nRT =(m/M)RT {where m is the mass and M is the molecular weight of the gas.}

→pM=(m/V)RT =ρRT

Let the air density at Shimla =ρ and at Kalka = ρ'. At Shimla p=72 cm of mercury, T=15+273=288 K

At Kalka, p' =76 cm of mercury and T'=35+273=308 K

Hence, pM/p'M =ρRT/ρ'RT'

→p/p' =ρT/ρ'T'

→ρ'/ρ =p'T/pT'

→ρ'/ρ=76*288/72*308=0.987

11. Figure (24-E1) shows a cylindrical tube with adiabatic walls and filled with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium in the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts of the vessel.

The figure for Q-11

Answer: Since n = pV/RT. Here, when the separator is in the middle, V is the same for both sides and p and T are also the same (given). Hence n is also the same for both sides.

The total volume of both sides = 2V. When the separator divides the tube in 1:3, the volume of one side V' = 2V/4 and of the other side V" = 6V/4. Since the separator is diathermic, the final temperature will remain T.

Now for the one side, p' =nRT/V', and for the other side p" =nRT/V". Hence the ratio

p'/p" = V"/V' =(6V/4)/(2V/4) = 3:1

12. Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which rms speed is double the speed calculated in the previous part.

Answer: The rms speed is given as

v = √(3RT/M₀)

Here for hydrogen, M₀ = 2 g/mol =0.002 kg/mol, T = 300K and R = 8.3 J/mol-k, so

v = √(3*8.3*300/0.002)

= 1933 m/s

Since the square of the rms speed is proportional to the absolute temperature,

v²/v'² = T/T'

Here, v' = 2v, T = 300 K, T' = ?

so, v²/4v² = 300/T'

→T' = 300*4 = 1200 K

rms speed, v = √(3pV/M)

13. A sample of 0.177 g of an ideal gas occupies 1000 cm³ at STP. Calculate the rms speed of the gas molecules.

Answer: Given that M = 0.177 g =1.77x10⁻⁴ kg, V = 1000 cm³ =1x10⁻³ m³, T = 273 K, p = 1 atm =1x10⁵ N/m². Hence, the rms speed

v = √(3pV/M)

=√(31x10⁵1x10⁻³/1.77x10⁻⁴)

=√(1.69x10⁶)

=1.30x10³ m/s

=1300 m/s

14. The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6x10⁻¹⁹ J). Calculate the temperature of the air. Boltzmann constant k = 1.38x10⁻²³ J/K.

Answer: Since the average translational kinetic energy is given as

K =3kT/2

Here, K = 0.04 eV =0.04*1.6x10⁻¹⁹ J. T =?

From this, T = 2K/3k

= 20.041.6x10⁻¹⁹/(3*1.38x10⁻²³)

= 309 K

15. Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the Earth.

Answer: Let the average speed of oxygen molecules = v. Then

v =√(8RT/πM₀)

Here T = 300 K, for oxygen M₀ =32 g/mol =0.032 kg/mol and R=8.3 J/mol-K. Hence

v = √(88.3300/0.032π) = 445 m/s

The Earth's diameter is approximately 12800 km = 1.28x10⁷ m. So average time taken by an oxygen molecule to travel a distance equal to Earth's diameter = 1.28x10⁷/445 =28764 s =28764/3600 hr =8.0 hr

16. Find the average magnitude of the linear momentum of the helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64x10⁻²⁷ kg and Boltzmann constant = 1.38x10⁻²³ J/K.

Answer: Given m = 6.64x10⁻²⁷ kg, T =273 K, k =1.38x10⁻²³ J/K. The average speed is given as

v = √(8kT/πm)

Hence, the average momentum

mv =√(8kTm/π)

=√(81.38x10⁻²³273*6.64x10⁻²⁷/π)

=80x10⁻²⁵ kg-m/s

=8.0x10²⁴ kg-m/s

17. The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Answer: Mean speed or average speed of hydrogen molecules is

v =√(8RT/πM₀)

and for the helium molecules

v' =√(8RT'/πM₀')

Hence v/v' =√(TM₀'/T'M₀)

But here, v = v', M₀' =4 g/mol and M₀ =2 g/mol. Hence

1 = √(4T/2T') =√(2T/T')

→2T/T' = 1

→T/T' =1/2

→T:T' = 1:2

18. At what temperature the mean speed of the molecules of hydrogen gas equal the escape speed from the Earth?

Answer: Escape speed from the earth =√2gr

The mean speed of the molecules of hydrogen gas =√(8RT/πM₀).

