Hc Verma II for Class 11 Science Physics Chapter 24 - Kinetic Theory Of Gases

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Hc Verma II Solutions for Class 11 Science Physics Chapter 24 Kinetic Theory Of Gases are provided here with simple step-by-step explanations. These solutions for Kinetic Theory Of Gases are extremely popular among Class 11 Science students for Physics Kinetic Theory Of Gases Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma II Book of Class 11 Science Physics Chapter 24 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma II Solutions. All Hc Verma II Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 32:

Answer:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas's kinetic energy and there will be a rise in the temperature.

Page No 32:

Answer:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.

Page No 32:

Answer:

No, the gas won't obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation.

Page No 32:

Answer:

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities.

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules.

Page No 32:

Answer:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules.

Page No 32:

Answer:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size.

Page No 32:

Answer:

No, the pressure on gas won't increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.

Page No 32:

Answer:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule's collision with the walls of the container.

Page No 32:

Answer:

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time.

Page No 32:

Answer:

If the gas is ideal, there will be no temperature change. Moreover, Charles's law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles's law.

Page No 32:

Answer:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 100⁰C,. Thus, cooking process gets faster.

Page No 32:

Answer:

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.

Page No 32:

Answer:

Yes, it is possible to boil water at 30⁰C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 30⁰C will not be hot.

Page No 32:

Answer:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.

Page No 33:

Answer:

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

Hence, correct answer is (d).

Page No 33:

Answer:

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.
At high temperature, molecules move very fast. So, they tend to collide elastically and forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

Page No 33:

Answer:

According to the kinetic theory, molecules show straight line in motion (translational). So, the kinetic energy is essentially transitional.

Thus, (a) is the correct answer.

Page No 33:

Answer:

Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature.

Thus, (d) is the correct answer.

Page No 33:

Answer:

The rms speed of a gas is given by $\sqrt{\frac{3 R T}{M_{o}}}$ .
Since hydrogen has the lowest M_o compared to other molecules, it will have the highest rms speed.
Thus, (a) is the correct answer.

Page No 33:

Answer:

The straight line T₁ has greater slope than T₂. This means $\frac{P}{ρ}$ ratio is greater for T₁ than T₂. Now, rms velocity of a gas is given by $\sqrt{\frac{3 P}{ρ}}$ . This means rms velocity of gas with T₁ molecules is greater than T₂ molecules. Again, gas with higher temperature has higher rms velocity.

So, T₁ > T₂.

Thus, (a) is the correct answer.

Page No 33:

Answer:

Root mean squared velocity is given by
$v_{r m s} = \sqrt{\frac{3 R T}{M}} \Rightarrow {(v_{r m s})}^{2} = \frac{3 R T}{M} \Rightarrow {(v_{r m s})}^{2} α T$

Thus, (c) is the correct answer.

Page No 33:

Answer:

Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

Page No 33:

Answer:

(b) 2000 ms⁻¹

Given,
Molecular mass of hydrogen, M_H = 2
Molecular mass of oxygen, M_o = 32
RMS speed is given by,

$v_{r m s =} \sqrt{\frac{3 R T}{M}} \Rightarrow \sqrt{\frac{3 R T}{M_{O}}} = 500$
Now,
$\Rightarrow \frac{v_{O r m s}}{v_{H r m s}} = \frac{\sqrt{\frac{3 R T}{M_{O}}}}{\sqrt{\frac{3 R T}{M_{H}}}} \Rightarrow \frac{v_{O} r m s}{v_{H r m s}} = \frac{\sqrt{\frac{3 R T}{32}}}{\sqrt{\frac{3 R T}{2}}} \Rightarrow \frac{v_{O r m s}}{{v_{H}}_{r m s}} = \frac{1}{4} \Rightarrow \frac{500}{v_{H r m s}} = \frac{1}{4} \Rightarrow v_{H r m s} = 4 \times 500 = 2000 {ms}^{- 1}$

Hence, the correct answer is (b).

Page No 33:

Answer:

Let the number of moles in the gas be n.
Applying equation of state, we get

$P V = n R T \Rightarrow P = \frac{n R T}{V} \Rightarrow 2 \times 10^{5} = \frac{n R T}{V} . . . (1) When half of the gas is removed, number of moles left behind = \frac{n}{2} Let the pressure be P' . P' = \frac{n}{2} \frac{R T}{V} Now, P' = \frac{1}{2} \times 2 \times 10^{5} = 10^{5} [From eq . (1)]$
=100 kPa

Thus, (a) is the correct answer.

Page No 33:

Answer:

$Given : v = \sqrt{\frac{3 RT}{32}} Let the new rms speed be v' . Molecule dissociate, M = 16 v' = \sqrt{\frac{3 R (2 T)}{16}} = \sqrt{\frac{3 R (4 T)}{32}} = 2 \sqrt{\frac{3 R T}{32}} = 2 v$

Thus, (c) is the correct answer.

Page No 33:

Answer:

$Here, P V = n R T . . . (1) Also, k = \frac{R}{N} \Rightarrow R = k N . . . (2) Now, P V = n k N T [From eq . (1) and eq . (2)] \Rightarrow n N = \frac{P V}{k T} n N = Number of molecules \frac{P V}{k T} = Number of molecules$

Thus, (d) is the correct answer.

Page No 33:

Answer:

According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT
$\Rightarrow P = \frac{n R}{V} T Given : P α T This means \frac{n R}{V} is a constant . So, V is also a constant .$

Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

Page No 33:

Answer:

As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Thus, (b) is the correct answer.

Page No 33:

Answer:

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Thus, (a) is the correct answer.

Page No 33:

Answer:

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

Page No 33:

Answer:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume, both the vessels will have same pressure.

Thus, (a) is the correct answer.

Page No 33:

Answer:

According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision.

Thus, (c) and (d) are correct answers.

Page No 34:

Answer:

The average speed of molecules is given by $\sqrt{\frac{8 k T}{π m}}$ . We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

Page No 34:

Answer:

The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

Page No 34:

Answer:

Pressure of an ideal gas is given by PV = $\frac{1}{3} m n u^{2}$ . We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

Thus, (c) is the correct answer.

Page No 34:

Answer:

Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

Page No 34:

Answer:

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases.
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro's law.
Thus, (a) and (c) are correct answers.

Page No 34:

Answer:

$In an ideal gas, the equation of state is given by P V = n R T \Rightarrow P V = n N_{A} \frac{R}{N_{A}} T \Rightarrow P V = n N_{A} k T \Rightarrow \frac{1}{n N_{A}} = \frac{k T}{P V} Multiplying both sides by mass of the gas M, we get \frac{M}{n N_{A}} = \frac{M k T}{P V} Now, n N_{A} gives the total number of molecules of the gas . Also, \frac{M}{n N_{A}} gives the mass of a single molecule . Hence, \frac{M k T}{P V} is the mass of a single molecule of the gas, Molecular mass is a property of the gas .$

Thus, (d) is the correct answer.

