NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory

NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQ I

Question:13.1

A cubic vessel (with face 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 the 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_rms of the gas.
(b) remains the same because the 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$\frac{\left ( v^{2}rms +\left ( 500 \right )^{2} \right )}{v^{2}rms}$ where vrms was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.

Answer:

The answer is option (b)
In the rocket, the relative velocity of molecules does not change with respect to walls of a container as the mass of a molecule when compared to that of the whole system is negligible. So, the whole gas system moves as a single unit. The rocket moves at a constant speed which makes the acceleration zero, and hence the pressure remains the same as observed by us inside the gas vessel.

Question:13.2

1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH, at 300 K (figure). One face of the cube (EFGH) is made up of a material that 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 be 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 answer is option (d)
The rate of transfer of momentum on the wall exerts pressure on the wall due to the force exerted by the molecules. Elastic collision takes place, and hence, the molecules bounce back. The walls absorb the magnitude of momentum transferred to it by each molecule, which is equal to 2mv. So, the net rate of change of momentum is mv, and the pressure of EFGH is halved compared to ABCD.

Question:13.3

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

Answer:

The answer is the option (b)
At constant temperature, Boyle’s law is applicable
$PV = nRT$ (n, R, T are constant in this case)
So, PV = constant and $P\propto \frac{1}{V}$

Question:13.4

A cylinder containing an ideal gas is in a 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) the only P will increase according to Charles’ law
(c) V will change but not P
(d) P will change but not V

Answer:

The answer is option (c)
In case of an ideal gas, the pressure remains the same throughout irrespective of the initial or final position. Here, $P= \frac{Mg}{A}$ and no friction acts on piston and the cylinder
We have, $PV = nRT$
Since, P, n and R, are constant here, $V\propto T$. So, the temperature increases as volume increases and vice-versa at the same time, when pressure remains constant.

Question:13.5

Volume versus temperature graphs for a given mass of an ideal gas is shown in the figure. At two different values of constant pressure. What can be inferred about the relation between $P_{1}$ and $P_{2}$?

(a)$P_{1}>P_{2}$
(b) $P_{1}=P_{2}$
(c) $P_{1}<P_{2}$
(d) Data is insufficient

Answer:

The answer is the option (a)

The pressure and quantity of gas are constant here.
From $PV = nRT$, we can say that, $V \propto T$
So, $\frac{V_{1}}{T_{1}}$ = constant (slope = constant)
$V = \frac{nRT}{P}$
$\frac{dV}{dT}=n \frac{R}{P}$
hence its value decreases with pressure, so, $\frac{dV}{dT}\propto \frac{1}{P}$
so, slope of $P_{1}$ is smaller than $P_{2}$ and hence a is the right option.

Question:13.6

1 mole of $H_{2}$ gas is contained in a box of volume $V = 1.00 m^3$ 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:

The answer is option (d)
The transfer of the rate of change of momentum by the particles to the walls is what causes the exertion of pressure by the gas. As temperature rises and hydrogen molecules break into atoms, the mass of the particle becomes halved, and the number of particles becomes double. Since the velocity is strictly dependent on the temperature, at the same temperature, i.e., at 300K, the velocity of particles of $H_{2}$ as well as H remains the same. Also, the pressure does not change as the atomic form changes since the rate of transfer of momentum is also the same in the case of $H_{2}$ as well as H.

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 $f_{1}(v)dv$ denote the fraction of molecules with speed between v and (v + dv) with $f_2(v)dv$, and similarly for oxygen. Then,
(a) $f_{1}(v) + f_{2}(v) = f (v)$ obeys the Maxwell’s distribution law
(b) $f_{1}(v) , f_{2}(v)$ will obey the Maxwell’s distribution law separately
(c) neither $f_{1}(v) nor f_{2}(v)$will obey Maxwell’s distribution law
(d) $f_{1}(v) and f_{2}(v)$will be the same

Answer:

The correct answer is option (b)
F1(v): speed of n molecules = (v+dv)
The number of molecules remains the same, which is one mole for $f_{1}(v) and f_{2}(v)$. Since the molecules differ in mass, they will also differ in speed, and hence, both gases will separately obey Maxwell’s distribution.