M₀ =0.002 kg/mol

r =6.4x10⁶ m

g =9.8 m/s²

R = 8.3 J/mol-K

for the given condition

8RT/πM₀ = 2gr

→T = πgrM₀/4R

=3.149.86.4x10⁶0.002/(48.3)

≈ 11800 K

19. Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

Answer: For hydrogen, M₀ = 2 g/mol; for nitrogen, M₀' =28 g/mol. Since both are in a mixture, T is the same for both. Mean speed of Hydrogen

v =√(8RT/πM₀), and mean speed of nitrogen molecules,

v' =√(8RT/πM₀')

Hence the ratio

v/v' = √(M₀'/M₀) =√(28/2) =√14 =3.74

20. Figure (24-E2) shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.
The figure for Q-20

Answer: Due to the diathermic separator, the temperature in both parts is the same as T. The rms speed in the left part v = √(3kT/m). The mean speed in the right part v' =√(8kT/πm'), where m and m' are the masses of molecules of gases in the left and right parts, respectively. Here given that

v = v'

→√(3kT/m) =√(8kT/πm')

→3/m = 8/πm'

→m/m' = 3π/8 = 1.18.

21. Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38x10⁻⁵ cm.

Answer: We should know the average speed of the molecules. Here at STP, T=273 K, M₀ =2 g/mol =0.002 kg/mol. R =8.3 J/mol-K. Thus the average speed, v = √(8RT)/(πM₀)

→ v =√(88.3273)/(π*0.002)

→ v = 1698 m/s

Mean free path = 1.38x10⁻⁵ cm

=1.38x10⁻⁷ m

So the average time taken in one collision =(1.38x10⁻⁷)/1698 s

=8.13x10⁻¹¹ s

Thus collisions per second =1/(8.13x10⁻¹¹)

=1.23x10¹⁰.

22. Hydrogen gas is contained in a closed vessel at 1 atm (100kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed, making an average angle of 45° with it. How many molecules strike each square meter of the wall per second?

Answer: (a) T = 300 K, M₀ =2 g/mol =0.002 kg/mol. R = 8.3 J/mol-K. Mean speed, v =√(8RT)/(πM₀)

=√(88.3300)/(π*0.002)

=1780 m/s

(b) The momentum of one molecule =mv, where m is the mass of one molecule. Since the molecules collide at an angle of 45°, the change of momentum perpendicular to the wall = 2mv*cos45°

=√2mv.

If there are n molecules striking per second per m², then the change of momentum of molecules striking per second =√2mnv = Force/m² =pressure.

Here m = (0.002/6.02x10²³) =3.32x10⁻²⁷ kg

So √2mnv =10⁵ N/m²

→n = 10⁵/(√23.32x10⁻²⁷1780)

=1.2x10²⁸.

23. Air is pumped into an automobile tire's tube up to a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day, the temperature rises to 40°C, and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.

Answer: Let the volume of the tire in the morning =V. Temperature = 20°C =20+273 K =293 K. p =200 kPa .

For the daytime, T' =273+40 K =313 K. V' = 1.02V; let the pressure =p'.

Since pV/T =p'V'/T'

p' =pVT'/TV'

=200V313/(293*1.02V)
=209 kPa.

24. Oxygen is filled in a closed metal jar of volume 1.0x10⁻³ m³ at a pressure of 1.5x10⁵ Pa and temperature 400 K. The jar has a small leak in it. The atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and temperature inside the jar equalize with the surroundings.

Answer: p =1.5x10⁵ Pa, V = 1.0x10⁻³ m³, T = 400 K, n = initial moles of molecules. At the final stage, p' = 1x10⁵ Pa, T' = 300 K, n' = final moles of molecules. V' = V. Now,

pV =nRT and p'V' =n'RT'

→n =pV/RT and n' =p'V'/RT' =p'V/RT'

The moles of molecules escaped

=n - n'

=pV/RT - p'V/RT'

=(p/T - p'/T')V/R

=(1.5x10⁵/400 - 1x10⁵/300)*1x10⁻³/8.3

=(375-333.33)*1x10⁻³/8.3

=5.02x10⁻³ moles.

Hence the mass of oxygen gas escaped = 5.02x10⁻³*32 g

=0.16 g.