Page No 34:

Answer:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325 $\times$ 10⁵Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we get

PV = nRT

⇒ V = $\frac{R T}{P}$

⇒ V= $\frac{8.314 \times 273}{1.01325 \times 10^{5}} = 0.0224 m^{3}$

Page No 34:

Answer:

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022 $\times$ 10²³
Number of molecules in 22.4 $\times$ 10³ cm³of ideal gas at STP = 6.022 $\times$ 10²³

Now,
Number of molecules in 1 cm³ of ideal gas at STP = $\frac{6.022 \times 10^{23}}{22.4 \times 10^{3}} = 2.688 \times 10^{19}$

Page No 34:

Answer:

Given:
Volume of ideal gas, V = 1 cm³= 10^-6 m³
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10⁻⁸ m of Hg
Density of ideal gas, ρ = 13600 kgm^-3
Pressure $(P)$ is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get

$n = \frac{P V}{R T} \Rightarrow n = \frac{ρ g h \times V}{R T} \Rightarrow n = \frac{13600 \times 9.8 \times 10^{- 8} \times 10^{- 6}}{8.31 \times 273} \Rightarrow n = 5.87 \times 10^{- 13}$

Number of molecules = N × n
                                    = 6.023 × 10²³ × 5.874 × 10⁻¹³
                                    = 35.384 × 10¹⁰
                                    = 3.538 × 10¹¹

Page No 34:

Answer:

We know that 22.4 L of O₂ contains 1 mol O₂ at STP. Thus,

$22. 4 \times 10^{3} {cm}^{3} of O_{2} = 1 mol O_{2} 1 {cm}^{3} of O_{2} = \frac{1}{22. 4 \times 10^{3}} mol O_{2} 1 mol of O_{2} = 32 g \frac{1}{22. 4 \times 10^{3}} mol of O_{2} = \frac{32}{22. 4 \times 10^{3}} = 1.43 \times 10^{- 3} g = 1.43 mg$

Page No 34:

Answer:

Let the pressure and temperature for the vessels of volume V₀and 2V₀ be P₁, T₁ and P₂ , T₂, respectively.
Since the two vessels have the same mass of gas, n₁ = n₂= n.
$T_{1} = 300 K T_{2} = 600 K Using the equation of state for perfect gas, we get P V = n R T For the vessel of volume V_{o} : P_{1} V_{o} = n R T_{1} . . . (1) For the vessel of volume 2 V_{o} : P_{2} (2 V_{o}) = n R T_{2} . . . (2) Dividing eq . (2) by eq . (1), we get \frac{2 P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = \frac{600}{300} = 2 \Rightarrow \frac{P_{2}}{P_{1}} = 1 \Rightarrow P_{2} : P_{1} = 1 : 1$

Page No 34:

Answer:

Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27 + 273 = 300 K
Height of mercury, h = 10⁻³ mm
Density of mercury, $ρ$ = 13600 kgm⁻³
Avogadro constant, N = 6 × 10²³ mol⁻¹
Pressure $(P)$ is given by
P = $ρ g h$

Using the ideal gas equation, we get
$P V = n R T$

$P V = n R T \Rightarrow n = \frac{P V}{R T} \Rightarrow n = \frac{ρ g h V}{R T} \Rightarrow n = \frac{10^{- 6} \times 13600 \times 10 \times 250 \times 10^{- 6}}{8.314 \times 300} Now, number of molecules = n N = \frac{10^{- 6} \times 13600 \times 10 \times 250 \times 10^{- 6}}{8.314 \times 300} \times 6 \times 10^{23} = 8 \times 10^{15}$

Page No 34:

Answer:

Given:
Maximum pressure that the cylinder can bear, P_max = 1.0 × 10⁶ Pa
Pressure in the gas cylinder, P₁ = 8.0 × 10⁵ Pa
Temperature in the cylinder, T₁ = 300 K

Let T₂ be the temperature at which the cylinder will break.
Volume is constant. Thus, (Given)
V₁= V₂ = V

Applying the five variable gas equation, we get

$\frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} (∵ V_{1} = V_{2}) \Rightarrow \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow T_{2} = \frac{P_{2} \times T_{1}}{P_{1}} \Rightarrow T_{2} = \frac{1.0 \times 10^{6} \times 300}{8.0 \times 10^{5}} = 375 K$

Page No 34:

Answer:

Given:
Mass of hydrogen, m = 2 g
Volume of the vessel, V = 0.02 m³
Temperature in the vessel, T = 300 K
Molecular mass of the hydrogen, M = 2 u

No of moles, n = $\frac{m}{M} = \frac{2}{2}$ = 1 mole

Rydberg's constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT
$\Rightarrow P = \frac{n R T}{V} \Rightarrow P = \frac{1 \times 8.3 \times 300}{0.02}$
$\Rightarrow$ P = 1.24 × 10⁵ Pa

Page No 34:

Answer:

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas, $ρ$ = 1.25 × 10⁻³ gcm⁻³ =1.25 kgm⁻³
Pressure, P = 1.01325 $\times$ 10⁵ Pa (At STP)
Temperature, T = 273 K (At STP)

Using the ideal gas equation, we get

$P V = n R T . . . (1) n = \frac{m}{M} . . . (2) ∴ P V = \frac{m}{M} R T \Rightarrow M = \frac{m}{V} \frac{R T}{P} \Rightarrow M = ρ \frac{R T}{P} \Rightarrow M = 1.25 \times \frac{8.31 \times 273}{10^{5}} \Rightarrow M = 2.83 \times 10^{- 2} = 28.3 g - {mol}^{- 1}$

Page No 34:

Answer:

Here,
Temperature in Simla, T₁= 15 + 273 = 288 K
Pressure in Simla, P₁ = 0.72 m of Hg
Temperature in Kalka, T₂= 35+273 = 308 K
Pressure in Kalka, P₂ = 0.76 m of Hg
Let density of air at Simla and Kalka be $ρ$ ₁ and $ρ$ ₂, respectively. Then,
$P V = \frac{m}{M} R T \Rightarrow \frac{m}{V} = \frac{P M}{R T} \Rightarrow ρ = \frac{P M}{R T}$

Thus,
$ρ_{1} = \frac{P_{1} M}{R T_{1}}$
$ρ_{2} = \frac{P_{2} M}{R T_{2}}$

Taking ratios, we get

$\frac{ρ_{1}}{ρ_{2}} = \frac{P_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} \Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{0.72}{288} \times \frac{308}{0.76} \Rightarrow \frac{ρ_{2}}{ρ_{1}} = 0.987$

Page No 34:

Answer:

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
n₁= n₂= n

Volume of the first part = V
Volume of the second part =3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
T₁ = T₂ = T
Let pressure of first and second parts be P₁ and P₂, respectively.
$For first part : Applying equation of state, we get P_{1} V = n R T . . . (1) For s e c o n d part : Applying equation of state, we get P_{2} (3 V) = n R T . . . (2) Dividing eq . (1) by eq . (2), we get \frac{P_{1} V}{P_{2} (3 V)} = 1 \Rightarrow \frac{P_{1}}{P_{2}} = \frac{3}{1} \Rightarrow P_{1} : P_{2} = 3 : 1$

Page No 34:

Answer:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M₀ = 2 g/mol=0.002 kg /mol
We know,
$C = \sqrt{\frac{3 R T}{M_{0}}} \Rightarrow C = \sqrt{\frac{3 \times 8.3 \times 300}{0.002}} \Rightarrow C = 1932.6 {ms}^{- 1}$

In the second case, let the required temperature be T.
Applying the same formula, we get

$\sqrt{\frac{3 \times 8.3 T}{0.002}} = 2 \times 1932.6 \Rightarrow T = 1200 K$

Page No 34:

Answer:

Here,
V = 10^-3 m³
Density = 0.177 kgm^-3
P = 10⁵pa
$C = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times 10^{5}}{0.177}} = 1301.9 {ms}^{- 1}$

Page No 34:

Answer:

We know from kinetic theory of gases that the average translational energy per molecule is $\frac{3}{2} k T$ .
Now,
E_avg= 0.040 eV = $0.040 \times 1.6 \times 10^{- 19} = 6.4 \times 10^{- 21} J$
$6.40 \times 10^{- 21} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T \Rightarrow T = \frac{2}{3} \times \frac{6.40 \times 10^{- 21}}{1.38 \times 10^{- 23}} = 309.2 K$

Page No 34:

Answer:

$Here, V_{avg} = \frac{\sqrt{8 RT}}{\sqrt{πM}} = \frac{\sqrt{8 \times 8.83 \times 300}}{\sqrt{3.14 \times 0.032}} = 445.25 m / s We know, T = \frac{Distance}{Speed} = \frac{6400000 \times 2}{445.25} = \frac{28747.83 h}{3600} = 7.985 h = 8 h$