Question:13.8

An inflated rubber balloon contains one mole of an ideal gas and 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:

The correct answer is option (d)
According to the ideal gas equation, PV = nRT
Here, $\frac{PV}{T}$ = constant.
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$
So, $P_{2}= \frac{1.1}{1.05}(p)=(1.0476)(p)$
So, the value of $P_{2}$ is between p and 1.1 p and hence, option d is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQII

Question:13.9

ABCDEFGH is a hollow cube made of an insulator (figure) Face ABCD has a positive charge on it. Inside the cube, we have ionised 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:

The correct answer is option (b) and (d)
The wall of ABCD is positively charged. The presence of hydrogen ions and the charge on the walls lead to an electrostatic force which acts in contrast to the collision and makes the kinetic theory of gases invalid in this case. The isotropy also gets lost given the presence of hydrogen ions instead of hydrogen molecules. Hence, options b and d are correct.

Question:13.10

Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory $PV = \frac{2}{3}E$, 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:

The correct answer is option (c)
As per the kinetic theory of gases, the perpendicular forces exerted by the molecules on the walls while in motion are only responsible for the pressure exerted due to the gas molecules. So, for molecules striking at angles other than 90 degrees, no pressure will be exerted. Hence, in this case, only translational motion change leads to pressure on the wall. So, $PV = \frac{2}{3}E$, which represents the translational motion part and hence c is the correct option.

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 $\left (\frac{2}{3} \right ){rd}$ the translational kinetic energy for each molecule

Answer:

The correct answer is the options (a) and (d)
If we assume a diatomic molecule along the z-axis, its energy along that axis will be zero. The total energy of a diatomic molecule can be expressed as: $E = \frac{1}{2} mvx^{2} + \frac{1}{2} mvy^{2} + \frac{1}{2} mvz^{2} + \frac{1}{2} IxWx^{2} + \frac{1}{2} IxWy^{2}$
The number of independent terms in the expression =5. The above expression obeys Maxwell’s distribution as their velocities can be predicted with Maxwell’s findings. In this case, for each molecule, 3 translational and 2 rotational energies are associated. So, at any temperature, rotational energy = $\left (\frac{2}{3} \right ){rd}$ translational KE

Question:13.12

Which of the following diagrams (figure) depicts ideal gas behavior?

Answer:

The correct answer is options (a) and (c)
a) $PV = nRT$
where R and T are constant.
Since P is constant, $V \propto T$as given in the graph of option a.
Hence, the option is correct.
c) In case where V = constant
$P \propto T$, which is a straight line. Hence, option c is correct.

Question:10.13

When an ideal gas is compressed adiabatically, its temperature rises, and 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 get accelerated in their motion inside the volume
(d) because of the redistribution of energy amongst the molecules

Answer:

The correct answer is option (a)
The number of collisions per second between the molecules and walls increases, and the mean free path becomes smaller as an ideal gas is compressed. This, in turn, increases the temperature of the gas, which increases the overall kinetic energy of the gas molecules as KE depends on the temperature.

NCERT Exemplar Class 11 Physics Solutions Chapter 13: Very Short Answer

Question:13.14

Calculate the number of atoms in 39.4 g of gold. The molar mass of gold is 197 g mole^-1.

Answer:

Number of atoms on 197 g gold = $6.023 \times 10^{23}$
Number of atoms in 1g of gold, n
$=6.023 \times \frac{10^{23}}{197}$
Hence, the number of atoms in 34g of gold =
$34 \times \left (6.023 \times \frac{10^{23}}{197} \right ) = 1.2 \times 10^{23}$ atoms

Question:13.15

The volume of a given mass of a gas at $27^{\circ}C$, 1 atm is 100 cc. What will be its volume at $327^{\circ}C$?

Answer:

Let, $P_{1}$ = 1 atm, $P_{2}$ = 1 atm, $V_{1}$ $= 100cc$, $V_{2}$ = ?$T_{1}$ $= 300K$, $T_{2}$ $= 600K$
From the ideal gas equation, we have
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$
Hence,
$V_{2} = \frac{100 \times 600}{300}$$= 200 cc$

Question:13.16

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

Answer:

$V_{1}$ rms $= 100m/s$, $T_{1}$$= 300K$, $T_{2}$ $= 400K$, $V_{2}$ rms =?
We know that,
$V_{rms}=\sqrt{\frac{3RT}{m}}$
$V_{rms}\propto \sqrt{T}$
Hence,
$\frac{V_{1}rms}{V_{2}rms}=\sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{V_{2}rms}=\sqrt{\frac{300}{400}}$
$V_{2}rms = 115.4 \frac{m}{s}$

Question:13.17

Two molecules of a gas have speeds of $9 \times 10^{16} ms^{-1}$ and $1 \times 10^{16} ms^{-1}$ respectively. What is the root mean square speed of these molecules?