25. An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0x10⁵ Pa and density of water = 1000 kg/m³.

Answer: Pressure due to 3.3 m deep river water = ρgh

= 10009.83.3 Pa

= 32340 Pa

Total pressure at this depth =Atmospheric pressure + Water pressure

=1.0x10⁵+32340 Pa

=1.3234x10⁵ Pa

The radius of the air bubble, r = 2 mm
Volume, V = 4πr³/3
Near the surface,
p' = 1x10⁵ Pa
V' =4πr'³/3, where r' is the radius of the bubble near the surface. The temperature is constant. Hence, from Boyle's law,
p'V' = pV
→p'(4πr'³/3) = p(4πr³/3)
→r'³ =pr³/p'
→r' =r(p/p')¹/³ =2*(1.3234)¹/³
→r' =2.2 mm.

26. Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m³. One of the tubes gets punctured, and the volume of the tube reduces to 0.0005m³. How many moles of the air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.

Answer: Initial pressure, p = 2 atm, volume, V = 0.002 m³. Number of moles = n, T = 300 K.

Final pressure, p' = 1 atm (atmospheric pressure), V' = 0.0005 m³, Number of moles = n'. T' = 300 K.

So the number of moles of air leaked out = n - n'

=pV/RT - p'V'/RT

=(pV - p'V')/RT

=(2x10⁵0.002 - 1x10⁵0.0005)/(8.3*300)

=(400-50)/2490

=350/2490

=0.14 moles.

27. 0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.

Answer: Since the internal energy of a gas is the total kinetic energy of all the molecules, it is given as,

U = (3/2)nRT

n = 0.04/4 =0.01 mole

R =8.3 J/mol-K, T =273+100 =373 K.

So, U = (3/2)0.018.3*373

=46.44 J

For the required temperature T, U'=46.44+12 =58.44 J

So, 58.44 = 1.50.018.3*T

→T = 58.44/(0.1245) = 469 K =(469-273)°C = 196°C.

28. During an experiment, an ideal gas is found to obey an additional law pV² = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.

Answer: Since the gas is ideal,

pV =nRT

→p =nRT/V

Similarly, for the final stage, temperature = T', pressure p' and volume V',

p' = nRT'/V'.
Now it obeys an additional law
pV² = p'V'²
→(nRT/V)V² = (nRT'/V')V'²
→TV = T'V', but given that V' = 2V.
→TV = T'*2V
→T' = T/2.

29. A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K, and the volume of the vessel is 0.166 m³. Find the pressure of the mixture.

Answer: V = 0.166 m³, T = 300 K.

1.60 g of oxygen = 1.60/32 moles =0.05 moles.

2.80 g of nitrogen = 2.80/28 moles =0.10 moles.

Hence, the total number of moles of ideal gases in the vessel, n = 0.05+0.10 =0.15

From the ideal gas equation,

pV = nRT

→p = nRT/V

→p = 0.158.3300/0.166

→p = 2250 N/m².

30. A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

Answer: Let the maximum height of the mercury column on the piston = h.

Diagram for Q-30

The initial pressure of air in the container, p = 75 cm of mercury = 0.75ρg N/m², where ρ = density of mercury in kg/m³.

Height of the cylinder = 100 cm = 1 m. If the area of the base of the cylinder = A m², then volume, V = A*1 =A m³.

Finally, the volume of air in the cylinder, V' = A(1-h) m³.

Pressure, p' = (0.75+h)ρg.

Since the temperature and the number of moles remain constant,

pV = p'V'

→0.75ρgA = (0.75+h)ρgA((1-h)

→0.75 = (0.75+h)(1-h)

→75 = (75+100h)(1-h)

→75+100h-75h-100h² =75

→100h²-25h=0

→h(100h-25)=0

Since h is not zero,

100h-25 =0

→h = 25/100 = 0.25 m = 25 cm.

31. Figure (24-E3) shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature, and volume are pₐ, Tₐ, V in vessel A and pᵦ, Tᵦ, V in vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy

p/T =½{pₐ/Tₐ + pᵦ/Tᵦ}

when equilibrium is achieved.
The figure for Q-31

Answer: Let the number of moles of the ideal gas in A = n and in B = n'.

From the ideal gas equation

pₐV = nRTₐ

→n = pₐV/RTₐ. Similarly

n' = pᵦV/RTᵦ

When the equilibrium is achieved after connecting both vessels, the total number of molecules is n" = n + n'.

In this case, the volume becomes 2V.