Page No 34:

Answer:

Here,
m = 6.64 × 10⁻²⁷ kg
T = 273 K

$Average speed of the He atom is given by V_{avg} = \sqrt{\frac{8 k T}{π m}} = \sqrt{\frac{8 \times 1.38 \times 10^{- 23} \times 273}{3.14 \times 6.64 \times 10^{- 27}}} = 1202.31$

We know,
Momentum = m × V_avg
= 6.64 × 10⁻²⁷ × 1201.35
= 7.97 × 10⁻²⁴
= 8 × 10⁻²⁴ kg-m/s

Page No 34:

Answer:

Mean velocity is given by

$V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

Let temperature for H and He respectively be T₁ and T₂, respectively.
For hydrogen:
M_H = 2g = 2 $\times$ 10^-3kg
For helium:
M_He= 4 g = 4 $\times$ 10^-3kg

Now,
A/q $\sqrt{\frac{8 R T_{1}}{π M_{H}}} = \sqrt{\frac{8 R T_{2}}{π M_{H e}}} \Rightarrow \sqrt{\frac{8 R T_{1}}{2 \times 10^{- 3} π}} = \sqrt{\frac{8 R T_{2}}{π \times 4 \times 10^{- 3}}} \Rightarrow \sqrt{\frac{T_{1}}{2}} = \sqrt{\frac{T_{2}}{4}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{1}{2} \Rightarrow T_{1} : T_{2} = 1 : 2$

Page No 34:

Answer:

Mean speed of the molecule is given by

$\sqrt{\frac{8 RT}{πM}} For H molecule, M = 2 \times 10^{- 3} k g = \sqrt{\frac{4 RT \times 10^{3}}{π}}$

For escape velocity of Earth:

$Let r be the radius of Earth . v = \sqrt{\frac{2 G M}{r}} Multiplying neumerator and denominator by R, we get v_{c} = \sqrt{\frac{G M}{r^{2}} 2 r} g = \frac{G M}{r^{2}} v_{c} = \sqrt{2 g r}$

$\sqrt{\frac{4 RT \times 10^{3}}{π}} = \sqrt{2 g r} \Rightarrow \frac{2 \times 8.314 \times T \times 10^{3}}{3.142} = 9.8 \times 6.37 \times 10^{6} \Rightarrow T \approx 11800 K$

Page No 35:

Answer:

We know,

$V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

Molar mass of H₂ = M_H = 2 $\times$ 10^-3kg

Molar mass of N₂ = M_N = 28 $\times$ 10^-3kg
Now,
$< V >_{H} = \sqrt{\frac{8 R T}{π M_{H}}} < V >_{N} = \sqrt{\frac{8 R T}{π M_{N}}} \frac{< V >_{H}}{< V >_{N}} = \sqrt{\frac{M_{N}}{M_{H}}} = \sqrt{\frac{28}{2}} = \sqrt{14} = 3.74$

Page No 35:

Answer:

Let the temperature of gas in both the chambers be T.
Let the molar mass of gas in the left chamber and right chamber be M₁and M₂, respectively.
Let mass of gas in the left and right chamber be m₁ and m₂, respectively. Then,

$A / Q v_{r m s} = \sqrt{\frac{3 k T}{m_{1}}} = \sqrt{\frac{8 k T}{π m_{2}}} \Rightarrow 3 m_{1} = \frac{8 m_{2}}{3.14} \Rightarrow \frac{m_{2}}{m_{1}} = 1.18$

Page No 35:

Answer:

Here,
$λ = 1.38 \times 10^{- 8} m$
T = 273 K
M = $2 \times 10^{- 3} kg$

Average speed of the H molecules is given by

$v_{a v g} = \sqrt{\frac{8 R T}{πM}} = \sqrt{\frac{8 \times 8.31 \times 273}{3.14 \times 2 \times 10^{- 3}}} = 1700 m s^{- 1}$

The time between two collisions is given by

$t = \frac{λ}{v_{a v g}} \Rightarrow t = \frac{1.38 \times 10^{- 8}}{1700} \Rightarrow t = 8 \times 10^{- 12} s Number of collisions in 1 s = \frac{1}{8.11 \times 10^{- 12}} = 1.23 \times 10^{11}$

Page No 35:

Answer:

Here,
P = 10⁵ Pa
T = 300 K
For H₂
M = 2×10^-3 kg

$\begin{array}{l} (a) Mean speed is given by \\ < v > = \sqrt{\frac{8 R T}{π M}} \\ = \sqrt{\frac{8 \times 8.3 \times 300 \times 7}{2 \times 10^{- 3} \times 22}} \\ = 1780 {ms}^{- 1} \\ {Let us consider a cubic volume of 1 m}^{3} . \\ V = 1 m^{3} \\ Momentum of 1 molecule normal to the striking surface before collision = m u \sin 45^{0} \\ Momentum of 1 molecule normal to the striking surface after collision = - m u \sin 45^{0} \\ Change in momentum of the molecule = 2 m u \sin 45^{0} = \sqrt{2} m u \\ Change in momentum of n molecules = 2 m n u \sin 45^{0} = \sqrt{2} m n u \\ Let Δ t be the time taken in changing the momentum. \\ Force per unit area due to one molecule = \frac{\sqrt{2} m u}{Δ t} = \frac{\sqrt{2} m u}{Δ t} \\ Observed pressure due to collision by n molecules = \frac{\sqrt{2} m n u}{Δ t} = 10^{5} \\ n = \frac{\frac{\sqrt{2} m n u}{Δ t}}{\frac{\sqrt{2} m u}{Δ t}} = \frac{10^{5}}{\sqrt{2} m u} \\ 6.0 \times 10^{23} molecules = 2 \times 10^{- 3} kg \\ 1 molecule = \frac{2 \times 10^{- 3}}{6 \times 10^{23}} = 3.3 \times 10^{- 27} kg \\ \Rightarrow n = \frac{10^{5}}{\sqrt{2} \times 3.3 \times 10^{- 27} \times 1780} = 1.2 \times 10^{28} \end{array}$

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$\begin{array}{l} Here, \\ P_{1} = 2 \times 10^{5} Pa \\ P_{2} = ? \\ T_{1} = 293 K \\ T_{2} = 313 K \\ V_{2} = V_{1} + 0.02 V_{1} = V_{1} (1.02) \\ Now, \\ \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ \Rightarrow \frac{2 \times 10^{5} V_{1}}{293} = \frac{P_{2} V_{1} (1.02)}{313} \\ \Rightarrow P_{2} = \frac{2 \times 10^{5} \times 313}{293 \times 1.02} = 209 kPa \end{array}$

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$\begin{array}{l} Here, \\ V_{1} = 1.0 \times 10^{- 3} m^{3} \\ T_{1} = 400 K \\ P_{1} = 1.5 \times 10^{5} P a \\ P_{2} = 1.0 \times 10^{5} P a \\ T_{2} = 300 \\ M = 32 g \\ Number of moles in the jar before n_{1} = \frac{P_{1} V_{1}}{R T_{1}} \\ Volume of the gas when pressure becomes equal to external pressure is given by \\ \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ \Rightarrow V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} \\ \Rightarrow V_{2} = \frac{1.5 \times 10^{5} \times 1.0 \times 10^{- 3} \times 300}{1.0 \times 10^{5} \times 400} = 1.125 \times 10^{- 3} \\ Net volume of leaked gas = V_{2} - V_{1} \\ = 1.125 \times 10^{- 3} - 1.0 \times 10^{- 3} \\ = 1.25 \times 10^{- 4} m^{3} \\ {Let n}_{2} be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get \\ n_{2} = \frac{P_{2} V_{2}}{R T_{2}} = \frac{1.0 \times 10^{5} \times 1.25 \times 10^{- 4}}{8.3 \times 300} = 0.005 \\ Mass of leaked gas= 32 \times 0 .005=0 .16 g \end{array}$

