Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 13 - Kinetic Theory

Share with your friends

Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 13 Kinetic Theory are provided here with simple step-by-step explanations. These solutions for Kinetic Theory are extremely popular among class 11 Science students for Physics Kinetic Theory Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 11 Science Physics Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 90:

Question 13.1:

A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 ms^–1 in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because 500 ms⁻¹ is very much smaller than v_rmsof the gas.
(b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
(c) will increase by a factor equal to $\frac{(v_{r m s}^{2} + {(500)}^{2})}{v_{r m s}^{2}}$ where v_rmswas the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.

Answer:

Pressure of the gas depends on the collision of the gas molecules on the wall of the container of the vessel, i.e., the relative motion of the gas molecule and the wall of the container Since the motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls, so the pressure of the gas inside the vessel will remain the same.
Hence, the correct answer is option (b).

Page No 90:

Question 13.2:

1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K. One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,

(a) the pressure on EFGH would be zero.
(b) the pressure on all the faces will the equal.
(c) the pressure of EFGH would be double the pressure on ABCD.
(d) the pressure on EFGH would be half that on ABCD.

Answer:

The molecules bounce back due to elastic collision and magnitude of momentum transferred to the wall by each molecule is 2mv but wall EFGH absorbs those molecules which strike to it. Therefore the rate of change in momentum to it becomes only mv so the pressure of EFGH would be half of ABCD.
Hence, the correct answer is option (d).

Page No 91:

Question 13.3:

Boyle’s law is applicable for an
(a) adiabatic process.
(b) isothermal process.
(c) isobaric process.
(d) isochoric process.

Answer:

Boyle’s law is applicable at constant temperature, and temperature remains constant in isothermal process.
Hence, the correct answer is option (b).

Page No 91:

Question 13.4:

A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction (Figure). If the temperature is increased,

(a) both p and V of the gas will change.
(b) Only p will increase according to Charle’s law.
(c) V will change but not p.
(d) p will change but not V.

Answer:

The pressure on the ideal gas does not change from the initial to the final position. According to the given arrangement

P = P_{a t m o s p h e r e} + \frac{M g}{A}

which shows that pressure is constant. As piston and cylinder is frictionless so by ideal gas equation PV = nRT where p, n, and R are constants so

V \propto T

So on increasing the temperature of system, it’s volume will increase but P will remain constant.
Hence, the correct answer is option (c).

Page No 91:

Question 13.5:

Volume versus temperature graphs for a given mass of an ideal gas are shown in Figure at two different values of constant pressure. What can be inferred about relation between P₁and P₂?

(a) P₁ > P₂
(b) P₁ = P₂
(c) P₁ < P₂
(d) data is insufficient.

Answer:

From ideal gas equation,

P V = n R T \Rightarrow V = \frac{n R T}{P} Slope of volume vs temperature graph = \frac{n R}{P}

As we can see from the graph in the question

{Slope}_{P_{1}} < {Slope}_{P_{2}} Hence, \frac{n R}{P_{1}} < \frac{n R}{P_{2}} \Rightarrow P_{1} > P_{2}

Hence the correct answer is option (a).

Page No 92:

Question 13.6:

1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300 K. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)
(a) same as the pressure initially.
(b) 2 times the pressure initially.
(c) 10 times the pressure initially.
(d) 20 times the pressure initially.

Answer:

Ideal gas equation,
PV = nRT

n_{1} = 1, n_{2} = 2 (Since one H_{2} molecules contains two H atoms), T_{1} = 300 K, T_{2} = 3000 K, V_{1} = V_{2} = 1 m^{3} P_{1} = \frac{n_{1} R T_{1}}{V_{1}} P_{2} = \frac{n_{2} R T_{2}}{V_{2}} \Rightarrow \frac{P_{2}}{P_{1}} = \frac{n_{2} R T_{2}}{n_{1} R T_{1}} \times \frac{V_{1}}{V_{2}} \Rightarrow P_{2} = 2 \times \frac{3000}{300} \times P_{1} \Rightarrow P_{2} = 20 P_{1}

So, new pressure becomes 20 times the initial pressure.
Hence, the correct answer is option (d).