So the ideal gas equation for this volume is,

p(2V) = n"RT

→p/T =n"R/2V =(n+n')R/2V

→p/T = {(pₐV/RTₐ)+(pᵦV/RTᵦ)}R/2V

→p/T = ½{pₐ/Tₐ + pᵦ/Tᵦ}.

Proved.

32. A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to the atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.

Answer: Number of moles of air,

n = pV/RT

(a) p = 100 kPa =1x10⁵ N/m²

V = 50 cc = 50x10⁻⁶ m³ =5x10⁻⁵ m³

R = 8.3 J/mol-K

T = 0°C = 273 K

→n = 1x10⁵5x10⁻⁵/(8.3273)

= 2.207x10⁻³ moles

= 2.207x10⁻³*28.8 g

= 0.0635 g.

(b) In the boiling water, T = 100°C =373 K

Now, n = pV/RT

→n = 1x10⁵*5x10⁻⁵/(8.3*373)

=1.615x10⁻³ moles

=1.615x10⁻³*28.8 g

=0.0465 g.

(c) When the container is closed, the number of moles of air in the container n=1.615x10⁻³, so the pressure in the melting ice bath,

p = nRT/V

= 1.615x10⁻³*8.3*273/5x10⁻⁵ Pa

= 73.2 kPa.

33. A uniform tube closed at one end contains a pallet of mercury 10 cm long. When the tube is kept vertically with the closed end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure = 75 cm of mercury.

Answer: Let the cross-section of the uniform tube = A cm². The volume of the air column trapped = 20A cm³ =20A/10⁶ m³. The pressure = 75-10 cm =65 cm of mercury =65ρg/100 N/m² (ρ is the density of air in kg/m³). When the tube is inverted, the pressure p' = 75+10 cm = 85 cm of mercury =85ρg/100 N/m². If the length of the air column now is L cm, then its volume V'=LA/10⁶ m³. Since the temperature is constant,

pV = p'V'

→(65ρg/100)20A/10⁶ = (85ρg/100)LA/10⁶

→L = 1300/85 cm = 15.3 cm.

34. A glass tube sealed at both ends is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer:

Let the pressure on both sides in the second case = p. If the length of the air column on the cooler side = L cm, then the length of the air column on the hotter side =90-L cm. The volume of air on the cooler side, V =LA/10⁶ m³, and temperature T = 273 K.

So, pV =nRT

→p(LA/10⁶) =nR*273 -------- (i)

On the other side, pressure = p, Volume, V' =(90-L)A/10⁶ m³, temperature T' = 127+273 =400 K

So, pV' =nRT'

→.p(90-L)A/10⁶ =nR*400 -----(ii)

Dividing (i) by (ii)

L/(90-L) = 273/400

→400L =273*90 - 273L

→(400+273)L =273*90

→673L = 273*90

→L = 273*90/673 = 36.5 cm.

35. An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of a uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm, respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.

Answer: The pressure of the trapped air column in the vertical tube, p = 76+20 =96 cm of mercury =96ρg/100 Pa. If the cross-sectional area of the tube is A cm², volume V = 43A/10⁶ m³.

Diagram for Q-35

When the tube is tilted through 60°, the pressure of the air column, p' = 76+20*cos 60° cm =76+10 cm =86 cm of mercury = 86ρg/100 Pa. If the length of the air column now is = L cm, then volume, V' = LA/10⁶ m³.

Since the amount of air and the temperature is the same,

pV = p'V'

→(96ρg/100)43A/10⁶ = (86ρg/100)LA/10⁶

→9643 = 86L

→L = 96*43/86 = 48 cm.

36. Figure (24-E4) shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in parts A and B are 400 K and 100 K, respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.
The figure for Q-36

Answer: Since the separator is weakly conducting, after a long time the temperature on both sides will be the same, say T. Also, the separator can freely slide; hence the pressure on both sides will be equal. Take initial pressure =p and final pressure = p'.

If the area of cross-section of the tube is S cm², the volume of the part A, V = 20*S/10⁶ m³ =20F m³ (Where F =S/10⁶). And of part B, V' =10F m³.

Assume that after a long time the separator's position is x cm from the left. Now the volume of part A, Vₐ' =xF m³, and that of part B, Vᵦ' =(30-x)F m³.