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$\begin{array}{l} Here, \\ V_{1} = \frac{4}{3} π {(2.0 \times 10^{- 3})}^{3} \\ h = 3.3 m \\ P_{1} = P_{o} + ρ g h \\ \Rightarrow P_{1} = 1.0 \times 10^{5} + 1000 \times 9.8 \times 3.3 \\ \Rightarrow P_{1} = 1.32 \times 10^{5} P a \\ P_{2} = 1.0 \times 10^{5} P a \\ Since temperature remains the same, applying Boyle's law we get \\ P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow V_{2} = \frac{P_{1} V_{1}}{P_{2}} \\ \Rightarrow V_{2} = \frac{1.32 \times 10^{5} \times \frac{4}{3} π {(2.0 \times 10^{- 3})}^{3}}{1.0 \times 10^{5}} \\ {Let R}_{2} be the new radius. Then, \\ \frac{4}{3} π R_{2}^{3} = \frac{1.32 \times 10^{5} \times \frac{4}{3} π {(2.0 \times 10^{- 3})}^{3}}{1.0 \times 10^{5}} \\ \Rightarrow R_{2}^{3} = \frac{1.32 \times 10^{5} \times {(2.0 \times 10^{- 3})}^{3}}{1.0 \times 10^{5}} \\ \Rightarrow R_{3} = \sqrt[3]{\frac{1.32 \times 10^{5} \times {(2.0 \times 10^{- 3})}^{3}}{1.0 \times 10^{5}}} \\ \Rightarrow R_{3} = 2.2 \times 10^{- 3} m \end{array}$

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$\begin{array}{l} Here, \\ P_{1} = 2 \times 10^{5} p a \\ V_{1} = 0.002 m^{3} \\ V_{2} = 0.0005 m^{3} \\ T_{1} = T_{2} = 300 K \\ Number of moles initially, n_{1} = \frac{P_{1} V_{1}}{R T_{1}} \\ \Rightarrow n_{1} = \frac{2 \times 10^{5} \times 0.002}{8.3 \times 300} \\ \Rightarrow n_{1} = 0.16 \\ Applying equation of state, we get \\ P_{2} V_{2} = n_{2} R T \\ Assuming the final pressure becomes equal to the atmospheric pressure, we get \\ P_{2} = 1.0 \times 10^{5} p a \\ \Rightarrow n_{2} = \frac{P_{2} V_{2}}{R T} \\ \Rightarrow n_{2} = \frac{1.0 \times 10^{5} \times 0.0005}{8.3 \times 300} \\ \Rightarrow n_{2} = 0.02 \\ Number of leaked moles = n_{2} - n_{1} \\ = 0.16 - 0.02 \\ = 0.14 \end{array}$

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$\begin{array}{l} Here, \\ m = 0.040 g \\ M = 4 g \\ n = \frac{0.040}{4} = 0.01 \\ T_{1} = (100 + 273) K = 373 K \\ He is a monoatomic gas. Thus, \\ C_{v} = 3 \times (\frac{1}{2} R) \\ \Rightarrow C_{v} = 1.5 \times 8.3 = 12.45 \\ {Let the initial internal energy be U}_{1} . \\ {Let the final internal energy be U}_{2} . \\ U_{2} - U_{1} = n C_{v} (T_{2} - T_{1}) \\ \Rightarrow 0.01 \times 12.45 (T_{2} - 373) = 12 \\ \Rightarrow T_{2} = 469 K \\ {The temperature in}^{0} C can be obtained as follows: \end{array} {469 - 273 = 196}^{0} C$

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$\begin{array}{l} Applying equation of state of an ideal gas, we get \\ P V = n R T \\ \Rightarrow P = \frac{n R T}{V} . . . 1 \\ Taking differentials, we get \\ \Rightarrow P d V + V d P = n R d T . . . 2 \\ Applying the additional law, we get \\ P V^{2} = c \\ V^{2} d P + 2 V P d V = 0 \\ \Rightarrow V d P + 2 P d V = 0 . . . 3 \\ Subtracting eq. (3) from eq. (2), we get \\ P d V = - n R d T \\ \Rightarrow d V = - \frac{n R}{P} d T \\ Now, \\ ⇒ d V = - \frac{V}{T} d T [From eq . (1)] \\ \Rightarrow \frac{d V}{V} = - \frac{d T}{T} \\ {Integrating between T}_{2} {and T}_{1}, we get \\ \Rightarrow \int_{V_{1}}^{2 V} = - \int_{T_{1}}^{T_{2}} \\ \Rightarrow \ln (2 V) - l n (V) = \ln (T_{1}) - \ln (T_{2}) \\ \Rightarrow \ln (\frac{2 V}{V}) = \ln (\frac{T_{1}}{T_{2}}) \\ \Rightarrow T_{2} = \frac{T_{1}}{2} \end{array}$

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$\begin{array}{l} Here, \\ V = 0 {.166 m}^{3} \\ T = 300 K \\ {Mass of O}_{2} = 1.60 g \\ M_{O} = 32 g \\ n_{O} = \frac{1.60}{32} = 0.05 \\ {Mass of N}_{2} = 2.80 g \\ M_{N} = 28 g \\ n_{N} = \frac{2.80}{28} = 0.1 \\ {Partial pressure of O}_{2} is given by \\ P_{O} = \frac{n_{O} R T}{V} = \frac{0.05 \times 8.3 \times 300}{0.166} = 750 \\ {Partial pressure of N}_{2} is given by \\ P_{N} = \frac{n_{N} R T}{V} = \frac{0.1 \times 8.3 \times 300}{0.166} = 1500 \\ Total pressure is sum of the partial pressures. \\ \Rightarrow P = P_{N} + P_{O} = 750 + 1500 = 2250 P a \end{array}$

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$\begin{array}{l} Here, \\ h = 1 m \\ P_{1} = 0.75 mHg = 0 .75 ρ g Pa \\ ρ {=13500 kg/m}^{3} \\ Let h be the height of the mercury above the piston. \\ P_{2} = P_{1} + h ρ g \\ Let the CSA be A. \\ V_{1} = A h = A \\ V_{2} = (1 - h) A \\ Applying Boyle's law, we get \\ P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow 0.75 ρ g A = P_{2} (1 - h) A \\ \Rightarrow 0.75 ρ g = (0.75 ρ g + h ρ g) (1 - h) \\ \Rightarrow 0.75 = (0.75 + h) (1 - h) \\ \Rightarrow h = 0.25 m \\ h = 25 cm \end{array}$

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$\begin{array}{l} {Let the partial pressure of the gas in chamber A and B be P}_{A}^{'} {and P}_{B}^{'}, respectively. \\ Applying equation of state for gas A, we get \\ \frac{P_{A} V}{T_{A}} = \frac{P_{A}^{'} 2 V}{T} \\ \Rightarrow P_{A}^{'} = \frac{P_{A} T}{2 T_{A}} \\ Similarly, for gas B: \\ P_{B}^{'} = \frac{P_{B} T}{2 T_{B}} \\ Total pressure is the sum of the partial pressures. It is given by \\ P = P_{A}^{'} + P_{B}^{'} \\ = \frac{P_{A} T}{2 T_{A}} + \frac{P_{B} T}{2 T_{B}} \\ \Rightarrow P = \frac{T}{2} (\frac{P_{A}}{T_{A}} + \frac{P_{B}}{T_{B}}) \\ \Rightarrow \frac{P}{T} = \frac{1}{2} (\frac{P_{A}}{T_{A}} + \frac{P_{B}}{T_{B}}) \end{array}$