Page No 92:

Question 13.7:

A vessel of volume V contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let f₁(v)dv, denote the fraction of molecules with speed between v and (v + dv) with f₂(v)dv, similarly for oxygen. Then
(a) f₁(v) + f₂(v) = f (v) obeys the Maxwell’s distribution law.
(b) f₁(v), f₂(v) will obey the Maxwell’s distribution law separately.
(c) Neither f₁(v), nor f₂(v) will obey the Maxwell’s distribution law.
(d) f₂(v) and f₁(v) will be the same.

Answer:

In a mixture of two gas, the average kinetic energy are equating. But the distribution in velocity are quite different because the mass and the number of moles of individual gas molecules may vary. So, both gases will obey Maxwell’s distribution law separately.
Hence, the correct answer is option (b).

Page No 92:

Question 13.8:

An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be
(a) 1.1 p
(b) p
(c) less than p
(d) between p and 1.1.

Answer:

As per the Ideal gas equation $P V = n R T$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{P V}{T} = \frac{P_{2} (1.05 V)}{1.1 T} P_{2} = \frac{P V (1.1 T)}{T (1.05 V)} P_{2} = \frac{1.1 P}{1.05} = 1.0476 P$
Which lies in between P and 1.1P.
Hence the correct answer is option (d).

Page No 92:

Question 13.9:

ABCDEFGH is a hollow cube made of an insulator Figure. Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure

(a) will be valid.
(b) will not be valid since the ions would experience forces other than due to collisions with the walls.
(c) will not be valid since collisions with walls would not be elastic.
(d) will not be valid because isotropy is lost.

Answer:

Due to the presence of hydrogen ions and positive charges on wall ABCD, there will be an electrostatic force that acts apart of the collision, so the kinetic theory of gas will not be valid. Due to the presence of ions in place of hydrogen molecules, the isotropy will also be lost.

Hence, the correct options are (b) and (d).

Page No 92:

Question 13.10:

Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory $p V = \frac{2}{3} E$ , E is
(a) the total energy per unit volume.
(b) only the translational part of energy because rotational energy is very small compared to the translational energy.
(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.
(d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero.

Answer:

According to the postulate of the kinetic theory of gases, it is assumed that pressure due to gaseous molecules is due to only perpendicular forces exerted by molecules on the wall during the motion of molecules, i.e. molecules striking at angles other than at

90 °

will not exert pressure. So, the pressure on the wall is due to the change in translational motion only.

Hence in the expression

P V = \frac{2}{3} E

, E represents only the translational part of energy.
Hence, the correct answer is option (c).

Page No 93:

Question 13.11:

In a diatomic molecule, the rotational energy at a given temperature
(a) obeys Maxwell’s distribution.
(b) have the same value for all molecules.
(c) equals the translational kinetic energy for each molecule.
(d) is (2/3)rd the translational kinetic energy for each molecule.

Answer:

Consider a diatomic molecule along the z-axis, so its rotational energy about the z-axis is zero.
So the energy of the diatomic molecule

= \frac{1}{2} {m v_{x}}^{2} + \frac{1}{2} {m v_{y}}^{2} + \frac{1}{2} {m v_{z}}^{2} + \frac{1}{2} I_{x} {ω_{x}}^{2} + \frac{1}{2} I_{y} {ω_{y}}^{2}

The independent terms in the above expression are 5, so the degree of freedom for such diatomic molecules is 5. As we can predict velocities of molecules by Maxwell’s distribution, hence the above expression obeys Maxwell’s distribution. 2 rotational and 3 translational degrees of freedom are associated with each molecule. So the rotational energy at a given temperature is

\frac{2}{3}

of its translational kinetic energy of each molecule.
In other words, translational kinetic energy =

3 \times \frac{R T}{2} = \frac{3 R T}{2}

, and the rotational kinetic energy =

2 \times \frac{R T}{2} = R T

. The fact that the distribution of the two is independent of each other is not emphasized. They are independently follows the Maxwell's distribution.
Hence, the correct options are (a) and (d).

Page No 93:

Question 13.12:

Which of the following diagrams (Fig. 13.5) depicts ideal gas behaviour?