Now the ideal gas equation for part A,

pV/400 = p'Vₐ'/T

→p*20F/400 = p'*xF/T

→p/p' = 20x/T --------------- (i)

Similarly for part B,

pV'/100 = p'Vᵦ'/T

→p*10F/100 = p'*(30-x)F/T

→p/p' = 10(30-x)/T ----------(ii)

Equating these two values of p/p' we get

20x/T = 10(30-x)/T

→20x = 300 - 10x

→30x = 300

→x = 10 cm from the left end.

37. A vessel of volume V₀ contains an ideal gas at pressure p₀ and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt=r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

Answer: Let at time t the pressure of the gas in the vessel = p, and the volume given Vₒ. If there are n moles in the vessel and in a time dt the number of moles of gas pumped out = dn, the decrease in pressure of the vessel = dV. From the ideal gas equation,

(p-dp)Vₒ = (n-dn)RT

→pVₒ - Vₒdp = nRT - dn*RT

But pVₒ = nRT, so

Vₒdp = dn*RT --------------- (i)

Now the pressure of the gas taken out is equal to the internal pressure, hence

(p-dp)*dV = dnRT

→pdV - dp*dV = dnRT

→pdV = dnRT, -------------- (ii)

(since dp*dV is negligible.)

Equating (i) and (ii)

Vₒdp = pdV

→dp/p = dV/Vₒ

→dp/p = rdt/Vₒ

(since dV/dt = r, dV =-rdt, since volume decreases with time)

Integrating between pₒ and p and t = 0 to t,

→∫dp/p = -∫rdt/Vₒ

→[ln p] = -r[t]/Vₒ

→(ln p- ln pₒ) = -rt/Vₒ

→ln (p/pₒ) = -rt/Vₒ

→p/pₒ = e^-(rt/V0)

→p = pₒe^-(rt/V0)

(b) Half of the gas has been pumped out, so the number of moles inside the vessel = n/2. If the pressure now is p', then

p'Vₒ =(n/2)RT = nRT/2 = pVₒ/2

→p' = p/2, so the pressure now is half the original pressure.

From the equation derived in part (a).

e^-(rt/V0) = 1/2

→-(rt/Vₒ) = ln(1/2) = -ln 2

→t = (Vₒ.ln 2)/r

38. One mole of an ideal gas undergoes a process

p = p₀/{1+(V/V₀)²}

where p₀ and V₀ are constants. Find the temperature of the gas when V = V₀.

Answer: Given n = 1 mole. So

pV = nRT, and for n =1 and V = Vₒ

→pVₒ = RT -------- (i)

But for the given process,

p = pₒ/{1+(V/Vₒ)²}

putting V = Vₒ

→p = pₒ/2

From (i)

(pₒ/2)Vₒ = RT

→pₒVₒ = 2RT

→T = pₒVₒ/2R mol⁻¹

39. Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surroundings through the windows, etc.

Answer: The internal energy U of an ideal gas is equal to the total translational kinetic energy of all the molecules, and it is given as

U = (3/2)nRT

But from the ideal gas equation

pV = nRT, hence

U = (3/2)pV

Since p and V are constant, U is also constant.

40. Figure (24-E5) shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated, and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate dN/dl.
The figure for Q-40

Answer: For constant volume,

p/T = p'/T'

here T = 300 K, T' = 600 K, p = 1 atm.

→p' = (600/300)*1 atm = 2 atm.

The radius r = 5 cm = 0.05 m

Circumference of the cork C =2πr

→C = 2π(0.05) =0.10π

The normal force on length dl = dN

The force of friction on dl =µdN

Hence, the total force of friction between cork and tube, F = C.µdN/dl

→F = 0.10π0.20dN/dl, (since µ = 0.20)

→F = 0.02π dN/dl

When the cork pops out, the net pressure on the cork area =p' - p =1 atm, because the outside pressure is still p.

Area of the cork, A = πr² =π(0.05)²

The force on the cork, F = (p'-p)A

→F = (1x10⁵ Pa)*{π(0.05)² m²}

Equating the force of friction and the force by gas on the cork, we get,

0.02πdN/dl = (1x10⁵){π(0.05)²}

→dN/dl = (1x10⁵)*0.0025/0.02

→dN/dl = 0.125x10⁵ = 1.25x10⁴ N/m

41. Figure (24-E6) shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T₀ and its pressure is p₀, which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T₀?
The figure for Q-41

Answer: (a) Since the pressure on both sides of the pistons is the same, there is no net pressure on the pistons; hence no net force on the pistons. Thus the tension in the wire is zero.

