Chapter13 Kinetic Theory
Chapter 13
KINETIC THEORY
MCQ I
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 $500 \mathrm{~m} \mathrm{~s}^{-1}$ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because $500 \mathrm{~m} \mathrm{~s}^{-1}$ is very much smaller than $v_{r m s}$ of 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 $\left(v^2_{r m s}+(500)^2\right) / v^2_{r m s}$ where $v_{r m s}$ was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.
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Answer: (b)
Comment for discussion: This brings in concepts of relative motion and that when collision takes place, it is the relative velocity which changes.
13.2 1 mole of an ideal gas is contained in a cubical volume $V$, ABCDEFGH at $300 \mathrm{~K}$ (Fig. 13.1). 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 $\mathrm{ABCD}$.
(d) the pressure on $\mathrm{EFGH}$ would be half that on $\mathrm{ABCD}$.
Fig. 13.1
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Answer: (d)
Comment for discussion: In the ideal case that we normally consider, each collision transfers twice the magnitude of its normal momentum. On the face EFGH, it transfers only half of that.
13.3 Boyle’s law is applicable for an
(a) adiabatic process.
(b) isothermal process.
(c) isobaric process.
(d) isochoric process.
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Answer: (b)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 (Fig. 13.2). If the temperature is increased,
Fig. 13.2
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Answer: (c) This is a constant pressure $(p=M g / A)$ arrangement.13.5 Volume versus temperature graphs for a given mass of an ideal gas are shown in Fig. 13.3 at two different values of constant pressure. What can be inferred about relation between $P_1$ & $P_2$?
Fig. 13.3
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Answer: (a)13.6 1 mole of $\mathrm{H}_{2}$ gas is contained in a box of volume $V=1.00 \mathrm{~m}^{3}$ at $T=300 \mathrm{~K}$. The gas is heated to a temperature of $T=3000 \mathrm{~K}$ 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.
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Answer: (d)
Comment for discussion: The usual statement for the perfect gas law somehow emphasizes molecules. If a gas exists in atomic form (perfectly possible) or a combination of atomic and molecular form, the law is not clearly stated.
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 $f_{1}(v) d v$, denote the fraction of molecules with speed between $v$ and $(v+d v)$ with $f_{2}(v) d v$, 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 the Maxwell’s distribution law.
(d) $f_{2}(v)$ and $f_{1}(v)$ will be the same.
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Answer: (b)
Comment: In a mixture, the average kinetic energy are equating. Hence, distribution in velocity are quite different.
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 $1.1 \mathrm{~T}$, and the volume is increaset to $1.05 \mathrm{~V}$, the final pressure will be
(a) $1.1 p$
(b) $p$
(c) less than $p$
(d) between $p$ and 1.1 .
MCQ II
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Answer: $(\mathrm{~d})$
Comment for discussion: In this chapter, one has discussed constant pressure and constant volume situations but in real life there are many situations where both change. If the surfaces were rigid, $p$ would rise to $1.1 p$. However, as the pressure rises, $V$ also rises such that $p v$ finally is $1.1 R T$ with $p_{\text {final }}>p$ and $V_{\text {final }}>V$. Hence (d).
13.9 ABCDEFGH is a hollow cube made of an insulator (Fig. 13.4). Face $A B C D$ has positve charge on it. Inside the cube, we have ionized hydrogen.
The usual kinetic theory expression for pressure
Fig. 13.4
(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.
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Answer: (b),(d)13.10 Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory $p V=\frac{2}{3} E, \mathrm{E}$ 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.
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Answer: (c)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.
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Answer: (a), (d)
Comment : The equation <K.E. of translation > = $(3 / 2) R T,<$ Rotational energy $>=R T$ is taught. The fact that the distribution of the two is independent of each other is not emphasized. They are independently Maxwellian.
13.12 Which of the following diagrams (Fig. 13.5) depicts ideal gas behaviour?
(a)
(c)
(b)
(d)
Fig. 13.5
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Answer: (a), (c)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.
VSA
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Answer: (a)
Comment : Conceptually, it is not often clear to the students that elastic collisions with a moving object leads to change in its energy.
13.14 Calculate the number of atoms in $39.4 \mathrm{~g}$ gold. Molar mass of gold is $197 \mathrm{~g} \mathrm{~mole}^{-1}$.