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$\begin{array}{l} (a) Here, \\ V_{1} = 5 \times 10^{- 5} m^{3} \\ P_{1} = 10^{5} Pa \\ T_{1} = 273 K \\ M = 28 .8 g \\ P_{1} V_{1} = n R T_{1} \\ \Rightarrow n = \frac{P_{1} V_{1}}{R T_{1}} \\ \Rightarrow \frac{m}{M} = \frac{10^{5} \times 5 \times 10^{- 5}}{8.3 \times 273} \\ \Rightarrow m = \frac{10^{5} \times 5 \times 10^{- 5} \times 28.8}{8.3 \times 273} \\ \Rightarrow m = 0.0635 g \end{array}$

$\begin{array}{l} (b) Here, \\ V_{1} = 5 \times 10^{- 5} m^{3} \\ P_{1} = 10^{5} Pa \\ P_{2} = 10^{5} Pa \\ T_{1} = 273 K \\ T_{2} = 373 K \\ M = 28 .8 g \\ \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ \Rightarrow \frac{5 \times 10^{- 5}}{273} = \frac{V_{2}}{373} \\ \Rightarrow V_{2} = \frac{5 \times 10^{- 5} \times 373}{273} \\ \Rightarrow V_{2} = 6.831 \times 10^{- 5} \\ Volume of expelled air = 6.831 \times 10^{- 5} - 5 \times 10^{- 5} \\ = 1.831 \times 10^{- 5} \\ Applying equation of state, we get \\ P V = n R T \\ \Rightarrow \frac{m}{M} = \frac{P V}{R T} = \frac{10^{5} \times 1.831 \times 10^{- 5}}{8.3 \times 373} \\ \Rightarrow m = \frac{28.8 \times 10^{5} \times 1.831 \times 10^{- 5}}{8.3 \times 373} = 0.017 \\ Thus, mass of expelled air = 0 .017 g \\ Amount of air in the container = 0 .0635 - 0 .017 \\ = 0.0465 g \end{array}$

$\begin{array}{l} (c) Here, \\ T = 273 K \\ P = 10^{5} Pa \\ V=5 \times 10^{- 5} m^{3} \\ Applying equation of state, we get \\ P V = n R T \\ \Rightarrow P = \frac{n R T}{V} = \frac{0.0465 \times 8.3 \times 273}{28.8 \times 5 \times 10^{- 5}} \\ \Rightarrow P = 0.731 \times 10^{5} \approx 73 kPa \end{array}$

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$\begin{array}{l} Let the CSA of the tube be A. \\ {Initial volume of air, V}_{1} = 20A cm = 0 .2A \\ Length of mercury, h = 0 .1 m \\ Let the pressure of the trapped air w hen the tube is inverted and vertical {be P}_{1} . \\ Now, pressure of the mercury and trapped air balances the atmospheric pressure . Thus, \\ P_{1} + 0.1 ρ g = 0.75 ρ g \\ \Rightarrow P_{1} = 0.65 ρ g \\ When the tube is inverted with the closed end down, the pressure acting upon the trapped air is given by \\ Atmospheric pressure + Mercury column pressure \\ Now, \\ Pressure of trapped air = Atmospheric pressure + Mercury column pressure [In equilibrium] \\ P_{2} = 0.75 ρ g + 0.1 ρ g = 0.85 ρ g \\ Applying the Boyle's law when the t emperature remains constant, we get \\ P_{1} V_{1} = P_{2} V_{2} \\ Let the new height of the trapped air be x. \\ \Rightarrow 0.65 ρ g 0 .2A = 0.85 ρ g x A \\ \Rightarrow x = 0.15 m = 15 cm \end{array}$

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$\begin{array}{l} Let CSA of the tube be A. \\ On the colder side: \\ P_{1} = 0.76 m Hg \\ T_{1} = 300 K \\ V_{1} = V \\ T_{2} = 273 K \\ V_{2} = A x \\ \frac{P_{1} V}{T_{1}} = \frac{P_{2} A x}{T_{2}} \\ \Rightarrow P_{2} = \frac{P_{1} V T_{2}}{T_{1} A x} \\ On the hotter side: \\ P_{1} = 0.76 m Hg \\ T_{1} = 300 K \\ {V_{1}}^{'} = V \\ {T_{2}}^{'} = 400 K \\ {V_{2}}^{'} = A y \\ \frac{{P_{1}}^{'} V}{T_{1}} = \frac{{P_{2}}^{'} A y}{{T_{2}}^{'}} \\ \Rightarrow {P_{2}}^{'} = \frac{P_{1} V {T_{2}}^{'}}{T_{1} A y} \\ In equilibrium, the pressures on both side will balance each other. \\ \Rightarrow {P_{2}}^{'} = P_{2} \\ \Rightarrow \frac{P_{1} V {T_{2}}^{'}}{T_{1} A y} = \frac{P_{1} V T_{2}}{T_{1} A x} \\ \Rightarrow \frac{{T_{2}}^{'}}{y} = \frac{T_{2}}{x} \\ From the length of the tube, we get \\ x + y + 0.1 = 1 \\ \Rightarrow y = 0.9 - x \\ \frac{400}{(0.9 - x)} = \frac{273}{x} \\ \Rightarrow x = 0.365 m \\ \Rightarrow x = 36.5 cm \end{array}$

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$\begin{array}{l} Here, \\ Initial pressure = Atmospheric pressure + Pressure due to mercury \\ ⇒ P_{1} = P_{o} + P_{H g} \\ Let the CSA of the tube be A. \\ P_{1} = 0.76 + 0.2 =0 .96 m Hg \\ T_{1} = T_{2} = T \\ V_{1} = 0.43 A \\ {If the tube is slanted, then the atmospheric pressure P}_{o} {remains the same. Only the P}_{H g} changes . \\ P_{2} = P_{o} + P_{H g} \cos 60^{0} = 0.76 + 0.2 \times 0.5 = 0.86 \\ P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{0 .96 \times 0.43 A}{0.86} \\ Let the length of the air column be l . \\ \Rightarrow A l = \frac{P_{1} V_{1}}{P_{2}} = \frac{0 .96 \times 0.43 A}{0.86} \\ \Rightarrow l = 0.48 m \\ \Rightarrow l = 48 cm \end{array}$

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Let the initial pressure of the chambers A and B be P_A1 and P_B1, respectively.
Let the final pressure of chambers A and B be P_A2 and P_B2, respectively.

$\begin{array}{l} \\ Let the CSA be A. \\ V_{A 1} = 0.20 A \\ T_{A 1} = 400 K \\ V_{B 1} = 0.1 A \\ T_{B 1} = 100 K \\ At first equilibrium, both side pressures will be the same. \\ \Rightarrow P_{A 1} = P_{B 1} \\ Let the final temperature at equilibrium be T. Then, \\ \frac{P_{A 1} V_{A 1}}{T_{A 1}} = \frac{P_{A 2} V_{A 2}}{T} \\ \Rightarrow \frac{P_{A 1} 0.2 A}{400} = \frac{P_{A 2} V_{A 2}}{T} \\ \Rightarrow P_{A 2} = \frac{P_{A 1} 0.2 A T}{400 V_{A 2}} . . . (1) \\ For second chamber: \\ \frac{P_{B 1} V_{B 1}}{T_{B 1}} = \frac{P_{B 2} V_{B 2}}{T} \\ \Rightarrow \frac{P_{B 1} 0.1 A}{100} = \frac{P_{B 2} V_{B 2}}{T} \\ \Rightarrow P_{B 2} = \frac{P_{B 1} 0.1 A T}{100 V_{B 2}} . . . (2) \\ At second equilibrium, pressures on both sides will be the same. \\ \Rightarrow P_{A 2} = P_{B 2} \\ \Rightarrow \frac{P_{A 1} 0.2 A T}{400 V_{A 2}} = \frac{P_{B 1} 0.1 A T}{100 V_{B 2}} \\ \Rightarrow \frac{P_{A 1}}{2 V_{A 2}} = \frac{P_{B 1}}{V_{B 2}} \\ ⇒ V_{B 2} = 2 V_{A 2} . . . (3) \\ Now, \\ V_{B 2} + V_{A 2} = 0.3 A \\ \Rightarrow 3 V_{A 2} = 0.3 A \\ \Rightarrow V_{A 2} = 0.1 A \\ Let V_{A 2} be l A . \\ \Rightarrow l = 0.1 m \\ ⇒ l = 10 cm \end{array}$