(b) Since the volume is constant,

p/T = p'/T'

Here p = p₀, T = T₀ and T' = 2T₀, so

p₀/T₀ = p'/2T₀

→p' = 2p₀

So the pressure inside the tube is 2p₀ and outside it is p₀. So net pressure on a piston = 2p₀ - p₀ =p₀. Area of a piston = A. The force on each of the pistons = tension in the wire = p₀A.

42. Figure (24-E7) shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height h₀ and pressure 2p₀, where p₀ is the atmospheric pressure. There is a hole in the wall of the tank at a depth h₁ below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h₂ of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.
The figure for Q-42

Answer: (a) Let us consider the pressure at the water level in the tank. It is given as 2p₀, but through the vertical tube, the water height, it comes to

p₀+(h₀+h2)ρg. Equating these two, we get,

p₀+(h₀+h₂)ρg = 2p₀

→(h₀+h₂)ρg = p₀

→h₀+h₂ =p₀/ρg

→h₂ = p₀/ρg - h₀

(b) Consider the exit level of the water as the reference line. If the velocity of outgoing water is v and the velocity inside the tank at this level =v' then, according to Bernoulli's theorem, the sum of the three heads inside and outside the exit should be constant,

p₀ +ρgh +½ρv² = p + ρgh' +½ρv'²

(outside ↑) (inside ↑)

Since the reference line is at the exit level, h = h' = 0. Velocity inside the tank at this level is negligible, so v' = 0. The pressure at this level inside,

p = 2p₀ + ρg(h₁-h₀), putting these values we get,

p₀+½ρv² = 2p₀ +ρg(h₁-h₀)

→½ρv² = p₀ + ρg(h₁-h₀)

→ρv² = 2{p₀+ρg(h₁-h₀)}

→v² = 2{p₀+ρg(h₁-h₀)}/ρ

→v = √[2{p₀+ρg(h₁-h₀)}/ρ]

(c) Let the level of water in the tube above the top of the tank = x when water stops coming out. Then pressure at the exit level = (x+h₁)ρg. But this should be zero because water stops coming out; hence

(x+h₁)ρg = 0

→x+h₁ = 0

→x = -h₁

43. An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of a cross-sectional area 10 cm² and weight 1 kg (figure 24-E8). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature remains constant throughout the process.
The figure for Q-43

Answer: Pressure, p, on the gas column will be equal to atmospheric pressure plus the pressure due to the weight of the piston. So,

p = p₀+mg/A

m = 1 kg, A =10 cm² =1x10⁻³ m², p₀ =100 kPa = 1x10⁵ N/m².

p = 1x10⁵ + 1x10/(1x10⁻³) Pa

= 1x10⁵ + 1x10⁴ Pa

= 1.1x10⁵ Pa

Volume V = (20/100)A =0.20*1x10⁻³ =2x10⁻⁴ m³

After the chamber is completely evacuated, the pressure on the gas column is only due to the weight of the piston.

p' = mg/A =1x10/10⁻³ =1x10⁴ Pa

If the length of the gas column now = X cm, then volume V' = (X/100)*(10⁻³) =X*10⁻⁵ m³

Since temperature is constant,

pV = p'V'

→(1.1x10⁵)*(2x10⁻⁴) =(1x10⁴)*(X*10⁻⁵)

→X = 2.2x10/10⁻¹ =2.2x100 cm

→X = 2.2 m.

44. An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of a cross-sectional area 10 cm² and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving around the Earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder.

Answer: In the laboratory on Earth,

p =1.1x10⁵ Pa, Volume V = 2x10⁻⁴ m³.

In the spaceship, the value of g = 0; hence, the piston is weightless, and it has no pressure on the gas column. Hence, the pressure now, p' = 1x10⁵ Pa. If the length of the gas column = x cm, then volume V' = (X/100)(10⁻³) m³ =10⁻⁵X m³. Since the temperature is constant,

pV = p'V'

→1.1x10⁵2x10⁻⁴ =1x10⁵10⁻⁵*X

→X = 2.2*10/1 =22 cm.

45. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice, and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulb? The volume of the connecting tube is negligible.

Answer: Let the volume of each bulb = V. Initial pressure = p and temperature, T = 273 K. If each bulb has n moles of gas initially, then the ideal gas equation for each of the bulbs is,

pV =nRT

→n = pV/RT.