Answer:

As per the ideal gas equation

P V = n R T

, where n and R are constant for a system.

(a) As P is constant

V = \frac{n R}{P} T i . e V \propto T

Hence (a) depicts an ideal gas behavior.

(b) As T is constant

P V = n R T i . e P V = constant P \propto \frac{1}{V}

As per the above proportionality, P vs V graph should be a rectangular hyperbola.
Hence (b) doesn't depict an ideal gas behavior.

P V = n R T i . e P = \frac{n R}{V} T P \propto T

As per the above proportionality, P vs T graph should be a straight line passing through the origin.
Hence (c) depicts an ideal gas behavior.

(d) From ideal gas equation

P V = n R T

PV vs T graph should be a straight line passing through the origin.
Hence (d) doesn't depict an ideal gas behavior.

Hence, the correct options are (a) and (c).

Page No 93:

Question 13.13:

When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases,
(a) because of collisions with moving parts of the wall only.
(b) because of collisions with the entire wall.
(c) because the molecules gets accelerated in their motion inside the volume.
(d) because of redistribution of energy amongst the molecules.

Answer:

As the ideal gas is compressed the mean free path becomes smaller, so the number of collisions per second between the molecules and walls increases which increases the temperature of the gas, in turn, kinetic energy of gas molecule increases. Kinetic energy depends on temperature.
â€‹Hence, the correct answer is option (a).

Page No 94:

Question 13.14:

Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g mole^–1.

Answer:

197 g gold contains one mole of gold atoms i.e Avagadro no of gold atoms.

$Weight of 6.023 \times 10^{23} gold atoms = 197 g Number of atoms in 1 g gold = \frac{6.023 \times 10^{23}}{197} Number of atoms in 39.4 g gold = \frac{6.023 \times 10^{23} \times 39.4}{197} = \frac{6.023 \times 10^{23}}{5} = 1.2046 \times 10^{23} ≃ 1.2 \times 10^{23} gold atoms$

Page No 94:

Question 13.15:

The volume of a given mass of a gas at 27°C, 1 atm is 100 cc. What will be its volume at 327°C?

Answer:

$P_{1} = 1 atm, P_{2} = 1 atm, V_{1} = 100 cc and V_{2}$ needs to be found out.
$T_{1} = 273 + 27 = 300 K and T_{2} = 327 + 273 = 600 K$
As per the ideal gas equation
$P V = n R T \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{1 \times 100}{300} = \frac{1 \times V_{2}}{600} \Rightarrow V_{2} = 200 cc$
So the volume at 327°C is 200 cc.

Page No 94:

Question 13.16:

The molecules of a given mass of a gas have root mean square speeds of 100 ms⁻¹ at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

Answer:

Root mean square speeds of the molecules of a gas,
$V_{r m s} = \sqrt{\frac{3 R T}{M}} V_{r m s} \propto \sqrt{T} \frac{V_{r m s_{1}}}{V_{r m s_{2}}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{100}{V_{r m s_{2}}} = \sqrt{\frac{300}{400}} (T_{1} = 273 + 27 = 300 K and T_{2} = 273 + 127 = 400 K) V_{r m s_{2}} = 100 \sqrt{\frac{400}{300}} = 200 \sqrt{\frac{1}{3}} = 115.47 {ms}^{- 1} ≃ 115.5 {ms}^{- 1}$

Page No 94:

Question 13.17:

Two molecules of a gas have speeds of 9 × 10⁶ ms⁻¹and 1 × 10⁶ ms⁻¹, respectively. What is the root mean square speed of these molecules.