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Answer: $\therefore$ Molar mass of gold is $197 \mathrm{~g} \mathrm{~mole}^{-1}$, the number of atoms $=6.0 \times$ $10^{23}$
$\therefore$ No. of atoms in $39.4 \mathrm{~g}=\frac{6.0 \times 10^{23} \times 39.4}{197}=1.2 \times 10^{23}$
13.15 The volume of a given mass of a gas at $27^{\circ} \mathrm{C}, 1 \mathrm{~atm}$ is $100 \mathrm{cc}$. What will be its volume at $327^{\circ} \mathrm{C}$ ?
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Answer: Keeping $P$ constant, we have
$V_{2}=\frac{V_{1} T_{2}}{T_{1}}=\frac{100 \times 600}{300}=200 \mathrm{cc}$
13.16 The molecules of a given mass of a gas have root mean square speeds of $100 \mathrm{~m} \mathrm{~s}^{-1}$ at $27^{\circ} \mathrm{C}$ and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at $127^{\circ} \mathrm{C}$ and 2.0 atmospheric pressure?
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Answer: $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$
$\frac{V_{1}}{V_{2}}=\frac{P_{2} T_{1}}{P_{1} T_{2}}=\frac{2 \times 300}{400}=\frac{3}{2}$
$P_{1}=\frac{1}{3} \frac{M}{V_{1}} c_{1}^{-2} ; \quad P_{2}=\frac{1}{3} \frac{M}{V_{2}} c_{2}^{-2}$
$\therefore c_{2}{ }^{2}=c_{1}{ }^{2} \times \frac{V_{2}}{V_{1}} \times \frac{P_{2}}{P_{1}}$
$=(100)^{2} \times \frac{2}{3} \times 2$
$c_{2}=\frac{200}{\sqrt{3}} \mathrm{~m} \mathrm{~s}^{-1}$
13.17 Two molecules of a gas have speeds of $9 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$ and $1 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, respectively. What is the root mean square speed of these molecules.
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Answer: $v_{r m s}=\sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2}}$
$=\sqrt{\frac{\left(9 \times 10^{6}\right)^{2}+\left(1 \times 10^{6}\right)^{2}}{2}}$
$=\sqrt{\frac{(81+1) \times 10^{12}}{2}}=\sqrt{41} \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$.
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.)
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Answer: $\mathrm{O}_{2}$ has 5 degrees of freedom. Therfore, energy per mole $=\frac{5}{2} R T$
$\therefore$ For 2 moles of $\mathrm{O}_{2}$, energy $=5 R T$
Neon has 3 degrees of freedom $\therefore$ Energy per mole $=\frac{3}{2} R T$
$\therefore$ For 4 mole of neon, energy $=4 \times \frac{3}{2} R T=6 R T$
$\therefore$ Total energy $=11$ RT.
Exemplar Problems-Physics
13.19 Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters $1 \mathrm{~A}$ and $2 \mathrm{~A}$. The gases may be considered under identical conditions of temperature, pressure and volume.
| $V_{1}$ | $V_{2}$ |
|---|---|
| $\mu_{1}, p_{1}$ | $\mu_{2}$, |
| $p_{2}$ |
Fig 13.6
SA
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Answer: $l \alpha \frac{1}{d^{2}}$
$d_{1}=1 \stackrel{o}{A} \quad \alpha_{2}=2 \stackrel{o}{\AA}$
$l_{1}: l_{2}=4: 1$
13.20 The container shown in Fig. 13.6 has two chambers, separated by a partition, of volumes $V_{1}=2.0$ litre and $V_{2}=3.0$ litre. The chambers contain $\mu_{1}=4.0$ and $\mu_{2}=5.0$ moles of a gas at pressures $p_{1}=1.00 \mathrm{~atm}$ and $p_{2}=2.00 \mathrm{~atm}$. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
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Answer: $V_1=2.0$ litre $V_2=3.0$ litre
$\mu_{1}=4.0$ moles $\mu_{2}=5.0$ moles
$P_1=1.00 atm P_2=2.00 atm$
$P_{1} V_{1}=\mu_{1} R T_{1} \quad P_{2} V_{2}=\mu_{2} R T_{2}$
$\mu=\mu_{1}+\mu_{2} \quad V=V_{1} V_{2}$
For 1 mole $P V=\frac{2}{3} E$
For $\mu_{1}$ moles $\quad P_{1} V_{1}=\frac{2}{3} \mu_{1} E_{1}$
For $\mu_{2}$ moles $\quad P_{2} V_{2}=\frac{2}{3} \mu_{2} E_{2}$
Total energy is $\left(\mu_{1} E_{1}+\mu_{2} E_{2}\right)=\frac{3}{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)$
$P V=\frac{2}{3} E_{\text {total }}=\frac{2}{3} \mu E_{\text {per mole }}$
$P\left(V_{1}+V_{2}\right)=\frac{2}{3} \times \frac{3}{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)$
$P=\frac{P_{1} V_{1}+P_{2} V_{2}}{V_{1}+V_{2}}$
$=\left(\frac{1.00 \times 2.0+2.00 \times 3.0}{2.0+3.0}\right) \mathrm{atm}$
$=\frac{8.0}{5.0}=1.60 \mathrm{~atm}$.