In the second stage, let the number of moles in the bulb in melting ice = n' and in the other = n". The pressure in each of the bulbs = p'.

So for the first bulb in melting ice

p'V =n'RT, →n' = p'V/RT

and for the second bulb at T' = 62°C i.e. 335 K,

p'V = n"RT', →n" = p'V/RT'

Since the total number of moles in both bulbs does not change,

n'+n" = 2n

→p'V/RT + p'V/RT' = 2pV/RT

→p'(1/T + 1/T') = 2p/T

{Here T = 273 K, T' =335 K and p =76ρg/100 Pa.}

→p'(1/273 + 1/335) = 2*(76ρg/100)/273

→p'(335+273)/(273335)=152ρg/100273

→p' =152ρg335/(100608)

=0.84ρg Pa

=0.84 m of mercury column

=84 cm of mercury column.

46. The weather report reads, "Temperature 20°C: Relative humidity 100%." What is the dew point?

Answer: 100% Relative humidity means that the air is fully saturated with water vapor. It can not hold more water vapor at this temperature of 20°C. If the temperature is reduced further, some of the water vapor will condense. Hence the dew point here is 20°C.

47. The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60°C, pressure = 104 kPa. If all the water vapor is removed from the room without changing the temperature, what will be the new pressure? The saturation vapor pressure at 25°C = 3.2 kPa.

Answer:

Relative humidity

= vapor pressure of air/Saturation vapor pressure at the same temperature

Here RH = 60%, i.e., 0.6

Saturation vapor pressure at 25°C =3.2 kPa. So,

VP/SVP = 0.6

→VP =0.6*SVP =0.6*3.2 =1.92 kPa ≈2 kPa

When all the water vapor is removed, the vapor pressure in the air becomes zero. Hence the pressure of air in the room = 104 kPa - 2 kPa = 102 kPa.

48. The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point?

Answer: If the dew point is 10°C, it means the air has saturation vapor pressure at this temperature. Even if the room temperature drops, the saturation vapor pressure for the given air will not change, and the dew point will remain 10°C.

49. Pure water vapor is trapped in a vessel of volume 10 cm³. The relative humidity is 40%. The vapor is compressed slowly and isothermally. Find the volume of the vapor at which it will start condensing.

Answer: When vapor pressure = Saturation vapor pressure, the vapor starts condensing. Given that relative humidity = 40%.

But, RH = VP/SVP

→0.40 = VP/SVP

→VP = 0.40*SVP = p (say)

V = 10 cm³.

Let the volume of vapor at which it starts condensing = V'. Pressure here = SVP = p' (say). Since the process is isothermal,

p.V = p'V'

→(0.40SVP)10 = (SVP)*V'

→V' = 0.40*10 cm³ = 4.0 cm³.

50. A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced into the tube, and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapor pressure at the room temperature is 1.0 cm.

Answer: The atmospheric pressure = 76 cm of mercury column.

Diagram for Q-50

When a small amount of water is introduced in the tube, it turns into water vapor above the mercury column due to low pressure. This water vapor exerts pressure above the mercury level in the tube and pushes it down. Since the mercury column is pushed down by 76-75.4 = 0.6 cm by the vapor, the present vapor pressure, VP = 0.6 cm of mercury.

Given that SVP = 1.0 cm, hence the relative humidity

RH = VP/SVP = 0.6/1.0 = 0.60 = 60%.

51. Using Figure (24.6) of the text, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm.

Answer: The boiling point of a liquid is the temperature at which the SVP equals the external pressure. We draw lines parallel to 760 mm (1 atm) and 380 mm (0.5 atm) of mercury in the graph to meet the Methyl alcohol curve. From these points, draw perpendiculars to the temperature axis.
The boiling point of Methyl alcohol at 1 atm and 0.5 atm
From the figure, we get the boiling point of methyl alcohol 65°C at 1 atm and 48°C at 0.5 atm.

52. The human body has an average temperature of 98°F. Assume that the vapor pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure (24.6) of the text for the vapor pressures.

Answer: 98°F = (98-32)*5/9

=66*5/9 =330/9 =36.7°C

Given that the vapor pressure of blood behaves like water, we find the pressure corresponding to 36.7°C for the water curve. It is 50 mm of mercury. Hence, it is the required answer.

53. A glass contains some water at room temperature 20°C. Refrigerated water is added to it slowly. When the temperature of the glass reaches 10°C, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20°C, and at 8.9 mm of mercury, it is 10°C.