Answer:

$V_{rms}=\sqrt{\frac{V1^{2}+V2^{2}...........Vn^{2}}{n}}$
We know that, $V_{1}=9 \times 10^{16} ms^{-1}$ and $V_{2}=1 \times 10^{16} ms^{-1}$
Hence, $V_{rms}=\sqrt{\frac{\left ( 9 \times 10^{6} \right )^{2}+\left ( 1\times 10^{6} \right )^{2}}{2}}$
$V_{rms}=10^{6}\times \sqrt{\frac{82}{2}}=6.4 \times 10^{6} m/s$

Question:13.18

A gas mixture consists of 2.0 moIes 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:

Degree of freedom is necessary to be known if we need to find the total energy of a particular molecule.
An oxygen molecule has 2 atoms, so the degree of freedom = 2 rotational + 3 translational = 5
Total internal energy for 2 moles of oxygen = $2\times \frac{5}{2} RT = 5RT$
Neon gas has a degree of freedom equal to 3, since it is a monoatomic gas-only translational degree of freedom is present. So, the total internal energy $= \frac{3}{2} RT$ per mole.
For 4 moles of neon gas, total internal energy = $4\times \frac{3}{2} RT = 6RT$
Hence total internal energy of 2 moles oxygen + 4 moles Neon = 5 + 6 = 11 RT

Question:13.19

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

Answer:

Mean free path formula, $\lambda =\frac{1}{\sqrt{2}}\pi d^{2}n$
Here, n = number of molecules per unit volume. In our case, n will be constant.
Here, d = diameter of molecules.
Now, $\lambda \propto \frac{1}{d^{2}}$
So,$\frac{\lambda _{1}}{\lambda _{2}}=\frac{d_2^{2}}{d_1^{2}}=\frac{4}{1}$
Hence, $\lambda _{1}:\lambda _{2}=4:1$

NCERT Exemplar Class 11 Physics Solutions Chapter 13: Short Answer

Question:13.20

The container shown in Figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain $\mu _{1}=4.0$ and $\mu _{2}=5.0$ moles of a gas at pressures $p_{1} = 1.00$ atm and $p_{2} = 2.00$ atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

Answer:

According to the ideal gas situation, we have PV = nRT
For chamber 1 and chamber 2,
$P_{1}V_{1} = n_{1}RT_{1}$ and $P_{2}V_{2} = n_{2}RT_{2}$
$P_{1}=1atm, P_{2}=2 atm, V_{1}=2L, V_{2}=3L$
$T_{1}=T, T_{2}=T, n_{1}=4, n_{2}=5$
When we remove the partition, $n = n_{1}+n_{2}$ and $V = V_{1} + V_{2}$
According to kinetic theory, Translational gas kinetic energy$=$$PV = \frac{2}{3}$. E per mole
So, the translational KE for both cases will be,
$P_{1}V_{1} = \frac{2}{3}. n_{1} E_{1}$ and $P_{2}V_{2} = \frac{2}{3}. n_{2} E_{2}$
Adding, $P_{1}V_{1}+P_{2}V_{2} =\frac{2}{3}. n_{1} E_{1}+ \frac{2}{3}. n_{2} E_{2}$
$n_{1}E_{1}+n_{2}E_{2}=\frac{3}{2}\left ( P_{1}V_{1}+P_{2} V_{2} \right )$
hence by combining the results we get,
$P(V_{1}+V_{2}) = \frac{2}{3}\left [ \frac{3}{2}(P_{1}V_{1}+P_{2}V_{2}) \right ]$
$P = \frac{P_1V_1 + P_{2}V_2}{V_{1} + V_{2}} = 1.6 atm$

Question:13.21

A gas mixture consists of molecules of A, B and C with masses. Rank the three types of molecules in decreasing order of (a) average KE, (b) RMS speeds.