Comment: This form of ideal gas law represented by Equation marked* becomes very useful for adiabatic changes.
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.
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Answer: The average K.E will be the same as conditions of temperature and pressure are the same
$v_{r m s} \alpha \frac{1}{\sqrt{m}}$
$m_{A}>m_{B}>m_{c}$
$v_{C}>v_{B}>v_{A}$
13.22 We have $0.5 \mathrm{~g}$ of hydrogen gas in a cubic chamber of size $3 \mathrm{~cm}$ kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of $100 \mathrm{~atm}$. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius $1 \stackrel{\circ}{\AA}$ ).
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Answer: We have $0.25 \times 6 \times 10^{23}$ molecules, each of volume $10^{-30} \mathrm{~m}^{3}$.
Molecular volume $=2.5 \times 10^{-7} \mathrm{~m}^{3}$
Supposing Ideal gas law is valid.
Final volume $=\frac{V_{i n}}{100}=\frac{(3)^{3} \times 10^{-6}}{100} \approx 2.7 \times 10^{-7} \mathrm{~m}^{3}$
which is about the molecular volume. Hence, intermolecular forces cannot be neglected. Therfore the ideal gas situation does not hold.
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?
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Answer: When air is pumped, more molecules are pumped in. Boyle’s law is stated for situation where number of molecules remain constant.13.24 A ballon has $5.0 \mathrm{~g}$ mole of helium at $7^{\circ} \mathrm{C}$. Calculate
(a) the number of atoms of helium in the balloon,
(b) the total internal energy of the system.
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Answer: $\mu=5.0$
$T=280 \mathrm{~K}$
No of atoms $=\mu N_{A}=5.0 \times 6.02 \times 10^{23}$
$$ =30 \times 10^{23} $$
Average kinetic energy per molecule $=\frac{3}{2} k T$
$\therefore$ Total internal energy $=\frac{3}{2} k T \times N$
$$ \begin{aligned} & =\frac{3}{2} \times 30 \times 10^{23} \times 1.38 \times 10^{-23} \times 280 \ & =1.74 \times 10^{4} \mathrm{~J} \end{aligned} $$
13.25 Calculate the number of degrees of freedom of molecules of hydrogen in $1 \mathrm{cc}$ of hydrogen gas at NTP.
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Answer: Volume occupied by lgram mole of gas at NTP $=22400 \mathrm{cc}$
$\therefore$ Number of molecules in 1cc of hydrogen
$$ =\frac{6.023 \times 10^{23}}{22400}=2.688 \times 10^{19} $$
As each diatomic molecule has 5 degrees of freedom, hydrogen being diatomic also has 5 degrees of freedom $\therefore$ Total no of degrees of freedom $=5 \times 2.688 \times 10^{19}$
13.26 An insulated container containing monoatomic gas of molar mass $m$ is moving with a velocity $v_{o}$. If the container is suddenly stopped, find the change in temperature.
LA
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Answer: Loss in K.E of the gas $=\Delta E=\frac{1}{2}(m n) v_{o}{ }^{2}$
$$ =1.344 \times 10^{20} $$
where $n=$ no: of moles.
If its temperature changes by $\Delta T$, then
$n \frac{3}{2} R \Delta T=\frac{1}{2} m n v_{o}^{2} . \quad \therefore \Delta T=\frac{m v_{o}{ }^{2}}{3 R}$
13.27 Explain why
(a) there is no atmosphere on moon.
(b) there is fall in temperature with altitude.
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Answer: The moon has small gravitational force and hence the escape velocity is small. As the moon is in the proximity of the Earth as seen from the Sun, the moon has the same amount of heat per unit area as that of the Earth. The air molecules have large range of speeds. Even though the rms speed of the air molecules is smaller than the escape 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.