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$\begin{array}{l} Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t . \\ Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP. \\ Applying equation of state to the gas inside the vessel, we get \\ (P - d P) V_{o} = (n - d n) R T \\ \Rightarrow P V_{o} - d P V_{o} = n R T - d n R T \\ But P V_{o} = n R T \\ \Rightarrow V_{o} d P = d n R T . . . (1) \\ The pressure of the gas taken out is equal to the inner pressure . \\ Applying equation of state, we get \\ (P - d P) d V = d n R T \\ \Rightarrow P d V = d n R T . . . (2) \\ From eq. (1) and eq. (2), we get \\ V_{o} d P = P d V \\ \Rightarrow \frac{d P}{P} = \frac{d V}{V_{o}} \\ \frac{d V}{d t} = r \\ \Rightarrow d V = r d t \\ \Rightarrow d V = - r d t . . . (3) [Since pressures decreases, rate is negative] \\ Now, \\ \frac{d P}{P} = \frac{- r d t}{V_{o}} [From eq . (3)] \\ (a) \\ {Integrating the equation P = P}_{0} to P = P and time t = 0 to t = t, we get \\ \int_{P_{o}}^{P} = \int_{0}^{t} \\ \Rightarrow \ln P - \ln P_{o} = - \frac{r t}{V_{o}} \\ \Rightarrow \ln (\frac{P}{P_{o}}) = - \frac{r t}{V_{o}} \\ \Rightarrow P = P_{o} e^{\frac{- r t}{V_{o}}} \end{array}$

$\begin{array}{l} (b) \\ P = \frac{P_{o}}{2} \\ \frac{P_{o}}{2} = P_{o} e^{\frac{- r t}{V_{o}}} \\ \Rightarrow e^{\frac{r t}{V_{o}}} = 2 \\ \Rightarrow \frac{r t}{V_{o}} = \ln 2 \\ \Rightarrow t = \frac{V_{o} \ln 2}{r} \end{array}$

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$\begin{array}{l} Given: \\ p = \frac{p_{o}}{1 + {(\frac{V}{V_{o}})}^{2}} \\ Multiplying both sides by V, we get \\ p V = \frac{p_{o} V}{1 + {(\frac{V}{V_{o}})}^{2}} \\ p V = R T [From eq . (1)] \\ Now, \\ R T = \frac{p_{o} V}{1 + {(\frac{V}{V_{o}})}^{2}} \\ T = \frac{1}{R} (\frac{p_{o} V_{o}}{1 + {(\frac{V_{o}}{V_{o}})}^{2}}) [{V=V}_{o}] \\ \Rightarrow T = \frac{p_{o} V_{o}}{2 R} \end{array}$

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$\begin{array}{l} We know internal energy at a particular temperature \\ U = n C_{v} T \\ Air in home is are chiefly diatomic molecules, so \\ C_{v} = \frac{5}{2} R \\ ∴ U = \frac{5 n}{2} R T \\ Now by eqn . of state \\ n R T = P V \\ U = \frac{5}{2} (n R T) \\ = > U = \frac{5}{2} P V \\ Now pressure P is constant also V of the room = constant \\ Thus, \\ U = constant \end{array}$

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$\begin{array}{l} Here, \\ P_{1} = 10^{5} Pa \\ A = π {(0.05)}^{2} \\ L = 0.2 m \\ V = A L = 0.0016 m^{3} \\ T_{1} = 300 K \\ T_{2} = 600 K \\ μ =0 .20 \\ Applying 5 variable equation of state, we get \\ \frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} \\ \Rightarrow \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \\ \Rightarrow P_{2} = \frac{T_{2}}{T_{1}} \times P_{1} = \frac{600}{300} \times 10^{5} \\ \Rightarrow P_{2} = 2 \times 10^{5} \\ Net pressure, P = P_{2} - P_{1} = 2 \times 10^{5} - 10^{5} = 10^{5} \\ Total force acting on the stopper = P A {=10}^{5} \times π \times {(0.05)}^{2} \\ Applying law of friction, we get \\ F = μ N = 0.2 N \\ \Rightarrow N = \frac{F}{μ} = \frac{10^{5} \times π \times {(0.05)}^{2}}{0.2} \\ \frac{d N}{d l} = \frac{N}{2 π r} = \frac{10^{5} \times π \times {(0.05)}^{2}}{0.2 \times 2 π \times (0.05)} = 0.125 \times 10^{5} \\ \Rightarrow \frac{d N}{d l} = 1.25 \times 10^{4} N/m \end{array}$

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$\begin{array}{l} (a) \\ Since pressure from outside and inside the cylinder is the same, there is no net pressure \\ acting on the pistons. So, tension will be zero . \\ (b) \\ T_{1} = T_{o} \\ T_{2} = 2 T_{o} \\ P_{1} = p_{o} = 10^{5} Pa \\ CSA = A \\ Let the pistons be L distance apart . \\ V = A L \\ Applying five variable gas equation, we get \\ \frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} \\ \Rightarrow \frac{10^{5}}{T_{0}} = \frac{P_{2}}{2 T_{o}} \\ \Rightarrow P_{2} = 2 \times 10^{5} = 2 P_{o} \\ Net force acting outside = 2 P_{0} - P_{0} = P_{0} \\ {Force acting on a piston F= P}_{o} A \\ By the free body diagram, we get \\ F-T =0 \\ {T = P}_{o} A \end{array}$

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$\begin{array}{l} (a) Pressure of water above the water level of the bigger tank is given by \\ P = (h_{2} + h_{o}) ρ g \\ {Let the atmospheric pressure above the tube be P}_{o} . \\ Total pressure above the tube = P_{0} + P \\ = (h_{2} + h_{o}) ρ g + P_{o} \\ {This pressure initially is balanced by pressure above the tank 2P}_{o} . \\ \Rightarrow 2 P_{o} = (h_{2} + h_{o}) ρ g + P_{o} \\ \Rightarrow h_{2} = \frac{P_{o}}{ρ g} - h_{o} \\ (b) Velocity of the efflux out of the outlet depends upon the total pressure above the outlet. \\ {Total pressure above the outlet = 2P}_{o} + (h_{1} - h_{o}) ρ g \\ Applying Bernouli's law, we get \\ {Let the velocity of efflux be v}_{1} {and the velocity with which the level of the tank falls be v}_{2} . { Pressure above the outlet is P}_{o} . Then, \\ \frac{{2P}_{o} + (h_{1} - h_{o}) ρ g}{ρ} + g z + \frac{v_{2}^{2}}{2} = \frac{P_{o}}{ρ} + g z + \frac{v_{2}^{2}}{2} \\ Now, let the reference point of the liquid be the level of the outlet. Thus, \\ z =0 \\ \Rightarrow \frac{P_{o} + (h_{1} - h_{o}) ρ g}{ρ} + \frac{v_{2}^{2}}{2} = \frac{v_{1}^{2}}{2} \\ Again, the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux . Thus, \\ v_{2} = 0 \\ \Rightarrow \frac{P_{o} + (h_{1} - h_{o}) ρ g}{ρ} = \frac{v_{1}^{2}}{2} \\ \Rightarrow v_{1} = {[\frac{2}{ρ} (P_{o} + (h_{1} - h_{o}) ρ g)]}^{\frac{1}{2}} \end{array}$
$\begin{array}{l} c) \\ Water maintains its own level, so height of the water of the tank \\ {will be h}_{1} when water will stop flowing \\ Thus height of water in the tube below the tank height will be = h_{1} \\ Hence height of the water above the tank height will be = - h_{1} \end{array}$