Answer:

We know,
$V_{r m s} = \sqrt{\frac{{V_{1}}^{2} + {V_{2}}^{2} + . . . + {V_{n}}^{2}}{n}}$
In the question $V_{1} = 9 \times 10^{6} {ms}^{- 1} and V_{2} = 1 \times 10^{6} {ms}^{- 1}$
$V_{r m s} = \sqrt{\frac{{V_{1}}^{2} + {V_{2}}^{2}}{2}} = \sqrt{\frac{{(9 \times 10^{6})}^{2} + {(1 \times 10^{6})}^{2}}{2}}$
$V_{r m s} = \sqrt{\frac{41 \times 10^{12}}{1}} = 6.4 \times 10^{6} {ms}^{- 1}$

Page No 94:

Question 13.18:

A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon at temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

Answer:

To find the total energy of a given molecule of gas we must find its degree of freedom. In a molecule of oxygen it has 2 atoms. So it has degree of freedom 5(translational = 3, Rotational = 2).
So total internal energy =

\frac{5}{2} R T

per mole.
As in the question, we have 2 moles of oxygen, so total internal energy of 2 moles oxygen =

2 \times \frac{5}{2} R T = 5 R T

.
Neon gas is a monoatomic, so its degree of freedom is only 3.
Hence total internal energy =

\frac{3}{2} R T

per mole.
So, the total internal energy of 4 moles neon =

4 \times \frac{3}{2} R T = 6 R T

.
Hence total internal energy of 2 moles of oxygen and 4 moles of neon=

5 R T + 6 R T = 11 R T

Page No 94:

Question 13.19:

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 $\overset{o}{A}$ and 2 $\overset{o}{A}$ . The gases may be considered under identical conditions of temperature, pressure, and volume.

Answer:

Mean free path of the molecules of a gas,
$l \propto \frac{1}{d^{2}} \Rightarrow \frac{l_{1}}{l_{2}} = \frac{{d_{2}}^{2}}{{d_{1}}^{2}} \Rightarrow \frac{l_{1}}{l_{2}} = {(\frac{2}{1})}^{2}$

Hence $λ_{1} : λ_{2} = 4 : 1$

Page No 94:

Question 13.20:

V₁
µ₁
P₁

V₂
µ₂
P₂

The container shown in the table has two chambers, separated by a partition, of volumes V₁ = 2.0 litre and V₂= 3.0 litre. The chambers contain µ₁= 4.0 and µ₂= 5.0 moles of a gas at pressures P₁ = 1.00 atm and P₂ = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

Answer:

As per the ideal gas equation,
$P V = μ R T$
For gas in the first container,
$P_{1} V_{1} = μ_{1} R T$
For gas in the second container,
$P_{2} V_{2} = μ_{2} R T$
Where,
$P_{1} = 1.00 atm, P_{2} = 2.00 atm V_{1} = 2.0 litres, V_{2} = 3.0 litres T_{1} = T_{2} = T μ_{1} = 4.0, μ_{2} = 5.0$

When partition between gases removed then,
$μ = μ_{1} + μ_{2} = 9.0$
$V = V_{1} + V_{2} = 5.0 litres$
By the kinetic theory of gases, the translational kinetic energy,
$= P V = \frac{2}{3} E$ per mole.
Hence for $μ_{1} and μ_{2}$ moles, the translational kinetic energy,
$= P_{1} V_{1} = \frac{2}{3} μ_{1} E_{1} and P_{2} V_{2} = \frac{2}{3} μ_{2} E_{2}$
Adding both we get:
$P_{1} V_{1} + P_{2} V_{2} = \frac{2}{3} μ_{1} E_{1} + \frac{2}{3} μ_{2} E_{2} μ_{1} E_{1} + μ_{2} E_{2} = \frac{3}{2} (P_{1} V_{1} + P_{2} V_{2})$

For the mixture of both the gases, after equilibrium, will be $P V_{total} = \frac{2}{3} E_{total} = \frac{2}{3} μ_{total} E_{per mole}$
Hence, $P (V_{1} + V_{2}) = \frac{2}{3} [\frac{3}{2} (P_{1} V_{1} + P_{2} V_{2})]$

$P = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{1} + V_{2}} = \frac{1.00 \times 2.0 + 2.00 \times 3.0}{2.0 + 3.0} = \frac{2 + 6}{5} = \frac{8}{5} = 1.6$ atm.

Page No 94:

Question 13.21:

A gas mixture consists of molecules of types A, B and C with masses m_A > m_B> m_C. Rank the three types of molecules in decreasing order of (a) average K.E., (b) rms speeds.