Answer:

$Vav =\sqrt{\frac{8 K_bT}{\pi m}} = \sqrt{\frac{8RT}{\pi m}} = \sqrt{\frac{8PV}{\pi m}}$
as the temperature and pressure are equal,
$Vav\propto \frac{1}{\sqrt{m}}$
hence, $Vc > Vb > Va$
now $KE \propto V^2$ and $KE \propto m$
hence KE in decreasing order will be,
$KE c > KE b > KE a$
b)
now, $V_{rms} =\sqrt{\frac{3K_bT}{m}}$
since it is given that, $m(a) > m(b) > m(c)$
$V_ {rms} (c) > V_{ rms} (b) > V_{ rms} (a)$

Question:13.22

We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm 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 considered as spheres of radius 1 A ).

Answer:

The volume of a sphere = the volume of one molecule
$=\frac{4}{3}\pi r^{3}$
Here $r= 10^{-10} m$
So, volume = $4 \times 1.05 \times 10^{-30} = 4.2 \times 10^{-30} m^{3}$
For 0.5 g of hydrogen gas, no of moles = $\frac{0.5}{2}=0.25 moles$
Hence the volume of hydrogen molecules = $0.25 \times 6.023 \times 10^{23} \times 4.2 \times 10^{-30} = 6.3 \times 10^{-7} m^{3}$
For a constant temperature, ideal gas à Pi Vi = Pf Vf
$Vf =\frac{P_i V_i}{P_f} = \frac{1}{100}\times (3\times 10^{-2})^{3}$
Hence, $V_f = 2.7 \times 10^{-7} m^{3}$
The gas here will not obey the ideal gas behaviour as the kinetic energy of the molecules will not interact with each other due to compression.

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:

The volume of a gas is inversely proportional to the pressure of the gas at a constant temperature as stated by Boyle’s law. When air is continuously pumped inside the tyre, the number of moles of gas (i.e. the mass of the gas) increases. Since Boyle’s law is only applicable for a constant mass of gas, it will not hold true in this case.

Question:13.24

A balloon has 5.0 mole of helium at $7^{\circ}C$. Calculate
(a) the number of atoms of helium in the balloon.
(b) the total internal energy of the system.

Answer:

Helium gas, n = 5 moles and T = 280 K
a) number of atoms of He = $5 \times 6.023 \times 10^{23} = 3.0115 \times 10^{24}$ atoms of He
b) the degree of freedom for a helium atom is 3 since it is monoatomic
so, average $KE =\frac{3}{2}K_b$T per molecule
average $KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 80 \times 3.0115 \times 10^{24}$
hence total energy = $1.74 \times 10^{ 4 }J$

Question:13.25

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

Answer:

Degree of freedom of H₂ molecule = 3 translational degree of freedom + 2 rotational degree of freedom = 5.
Number of molecules in 1cc of hydrogen gas at NTP à 22.4 L = 22400 cc H₂ gasà it has $6.023 \times 10^{23}$ molecules
So, 1cc of H₂ gas at STP has,$\frac{ 6.023}{22400}\times 10^{23} = 2.688 \times 10^{19}$ molecules
Hence the total degree of freedom =$5 \times 2.688 \times 10^{19 }= 1.344 \times 10^{20}$

Question:13.26

An insulated container containing monoatomic gas of molar mass m is moving with a velocity $v_{0}$. If the container is suddenly stopped, find the change in temperature.

Answer:

Since the gas is monoatomic, only translational degree of freedom will be applicable. KE per molecule = $\frac{3}{2} R T$. when the container which is insulated has suddenly stopped, a transfer of kinetic energy takes place to the gas molecules. Let $\Delta T$ be the slight increase in temperature and n be the number of moles of gas.
The resultant increase in translational KE can be presented as,
$KE = \frac{3}{2} n. R. \Delta T$
Increased KE due to velocity,
$v_{0} = \frac{1}{2} (m n) v_{0}^{2}$
Hence,
$\frac{mn}{2} v_{0}^{2}=\frac{3}{2} n. R. \Delta T$
So,
$\Delta T = \frac{m n v_{0}^{2} \times 2}{2 \times 3 \times n R} =\frac{ m v_{0}^{2}}{3R}$