At $300 \mathrm{~K} \quad V_{r m s}=\sqrt{\frac{3 k T}{m}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{7.3 \times 10^{-26}}}=1.7 \mathrm{~km} / \mathrm{s}$ $\mathrm{V}_{\text {esc }}$ for moon $=4.6 \mathrm{~km} / \mathrm{s}$
(b) 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.
13.28 Consider an ideal gas with following distribution of speeds.
| Speed $(\mathrm{m} / \mathrm{s})$ | % of molecules |
|---|---|
| 200 | 10 |
| 400 | 20 |
| 600 | 40 |
| 800 | 20 |
| 1000 | 10 |
(i) Calculate $V_{r m s}$ and hence T. $\left(m=3.0 \times 10^{-26} \mathrm{~kg}\right)$
(ii) If all the molecules with speed $1000 \mathrm{~m} / \mathrm{s}$ escape from the system, calculate new $V_{r m s}$ and hence $T$.
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Answer: (This problem is designed to give an idea about cooling by evaporation)
(i)
$$ \begin{aligned} & V_{r m s}^{2}=\frac{n_{i} v_{i}^{2}}{n_{i}} \ & =\frac{10 \times(200)^{2}+20 \times(400)^{2}+40 \times(600)^{2}+20 \times(800)^{2}+10 \times(1000)^{2}}{100} \ & =\frac{10 \times 100^{2} \times(1 \times 4+2 \times 16+4 \times 36+2 \times 64+1 \times 100)}{100} \ & =1000 \times(4+32+144+128+100)=408 \times 1000 \mathrm{~m}^{2} / \mathrm{s}^{2} \end{aligned} $$
$$ \begin{aligned} & \therefore v_{r m s}=639 \mathrm{~m} / \mathrm{s} \ & \frac{1}{2} m v_{r m s}^{2}=\frac{3}{2} k T \ & \therefore T=\frac{1}{3} \frac{m v_{r m s}^{2}}{k}=\frac{1}{3} \times \frac{3.0 \times 10^{-26} \times 4.08 \times 10^{5}}{1.38 \times 10^{-23}} \ & \quad=2.96 \times 10^{2} \mathrm{~K}=296 \mathrm{~K} \end{aligned} $$
(ii) $V_{r m s}^{2}=\frac{10 \times(200)^{2}+20 \times(400)^{2}+40 \times(600)^{2}+20 \times(800)^{2}}{90}$
$$ \begin{aligned} & =\frac{10 \times 100^{2} \times(1 \times 4+2 \times 16+4 \times 36+2 \times 64)}{90} \ & =10000 \times \frac{308}{9}=342 \times 1000 \mathrm{~m}^{2} / \mathrm{s}^{2} \ v_{r m s} & =584 \mathrm{~m} / \mathrm{s} \ T & =\frac{1}{3} \frac{\mathrm{m} V_{r m s}^{2}}{k}=248 \mathrm{~K} \end{aligned} $$
13.29 Ten small planes are flying at a speed of $150 \mathrm{~km} / \mathrm{h}$ in total darkness in an air space that is $20 \times 20 \times 1.5 \mathrm{~km}^{3}$ 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 saftey region around the plane can be approximated by a sphere of radius $10 \mathrm{~m}$.
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Answer: Time $t=\frac{\lambda}{v}$
$\lambda=\frac{1}{\sqrt{2} \pi d^{2} n}, d=$ diameter and $n=$ number density
$n=\frac{N}{V}=\frac{10}{20 \times 20 \times 1.5}=0.0167 \mathrm{~km}^{-3}$
$t=\frac{1}{\sqrt{2} \pi d^{2}(N / V) \times v}$
$=\frac{1}{1.414 \times 3.14 \times(20)^{2} \times 0.0167 \times 10^{-3} \times 150}$
$=225 \mathrm{~h}$
13.30 A box of $1.00 \mathrm{~m}^{3}$ is filled with nitrogen at $1.50 \mathrm{~atm}$ at $300 \mathrm{~K}$. The box has a hole of an area $0.010 \mathrm{~mm}^{2}$. How much time is required for the pressure to reduce by $0.10 \mathrm{~atm}$, if the pressure outside is $1 \mathrm{~atm}$.