Answer:

(a) Average kinetic energy of translation of molecule is given by,
${KE}_{avg} = \frac{3}{2} \times k_{B} T$
Where, T = temperature of molecule, k = Boltzmann constant
Thus, the average kinetic energy of translation depends only upon the temperature and is independent of mass of molecule.
Therefore, average kinetic energy of all molecules are same.

(b) Root mean square speed,

$V_{r m s} = \sqrt{\frac{3 k_{B} T}{m}} V_{r m s} \propto \frac{1}{\sqrt{m}}$
As m_A > m_B> m_C
â€‹So, $V_{r m s, A} < V_{r m s, B} < V_{r m s, C}$

Page No 95:

Question 13.22:

We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?
(Hydrogen molecules can be consider as spheres of radius 1 Å)

Answer:

As per the question
$V_{i} = {(3 \times 10^{- 2})}^{3} = 27 \times 10^{- 6} m^{3} P_{i} = 1 atm, P_{f} = 100 atm, and T = contsant$
We know,
$P V = n R T$ where n, R, and T are constant.
So, $\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} \Rightarrow \frac{100}{1} = \frac{27 \times 10^{- 6}}{V_{f}}$
$\Rightarrow V_{f} = 2.7 \times 10^{- 7} m^{3}$

Now let's find the volume occupied by 0.5g of H₂ molecules
0.5g H₂ molecules contain 0.5 moles of the hydrogen atom.
0.5 moles = $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ of the hydrogen atoms
The volume of each hydrogen atom,
= $\frac{4}{3} π r^{3} = \frac{4}{3} \times 3.14 {(10^{- 10})}^{3} m^{3} = 4.20 \times 10^{- 30} m^{3}$
The volume of $3.011 \times 10^{23}$ hydrogen atoms,
= $(4.20 \times 10^{- 30}) \times (3.011 \times 10^{23}) = 12.64 \times 10^{- 7} m^{3}$

Hence on compression, the volume of the gaseous molecules is of the order of the volume available for the molecules. Hence, the nuclear force of interaction will play the role, as in the kinetic theory of gas molecules do not interact with each other so gas will not obey the ideal gas behavior.

Page No 95:

Question 13.23:

When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?

Answer:

According to Boyle’s law at a constant temperature, the volume of gas is inversely proportional to pressure i.e. it is valid only for a constant mass of gas. But in this case when air is pumped continuously in the tyre, so the number of moles of air increases. So Boyle’s law is not valid in this case.

Page No 95:

Question 13.24:

A ballon has 5.0 g mole of helium at 7°C. Calculate
(a) the number of atoms of helium in the balloon,
(b) the total internal energy of the system.

Answer:

Average K.E per molecule = $\frac{3}{2} K_{B} T$ .
No. of moles of helium, n = 5-gram mole.
T = $7 °$ = $7 + 273 = 280 K$

(a) Hence, the number of atoms,
$N = n N_{A}$
$= 5 \times 6.023 \times 10^{23} = 30.015 \times 10^{23}$

(b) As KE= $\frac{3}{2} K_{B} T$
Therefore total internal energy,
$= (\frac{3}{2} K_{B} T) \times N = \frac{3 \times 1.38 \times 10^{- 23} \times 280 \times 30.015 \times 10^{23}}{2} = 1.74 \times 10^{4} J$

Page No 95:

Question 13.25:

Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.

Answer:

Hydrogen gas molecule is diatomic so it has 5 degrees of freedom.
Now we know at NTP, 22.4 L of gas contains 1 mole of the gas
Hence,
$22.4 L = 22400 cc = 6.023 \times 10^{23} molecules 1 cc = \frac{6.023 \times 10^{23}}{22400} = 2.688 \times 10^{19} molecules$
1 molecule has 5 degrees of freedom,
$2.688 \times 10^{19} molecules have 2.688 \times 10^{19} \times 5 = 1.344 \times 10^{20} degees of freedom$

Page No 95:

Question 13.26:

An insulated container containing monoatomic gas of molar mass m is moving with a velocity v₀. If the container is suddenly stopped, find the change in temperature.