Answer: It is clear that the dew point is at 10°C. Since at 8.9 mm of mercury, the boiling point of water is 10°C, the SVP of water at 10°C is 8.9 mm of mercury. Similarly, the SVP at 20°C is 17.5 mm of mercury.

Relative humidity

=SVP at Dew point/SVP at air temp.

=8.9/17.5

=0.51

=51%.

54. 50 m³ of saturated vapor is cooled down from 30°C to 20°C. Find the mass of the water condensed. The absolute humidity of saturated water vapor is 30 g/m³ at 30°C and 16 g/m³ at 20°C.

Answer: At both temperatures, the vapor is saturated. Amount of water in the given volume of vapor at 30°C = 50 m³*30 g/m³ =1500 g.

Amount of water in the given volume of vapor at 20°C =50 m³*16 g/m³ =800 g.

Hence the mass of water condensed =1500-800 =700 g.

55. A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapor pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value.

Answer: When the mercury column becomes constant, the vapor above is saturated. The saturated vapor pressure at atmospheric pressure is given as 0.80 cm of mercury. So the vapor above the mercury column applies a pressure of 0.80 cm of mercury, and the height of the column decreases by this height. Thus the final height of the mercury column =76 cm - 0.80 cm =75.2 cm.

56. 50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa, and the room temperature is 27°C. The water level in the jar is the same as the level outside. The saturation vapor pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

Answer: Atmospheric pressure =99.4 kPa. Since the water level in the jar is the same as the level outside, the combined pressure of oxygen and the water vapor is equal to the atmospheric pressure. SVP at 27°C = 3.4 kPa; hence the pressure of oxygen gas, p =99.4-3.4 =96 kPa =9.6x10⁴ N/m².

The volume of the oxygen gas, V =50 cc =50x10⁻⁶ m³ =5x10⁻⁵ m³. R=8.3 J/mol-K, T =273+27 K =300 K.

Hence, the number of moles in the oxygen gas, n =pV/RT

→n =9.6x10⁴*5x10⁻⁵/(8.3*300)

→n =1.93x10⁻³.

57. A faulty barometer contains a certain amount of air and saturated water vapor. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74 cm of mercury. Saturation vapor pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir.

Answer: Let the length of the barometer tube above the mercury level in the reservoir = L. Due to saturated vapor and air, the mercury height = 74 cm. The height of the tube above the mercury column = L-74 cm. Let the air pressure in the tube = P, then P+SVP+74 cm =Atmospheric pressure in cm of Hg.

→P+1+74=76

→P = 76-75 =1 cm.

When the atmospheric pressure is 74 cm, it shows 72.10 cm. If air pressure =P', So, P'+1+72.10 =74

→P' = 74-73.10 =0.90 cm

Height of tube above mercury column =L-72.1 cm

Let the area of cross-section of the tube =A cm².

Since pV =p'V'

1(L-74)A = 0.90(L-72.1)A

→L-74 =0.90L -0.90*72.1

→0.10L =74-64.89 =9.11

→L = 9.11/0.10 = 91.10 cm.

58. On a winter day, the outside temperature is 0°C and the relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapor pressure at 0°C is 4.6 mm of mercury, and at 20°C, it is 18 mm of mercury.

Answer: For the outside air,

Relative Humidity =VP/SVP

→40% =VP/(4.6 mm of mercury)

→VP = 0.40*4.6 mm of mercury

=1.84 mm of mercury.

Temperature, T =273 K.

Now in the room at 20°C, SVP =18 mm of mercury.

Temperature, T' =273+20 =293 K. Since the air comes into the room and is heated to 20°C, it must expand because the air pressure is the same. Consider a volume V outside which changes to volume V' inside. At constant pressure,

V/T = V'/T'

→V/273 = V'/293

→V'/V = 293/273 =1.07.

Hence, 1 m³ of air expands to 1.07 m³ in the room. The amount of vapor present in 1 m³ outside is now present in 1.07 m³ of air inside. Thus, in 1 m³ of air inside, the amount of vapor is 1/1.07 of the amount in 1 m³ outside. Since the vapor pressure is proportional to the amount of vapor present in the same volume of the gas, the VP in the room =(1/1.07) of 1.84 mm of mercury.

=1.72 mm of mercury.

Hence the relative humidity in the room =VP/SVP

=1.72/18 =0.095 =9.5%.