NCERT Exemplar Class 11 Physics Solutions Chapter 13: Long Answer

Question:13.27

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

Answer:

a) The acceleration on the moon due to gravity is one-sixth that of Earth
escape velocity on the moon can be written as: $V es = \sqrt{2gR} = 2.38 km/s$
mass of hydrogen, m = 1.67 x 10-24 kg
$v (rms) =\sqrt{ \frac{3 Kb T}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23}\times 300}{1.67 \times 10^{-24}}} = 2.72 km/s$
v(RMS) is larger than the escape velocity as the gravitational force is small. The sun-moon distance is fairly equal to that of the earth. The energy of the sun reaches in a larger intensity to the moon due to its lower density of the atmosphere. So, the molecules of sunlight have a larger RMS speed and hence can escape out fairly easily. Hence, the moon has lost its atmosphere over time.
b) The kinetic energy of the air molecule gives the characteristic temperature to the atmosphere. At greater heights, atmospheric pressure is low, which makes air molecules rise higher, which in turn leads to an increase in their potential energy and decrease in their kinetic energy. Since the overall kinetic energy decreases, the temperature also decreases. Also, the lower atmospheric pressure causes expansion of the gas, which also provide a cooling effect.

Question:13.28

Consider an ideal gas with the following distribution of speeds.

(i) Calculate V rms and hence T. ($m = 3.0 \times 10^{-26} kg$)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T.

Answer:

(i)
$\\v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2} + 10(1000)^{2}}{10+20+40+20+10}\\=4.08\times10^{5}$

$\frac{1}{2}m v^{2} (rms)=\frac{3}{2} Kb T$
$T = \frac{m v^{2} (rms)}{3 Kb} = \frac{3 \times 10^{-26} \times 10^{5} \times 4.08}{3\times 1.38\times 10^{-23}} = 296K$
(ii)
Molecules escape at rate of 1000m/s, we can calculate the v² (rms) as follows
$v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2}}{10+20+40+20}$
$v^2 (rms) = \frac{105[1\times 4 + 2 \times 16 + 4 \times 36 + 2 \times 64]}{90}$
$v(rms) = \frac{100}{3}\sqrt{ 308} \sim 585 m/s$
$T = \frac{1}{3} (\frac{m v^{2} (rms)}{Kb}) = \frac{3 \times 10-26 \times (585)2}{3 \times 1.28 \times 10^{-23}} = 284.04 K$

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 10m.

Answer:

The motion of molecules in a confined space can be considered a plane. Mean free path λ can be considered as distance travelled by molecules between two planes to avoid any collision, time = distance/speed
$=\frac{\lambda }{v}=\frac{1}{\sqrt{2n}\pi d^{2}.v}$
Number of particles per unit volume V = N/volume
$N =\frac{10}{20\times 20 \times 1.5} km^{3} = \frac{0.0167}{km^{3}}$
$d = 2 \times 10 = 20m = 150 km/hr$
so, time
$=\frac{1}{\sqrt{2n}\pi d^{2}. V} \\= \frac{1}{1.414 \times 0.0167\times 3.14\times 20 \times 20\times 10^{-6}\times 150 }\\\\= 225 hours.$

Question:13.3

A box of 1.00m³ is filled with nitrogen at 1.50 atm at 300K. 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:

Let v1 = volume of box = 1 m³
Let p1 be the initial pressure = 1.5 atm
Let p2’ be the final pressure =1.5 - 0.1 = 1.4 atm
t1 = initial temperature = 300K, t2 = final temperature = 300 K
let a be the area of hole = $10 ^{-8} m^{2}$
The pressure difference initially between the tyre and atmosphere, ΔP = (1.5 -1) = 0.5 atm
Mass of N2 molecule
$=\frac{0.028}{6.023 \times 10^{23}} = 46.5 \times 10^{-27} kg$
$Kb = 1.38 \times 10^{-23}$
Now since
$|v_{ix}|= | v_{iy} |= \left |v_{iz} \right |$
Hence, $v^2_{rms} = 3v_{ix}^2$
For a gas molecule, Kinetic energy = $\frac{3}{2}KbT$
Now,
$\frac{1}{2}mv^{2} = 3v_{ix}^{2}$
$v_{ix }= \sqrt{\frac{Kb T}{m}}$ (A)
The N2 molecules striking the wall in time $\Delta t$ outwards
$\frac{1}{2} P_{n1} \sqrt{\frac{Kb T}{m}}$ (A) outwards.
The temperature of the air and inside the box is equal to T.
The N2 molecules striking the wall in time $\Delta t$ inwards = $\frac{1}{2} P_{n2} \sqrt{\frac{Kb T}{m}}$ (A) inwards.
Here, Pn2 = density of air molecules
So, we can calculate the net number of molecules going outward as:
$\frac{1}{2} \left [ P_{n1}-P_{n2} \right ] \sqrt{\frac{K_b T}{m}}(A)\Delta t$
We know that,$PV = nRT, n= \frac{PV}{RT}$
Pn1 = total number of molecules in the box/ volume of the box
$=\frac{nNa}{V}= \frac{PNa}{RT}$ per unit volume.
After a time, T, the pressure reduces to (1.5-0.1) = 1.4 atm = P’2
The new density of the Na molecules $=P^{'}n1 = \frac{P2^{'}Na}{RT}$ per unit volume.
The number of molecules going out from volume $V = (P1 - P^{'}1) v = \frac{P1Na}{RT} (V) -\frac{ P2Na}{RT}(V)$
= $\frac{Na V}{RT} [P1 - P^{'}2]$ where P’2 = final pressure of the box
The total number of molecules which exit the hole in time t’ is:
$\frac{1}{2}[Pn1 - Pn2] \sqrt{\frac{Kb T}{m t^{'}}} (A)$
$Pn1 - Pn2 = \frac{P1Na}{RT} -\frac{ P2Na}{RT}= \frac{Na}{RT} [P1-P2]$
The net number of molecules exiting in time t' is
$\frac{Na}{2RT} [P1 - P2] \sqrt{\frac{Kb T}{m t^{'}}} (A)$
Now we can write from the above results,
$t^{'} = 2\frac{(P_1 - P^{'}_2)}{(P_1-P_2)}. \left ( \frac{V}{a} \right ).\sqrt{ \frac{m}{Kb T}}$
$t^{'} = \frac{2[1.5-1.4]}{ [1.5-1]}.\frac{1}{10^{-8}}. \sqrt{\frac{46.5 \times 10^{-27}}{1.38 \times 10^{-23}\times 300} }$
$t^{'}= 0.4 \times 10^{8}\sqrt{ \frac{775 \times 10^{-6}}{69}}$
$t^{'} = 0.4 \times 10^{5} \times 3.35 = 1.34 \times 10^{5}$seconds.

Question:13.31

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

Answer:

We assume ‘p’ to be the number of molecules per unit volume. Hence ‘p’ is the per unit volume molecular density. Let v be the velocity of gas molecules.
The molecules of the gas strike the front face and the back face of the box when it moves. The back face relative velocity = $(v-v_0)$
Change in momentum on the front face of the box = $2m(v+v_{0})$ and the change in momentum on the back face of the box = $2m(v-v _0).$
Nf =Total number of molecules striking the box’s front face = $\frac{1}{2}[A.(v+v_{0}) \Delta t] p$
$Nf = \frac{1}{2}(v+v_{0}) A.p. \Delta t$
Nb = Total number of molecules striking the box’s back face
$=\frac{1}{2}[A.(v+v_{0}) \Delta t] p$
Nb
$=\frac{1}{2}(v+v_{0}) A.p. \Delta t$
Total change in momentum at the front face can be calculated s below:
$Pf = 2m(v+v_{0}) \times Nf = 2m(v+v_{0}) \times \frac{1}{2}(v+v_{o}) A.p. \Delta t$
$Pf = - m(v+v_0)2 A.p. \Delta t$ in the backward direction.
Force on front end = Ff = Pf = $- m(v+v_0)2 A. p$ in the backward direction
In the same way, the force on the back face, Fb = $+ m(v-v_{0})2 A. p$
So, the net force can be calculated as:
$- m(v+v_{0})^{2} A. p+ m(v-v_{0})^{2} A. p$
It can be expanded and simplified as,
$= -m A p 4v.v_{0}$
The magnitude of dragging force= $4m v.v _{0}.A.p$
Kinetic energy for a molecule of gas can be calculated as:
$KE = \frac{1}{2}mv^{2} = \frac{3}{2} Kb T$
Hence, $v = \sqrt{\frac{Kb T}{m}}$
So, we can write the drag force as $4m.A. p. v_o \sqrt{\frac{Kb T}{m}}$