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$\begin{array}{l} Atmospheric pressure inside the cylinderical vessel, P_{0} = 10^{5} Pa \\ {A=10cm}^{2} = 10 \times 10^{- 4} m^{2} \\ Pressure due to the weight of the piston = \frac{m g}{A} = \frac{1 \times 9.8}{10 \times 10^{- 4}} \\ P_{1} = 10^{5} + 9.8 \times 10^{3} \\ V_{1} = 0.2 \times 10 \times 10^{- 4} = 2 \times 10^{- 4} \\ After evacution, external pressure above the piston =0 \\ P_{2} = 0 + 9.8 \times 10^{3} \\ Now, \\ P_{1} V_{1} = P_{2} V_{2} \\ Let L be the final length of the gas column. Then, \\ V_{2} = 10 \times 10^{- 4} L \\ \Rightarrow (10^{5} + 9.8 \times 10^{3}) \times 0.2 \times 10 \times 10^{- 4} = 9.8 \times 10^{3} \times 10 \times 10^{- 4} L \\ ⇒L=2 .2 m \end{array}$

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$\begin{array}{l} Here, \\ V = 50 m^{3} \\ T = 273 + 15 = 288 K \\ R H = 40 % \\ a) \\ Let x be the amount of water that evaporates. \\ P=RH \times SVP \\ ⇒P=0 .4 \times 1600=640 \\ P V = \frac{m}{M} R T \\ \Rightarrow m = \frac{P V M}{R T} = \frac{640 \times 50 \times 18}{8.3 \times 288} = 241 g \\ Water will evaporate until VP becomes equal to SVP. \\ \Rightarrow 1600 \times 50 = \frac{m + x}{M} R T \\ \Rightarrow 1600 \times 50 = \frac{241 + x}{18} \times 8.3 \times 288 \\ \Rightarrow x \approx 361 g \\ b) \\ T = 288 + 5 = 293 K \\ S V P = 2400 Pa \\ Difference in pressure = 2400 - 1600 = 800 Pa \\ Let x be the amount of water that evaporates. \\ P V = \frac{x}{M} R T \\ \Rightarrow 800 \times 50 = \frac{x}{18} \times 8.3 \times 293 \\ \Rightarrow x \approx 296 g \end{array}$

Page No 36:

Answer:

$\begin{array}{l} Here, \\ P_{1} = 0.76 m Hg \\ P_{2} = P \\ T_{1} = 273 K \\ T_{2} = 335 K \\ {Let each of the bulbs have n}_{1} moles initially. \\ {Let the number of moles left in second bulb after its pressure reached P be n}_{2} . \\ Applying equation of state, we get \\ \frac{P_{1} V}{n_{1} T_{1}} = \frac{P V}{n_{2} T_{2}} \\ \Rightarrow \frac{0.76}{273 n_{1}} = \frac{P}{335 n_{2}} \\ \Rightarrow n_{2} = \frac{273 P}{335 \times 0.76} n_{1} \\ Number of moles left in the second bulb after the temperature rose = n_{1} - n_{2} \\ = n_{1} - \frac{273 P}{335 \times 0.76} n_{1} \\ {Let n}_{3} moles be left when pressure reached P. Applying equation of state in the first bulb, we get \\ \frac{P_{1} V}{n_{1} T_{1}} = \frac{P V}{n_{3} T_{1}} \\ \Rightarrow \frac{0.76}{n_{1}} = \frac{P}{n_{3}} \\ \Rightarrow n_{3} = \frac{P n_{1}}{0.76} \\ n_{3} = {its own n}_{1} moles + the it recieved from the first \\ n_{3} = n_{1} + (n_{1} - n_{2}) \\ \Rightarrow \frac{P n_{1}}{0.76} = n_{1} + n_{1} - \frac{273 P}{335 \times 0.76} n_{1} \\ \Rightarrow \frac{P}{0.76} = 2 - \frac{273 P}{335 \times 0.76} \\ \Rightarrow P = 0.8375 \\ \Rightarrow P = 84 cm of Hg \end{array}$

Page No 37:

Answer:

Here,
Relative humidity = 100%
$\begin{array}{l} R H = \frac{Vapour pressure of air}{SVP at the same temperature} = 1 \\ \Rightarrow Vapour pressure of air = SVP at the same temperature \\ {So, the air is saturated at 20}^{0} C {. So, dew point is 20}^{0} C. \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ T = 298K \\ RH = 60% \\ P = 1 .04 \times 10^{5} pa \\ R H = \frac{Vapour pressure of water vapour}{Saturated vapour pressure} \\ = 0.6 \\ Saturated vapour pressure = 3 .2 \times 10^{3} Pa \\ \Rightarrow Vapour pressure of water vapour (VP) = 0 .6 \times 3 .2 \times 10^{3} =1 .92 \times 10^{3} Pa \\ If the water vapour is completely removed from the air, then net pressure =1 .04 \times 10^{5} - 1.92 \times 10^{3} \\ = 1.02 \times 10^{5} Pa \end{array} =102 kPa$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ Temperature = 20° C \\ Dew point = 10° C \\ Air becomes saturated at 10° C . But if the room temperature is lowered to 15°C, the dew point will still be at 10°C. \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ RH=40% \\ V_{1} = 10 \times 10^{- 6} m^{3} \\ RH= \frac{V P}{S V P} = 0.4 \\ {Let SVP = P}_{o} \\ {Condensation occurs when VP = P}_{o} . \\ \Rightarrow P_{1} = 0.4 P_{o} \\ \Rightarrow P_{2} = P_{o} \\ Since the process is isothermal, applying Boyle's law we get \\ P_{1} V_{1} = P_{2} V_{2} \\ \Rightarrow V_{2} = \frac{P_{1} V_{1}}{P_{2}} \\ \Rightarrow V_{2} = \frac{0.4 P_{o} \times 10 \times 10^{- 6}}{P_{o}} \\ \Rightarrow V_{2} = 4.0 \times 10^{- 6} \\ \Rightarrow V_{2} = 4.0 {cm}^{3} \\ Thus water vapour condenses at volume 4 {.0 cm}^{3} . \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ Atmospheric pressure, P = 0.76 m Hg \\ Pressure due to water vapour inside, P' =0 .754 mHg \\ Vapour pressure = P - P' = 0.76 - 0.754 = 0.006 mHg \\ SVH = 0 .01 mHg \\ R H = \frac{Vapour pressure}{SVH} \times 100 % \\ = \frac{0.006}{0 .01} \times 100 % = 60 % \end{array}$

Page No 37:

Answer:

We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 65⁰ of methyl alcohol.

For 0.5 atm pressure, corresponding pressure in mm Hg will be 375 mm. We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 375 mm. This gives the boiling point 48⁰ of methyl alcohol.

Page No 37:

Answer:

Here,
T = 98⁰F

$\begin{array}{l} {When we convert the temperature to}^{0} C, we get \\ \frac{5}{9} (F - 32) \\ = \frac{5}{9} (98 - 32) \\ = {36.7}^{0} C \end{array}$

We drop perpendicular corresponding to a temperature of 36.7⁰C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg.