Answer:

Degrees of freedom for a monoatomic gas is 3.
As per the question, final KE is zero.
Change in kinetic energy = $∆ K E = \frac{1}{2} {n m v_{o}}^{2}$
Let the change in temperature be $∆ T$
So change in internal energy will be $∆ U = n C_{v} ∆ T = n (\frac{3}{2} R) ∆ T$
Now as per the law of conservation of energy, $∆ K E = ∆ U$
i.e
$\frac{1}{2} {n m v_{o}}^{2} = n (\frac{3}{2} R) ∆ T ∆ T = \frac{{m v_{o}}^{2}}{3 R}$
Hence the change in temperature will be $\frac{{m v_{o}}^{2}}{3 R}$ .

Page No 95:

Question 13.27:

Explain why
(a) there is no atmosphere on moon.
(b) there is fall in temperature with altitude.

Answer:

(i) The moon has a small gravitational force and hence the escape velocity is small. The air molecules have a large range of speeds. Even though the RMS speed,

V_{r m s}

of the air molecules is smaller than escape,

V_{e}

velocity on the moon a significant number of molecules have speed greater than escape velocity and they escape. Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again a significant number of molecules escape as their speeds exceed escape speed. Hence, over a long time, the moon has lost most of its atmosphere.

(ii) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At greater height more volume is available and gas expands and hence some cooling takes place.

Page No 95:

Question 13.28:

Consider an ideal gas with following distribution of speeds.

Speed (m/s)	% of molecules
200	10
400	20
600	40
800	20
1000	10

(i) Calculate V_rmsand hence T. (m = 3.0 × 10^–26kg)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new V_rmsand hence T.

Answer:

(i) Root mean square speed,
$V_{r m s} = \sqrt{\frac{n_{1} {V_{1}}^{2} + n_{2} {V_{2}}^{2} + . . . + n_{n} {V_{n}}^{2}}{n_{1} + n_{2} + . . . + n_{n}}}$
$V_{r m s} = \sqrt{\frac{10 \times 200^{2} + 20 \times 400^{2} + 40 \times 600^{2} + 20 \times 800^{2} + 10 \times 1000^{2}}{100}} V_{r m s} = 639 {ms}^{- 1}$
$\frac{1}{2} m {(V_{r m s})}^{2} = \frac{3}{2} K T T = \frac{m {(V_{r m s})}^{2}}{3 K} = \frac{3 \times 10^{- 26} \times {(639)}^{2}}{3 \times 1.38 \times 10^{- 23}} = 296 K$

(ii) If all the molecules with speed 1000 m/s escape from the system.
$V_{r m s} = \sqrt{\frac{10 \times 200^{2} + 20 \times 400^{2} + 40 \times 600^{2} + 20 \times 800^{2}}{90}} V_{r m s} = 585 {ms}^{- 1}$
$\frac{1}{2} m {(V_{r m s})}^{2} = \frac{3}{2} K T T = \frac{m {(V_{r m s})}^{2}}{3 K} = \frac{3 \times 10^{- 26} \times {(585)}^{2}}{3 \times 1.38 \times 10^{- 23}} = 248.04 K$

Page No 96:

Question 13.29:

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km³ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.

Answer:

The velocity of the planes, $v = 150 {kmh}^{- 1}$
Number of planes, N = 10
The diameter of the planes, d = 20 m (as R=10 m) = $20 \times 10^{- 3} km$
Volume, V= 20 × 20 × 1.5 km³
Number of planes per unit volume, $n = \frac{N}{V} = \frac{10}{20 \times 20 \times 1.5} = 0.0167 {km}^{- 3}$
Now mean free path
$λ = \frac{1}{\sqrt{2} π d^{2} n}$
Time elapse before the collision of two planes randomly i.e., relaxation time,
$t = \frac{λ}{v} = \frac{1}{\sqrt{2} π d^{2} n v}$
$t = \frac{1}{1.41 \times 3.14 \times {(20)}^{2} \times 10^{- 6} \times (0.0167) \times (150)} = 224.74 h ≃ 225 h$

The time elapsed between the successive collision of the planes is 225 hours.