Page No 37:

Answer:

$\begin{array}{l} Here, \\ {Dew point = 10}^{0} C [{∵ Dew appears at 10}^{0} C] \\ At boiling point, SVP equals atmospheric pressure . \\ {At 20}^{0} C, SVP = 17 .5 mmHg \\ At dew point, SVP = 8 .9 mmHg \\ RH= \frac{SVP at dew point}{SVP at air temperature} \times 100 % \\ = \frac{8.9}{17.5} \times 100 % \\ = 51 % \end{array}$

Page No 37:

Answer:

$\begin{array}{l} We know that 1 m^{3} {of air contains 30 g of water vapour at 30}^{0} C . \\ {So, amount of water vapour in 50 m}^{3} {of air at 30}^{0} C = (30 \times 50) g=1500 g \\ Also, 1 m^{3} {of air contains 16 g of water vapour at 20}^{0} C . \\ {Amount of water vapour in 50 m}^{3} {of air at 20}^{0} C = (16 \times 50) g=800 g \\ Amount of water vapour condensed= (1500 - 800) g = 700 g \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Atmospheric pressure = 76 cm Hg \\ SVP = 0 .80 cm Hg \\ When water is introduced into the barometer, water evaporates. \\ Thus, it exerts its vapour pressure over the mercury meniscus. \\ As more and more water evaporates, the vapour pressure increases that \\ forces down the mercury level further . \\ Finally, when the volume is saturated with the vapour at the atmospheric temperature, the \\ highest vapour pressure, i.e. SVP is observed and the fall of mercury level \\ reaches its minimum. Thus, \\ Net pressure acting on the column = 76 - 0.80 cmHg \\ Net length of Hg column at SVG = 75 .2 cm \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ {Atmospheric pressure, P}_{o} = 99.4 \times 10^{3} Pa \\ {SVP at 27}^{0} C P_{w} = 3 .4 \times 10^{3} Pa \\ T = 300 K \\ V = 50 \times 10^{- 6} m^{3} \\ Now, \\ Pressure inside the jar = Pressure outside the jar [∵ Level of water is same inside and outside of the jar] \\ Pressure outside the jar = Atmospheric pressure \\ {Pressure inside the jar = VP of oxygen + SVP of water at 27}^{0} C \\ ⇒ P_{0} = P + P_{w} \\ \Rightarrow P = P_{o} - P_{w} = 99.4 \times 10^{3} - 3 .4 \times 10^{3} = 96 \times 10^{3} \\ Applying equation of state, we get \\ P V = n R T \\ \Rightarrow 96 \times 10^{3} \times 50 \times 10^{- 6} = n \times 8.3 \times 300 \\ \Rightarrow n = 1.9277 \times 10^{- 3} mol \approx 1 .93 \times 10^{- 3} mol \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Given: \\ Let the CSA be A. \\ Case 1: \\ V_{1} = (x - 74) A \\ SVP = 1 cm Hg \\ Atmospheric pressure, P_{o} = 76 cm Hg \\ Mercury column height = 74 .0 cm \\ Let P be the air pressure above the barometer. Then, \\ Atmospheric pressure = SVP + Air pressure above the barometer mercury level + Mercury column height \\ ⇒1+ P +74 =76 \\ \Rightarrow P = 1 cm \\ Case 2: \\ Atmospheric pressure, P_{0}' = 74.0 cm Hg \\ Let P' be the air pressure . Then, \\ P' + 72.10 + 1 = 76 \\ \Rightarrow P' = 0.9 \\ V_{2} = (x - 72.1) A \\ Applying Boyle's law, we get \\ P V_{1} = P' V_{2} \\ ⇒1 \times (x - 74) A = 0.9 \times (x - 72.1) A \\ \Rightarrow x = 91.1 cm \\ Length of the tube = 91 .1 cm \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Given: \\ {SVP at 0}^{o} C = 4 .6 mm Hg \\ RH =40% \\ \Rightarrow \frac{{VP of air (P}_{1})}{SVP at the same temperature} = 0.4 \\ \Rightarrow \frac{P_{1}}{4.6} = 0.4 \\ \Rightarrow P_{1} = 1.84 mmHg \\ Here, V is constant. \\ T_{1} = 273 K \\ T_{2} = 293 K \\ Applying equationof state, we get \\ \frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} \\ \Rightarrow P_{2} = \frac{P_{1} T_{2}}{T_{1}} \\ \Rightarrow P_{2} = \frac{1.84 \times 293}{273} \\ \Rightarrow P_{2} = 1.97 \\ RH inside home= \frac{{VP of air (P}_{2})}{SVP at the same temperature} \times 100 % \\ SVP at the same temperature = 18 mmHg \\ \Rightarrow R H = \frac{1.97}{18} \times 100 = 10.9 % \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ SVP=3600 Pa \\ T = 273 + 27 = 300K \\ {V = 1m}^{3} \\ M = 18 g for water \\ R H = 50 % \\ \Rightarrow \frac{V P}{S V P} = 0.5 \\ \Rightarrow V P = 0.5 \times 3600 = 1800 \\ {Let m}_{1} be the mass of water present in the 50% humid air. \\ PV = nRT \\ \Rightarrow P V = \frac{m_{1}}{M} R T \\ \Rightarrow 1800 = \frac{m_{1}}{18} \times 8.3 \times 300 \\ \Rightarrow m_{1} = 13 g \\ Required pressure for saturation=3600 Pa \\ {Let m}_{2} be the amount of water required for saturation. \\ \Rightarrow 3600= \frac{m_{2}}{M} R T \\ \Rightarrow m_{2} = \frac{3600 \times 18}{8.3 \times 300} = 26 g \\ {Total excess water vapour that has to be added = m}_{2} - m_{1} \\ = 36 - 13 = 13 g \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ T = 300 K \\ S V P = 3300 Pa at 300 K \\ R H = 20 % \\ \Rightarrow \frac{P}{S V P} = 0.2 \\ \Rightarrow P = 0.2 \times S V P = 0.2 \times 3300 = 660 \\ V {=50 m}^{3} \\ M = 18 g \\ Now, \\ PV = nRT \\ ⇒ PV = \frac{m}{M} R T \\ \Rightarrow 660 \times 50 = \frac{m}{18} \times 8.3 \times 300 \\ \Rightarrow m = 238.55 g≈238 g \end{array}$

Page No 37:

Answer:

$\begin{array}{l} Here, \\ M = 18 g for water \\ m = 500 g \\ V = 50 m^{3} \\ T = 300 K \\ S V P = 3300 P a \\ R H = 20 % \\ \frac{V P}{S V P} = 0.2 \\ \Rightarrow V P = P_{1} = 0.2 \times 3300 = 660 Pa \\ {Partial pressure P}_{2} for evaporated water is given by \\ P_{2} V = \frac{m}{M} R T \\ \Rightarrow P_{2} = \frac{500}{18 \times 50} \times 8.31 \times 300 \\ \Rightarrow P_{2} = 1385 P a \\ Total pressure, P = P_{1} + P_{2} = 1385 + 660 = 2045 Pa \\ RH= \frac{P}{S V P} \times 100= \frac{2045}{3300} \times 100 % = 61.9 \approx 62 % \end{array}$

Page No 37:

Answer:

(a) Relative humidity is given by

$\frac{VP}{SVP at 15 ° C} \Rightarrow 0.4 = \frac{VP}{1.6 \times 10^{3}}$
⇒ VP = 0.4 × 1.6 × 10³

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change = 1.6 × 10³ − 0.4 × 1.6 × 10³
                                 = (1.6 − 0.4 × 1.6) 10³
                                 = 0.96 × 10³

Let the mass of water evaporated be m. Then,

$\Rightarrow 0.96 \times 10^{3} \times 50 = \frac{m \times 8.3 \times 288}{18} \Rightarrow m = \frac{0.96 \times 50 \times 18 \times 10^{3}}{8.3 \times 288} = 361.45 \approx 361 g$

(b) At 20°C, SVP = 2.4 KPa
At 15°C, SVP = 1.6 KPa

Net pressure change = (2.4 − 1.6) × 10³ Pa
                                 = 0.8 × 10³ Pa

Mass of water evaporated is given by

$m = \frac{m' \times 8.3 \times 293}{18} \Rightarrow m' = \frac{0.8 \times 50 \times 18 \times 10^{3}}{8.3 \times 293}$

= 296.06 ≈ 296 g

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