Page No 96:

Question 13.30:

A box of 1.00 m³ is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area 0.010 mm². How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

Answer:

As, volume of the box, V = 1.00 m³
Area of the hole = a = 0.0010 mm²
= 8.01×10⁻⁶m²
=10⁻⁸ m²
Temperature outside = Temperature inside
Initial pressure inside the box = 1.5 atm
Final pressure inside the box = 0.10 atm
Let molecular speed of nitrogen inside the box,
$Let, molecular speed of the nitrogen molecule be v_{i x,} along x - axis and number of molecules per unit volume . In time ∆ t, no of particles striking the wall = \frac{1}{2} n_{i} (v_{i x} ∆ t) a In general, gas Is In equilibrium as the wall is very large as compared to hole, {v^{2}}_{r m s} = {v_{x}}^{2} + {v_{y}}^{2} + {v_{z}}^{2} (by symmetry, v_{x} = v_{y} = v_{z}) {v_{x}}^{2} = \frac{{v^{2}}_{r m s}}{3} = \frac{(\frac{3 k T}{m})}{3} = \frac{k T}{m} \Rightarrow ∆ t = \frac{1}{2} n_{i} (v_{i x} ∆ t) a = \frac{1}{2} n_{i} (\sqrt{\frac{k T}{m}} \times ∆ t) a If n_{1} particles enters and n_{2} particles ejected out from the hole, then, Net particles flow in time ∆ t, = \frac{1}{2} (n_{1} - n_{2}) (\sqrt{\frac{k T}{m}} \times ∆ t) a$
From ideal gas equation,
$From ideal gas equation P V = μ R T \frac{μ}{V} = \frac{(\frac{P V}{R T})}{V} = \frac{P}{R T} \Rightarrow n = \frac{μ N_{A}}{V} = \frac{P N_{A}}{R T} So, n_{1} V - n'_{1} V = number of particle gone out,$
$= \frac{1}{2} (n_{1} - n_{2}) (\sqrt{\frac{k T}{m}} \times ∆ t) a \Rightarrow \frac{P_{1} N_{A}}{R T} V - \frac{{P_{1}}^{'} N_{A}}{R T} V = \frac{1}{2} (P_{1} - P_{2}) \frac{N_{A}}{R T} (\sqrt{\frac{k T}{m}} \times ∆ t) a So, ∆ t = 2 (\frac{P_{1} - {P_{1}}^{'}}{P_{1} - P_{2}}) \frac{V}{a} \sqrt{\frac{m}{k T}} By, putting the required values in the above expression we get, ∆ t = 1.38 \times 10^{5} s$

Page No 96:

Question 13.31:

Consider a rectangular block of wood moving with a velocity v₀in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A. Show that the drag force on the block is $4 ρ A v_{0} \sqrt{\frac{k T}{m}}$ , where m is the mass of the gas molecule.

Answer:

Let, rms speed of the gas molecule be, v,
n be the number of molecules per unit volume and,
m be the mass of each molecule.
$Mass density = ρ = \frac{M}{V} = \frac{m n V}{V} = m n$
When the block is moving with speed v_o, relative speed of molecules with respect to the front face = v + v_o
Momentum transferred to block per collision
= 2m (v+v_o), where,
So, number of collisions in time $∆ t$ on cross-section of area A is,
= $\frac{1}{2} (v + v_{o}) n ∆ t A$
Momentum is transferred after the collision of molecules to the block,
Hence, momentum transferred in time $∆ t$ ,
$= m {(v + v_{o})}^{2} n A ∆ t from front surface$
Similarly, momentum transferred in time $∆ t$ ,
$= m {(v - v_{o})}^{2} n A ∆ t from back surface$
Hence, net drag force = rate of change of momentum
$= \frac{m n A [{(v + v_{o})}^{2} - {(v - v_{o})}^{2}] ∆ t}{∆ t} from front surface, = 4 m n A v v_{o} = 4 ρ A v v_{o} (as, ρ = m n)$
But,
$\frac{1}{2} m {v_{x}}^{2} = \frac{1}{2} k T (here, v = v_{x} along x - axis) \Rightarrow v_{x} = \sqrt{\frac{k T}{m}} So, the drag force = 4 ρ A \sqrt{\frac{k T}{m}} v_{0}$

View NCERT Solutions for all chapters of Class 11