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Answer: $V_{1 \mathrm{x}}=$ speed of molecule inside the box along $x$ direction
$n_{1}=$ number of molecules per unit volume
In time $\Delta t$, particles moving along the wall will collide if they are within $\left(V_{1 x} \Delta t\right)$ distance. Let $a=$ area of the wall. No. of particles colliding in time $\Delta t=\frac{1}{2} n_{i}\left(V_{i x} \Delta t\right) a$ (factor of $1 / 2$ due to motion towards wall).
In general, gas is in equilibrium as the wall is very large as compared to hole.
$\therefore V_{1x}^2+V_{1y}+V_1^2=V^2_r m s$
$\therefore V_{1 x}^{2}=\frac{V^{2}{ }_{r m s}}{3}$
$\frac{1}{2} m V^2_{r m s}=\frac{3}{2} k T \Rightarrow V^2_{r m s}=\frac{3 k T}{m}$
$\therefore V^{2}{ }_{1 x}=\frac{k T}{m}$
$\therefore$ No. of particles colliding in time $\Delta t=\frac{1}{2} n_{1} \sqrt{\frac{k T}{m}} \Delta t a$. If particles collide along hole, they move out. Similarly outer particles colliding along hole will move in.
$\therefore$ Net particle flow in time $\Delta t=\frac{1}{2}\left(n_{1}-n_{2}\right) \sqrt{\frac{k T}{m}} \Delta t a$ as temperature is same in and out.
$p V=\mu R T \Rightarrow \mu=\frac{P V}{R T}$
$n=\frac{\mu N_{A}}{V}=\frac{P N_{A}}{R T}$
After some time $\tau$ pressure changes to $p_{1}^{\prime}$ inside
$\therefore n_{1}^{\prime}=\frac{P_{1}^{\prime} N_{A}}{R T}$
$n_{1} V-n_{1}^{\prime} V=$ no. of particle gone out $=\frac{1}{2}\left(n_{1}-n_{2}\right) \sqrt{\frac{k T}{m}} \tau a$
$\therefore \frac{P_{1} N_{A}}{R T} V-\frac{P_{1}^{\prime} N_{A}}{R T} V=\frac{1}{2}\left(P_{1}-P_{2}\right) \frac{N_{A}}{R T} \sqrt{\frac{k T}{m}} \tau a$
$\therefore \tau=2\left(\frac{P_{1}-P_{1}^{\prime}}{P_{1}-P_{2}}\right) \frac{V}{a} \sqrt{\frac{m}{k T}}$
$=2\left(\frac{1.5-1.4}{1.5-1.0}\right) \frac{5 \times 1.00}{0.01 \times 10^{-6}} \sqrt{\frac{46.7 \times 10^{-27}}{1.38 \times 10^{-23} \times 300}}$
$=1.38 \times 10^{5} \mathrm{~s}$
13.31 Consider a rectangular block of wood moving with a velocity $v_{0}$ in a gas at temperature Tand mass density $\rho$. Assume the velocity is along $x$-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 $4 \rho A v_{o} \sqrt{\frac{k T}{m}}$, where $m$ is the mass of the gas molecule.
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Answer: $n=$ no. of molecules per unit volume
$v_{\text {rms }}=$ rms speed of gas molecules
When block is moving with speed $v_{o}$, relative speed of molecules w.r.t. front face $=v+v_{o}$
Coming head on, momentum transferred to block per collission
$$ =2 m\left(v+v_{o}\right) \text {, where } m=\text { mass of molecule. } $$
No. of collission in time $\Delta t=\frac{1}{2}\left(v+v_{o}\right) n \Delta t A$, where $\mathrm{A}=$ area of cross section of block and factor of $1 / 2$ appears due to particles moving towards block.
$\therefore$ Momentum transferred in time $\Delta t=m\left(v+v_{o}\right)^{2} n A \Delta t$ from front surface
Similarly momentum transferred in time $\Delta t=m\left(v-v_{o}\right)^{2} n A \Delta t$ from back surface
$\therefore$ Net force (drag force) $=\operatorname{mnA}\left[\left(v+v_{o}\right)-\left(v-v_{o}\right)^{2}\right]$ from front
$$ =m n A\left(4 v v_{o}\right)=(4 m n A v) v_{o} $$
$$ =(4 \rho A v) v_{o} $$
We also have $\frac{1}{2} m v^{2}=\frac{1}{2} k T \quad(v-$ is the velocity along $x$-axis $)$
Therefore, $v=\sqrt{\frac{k T}{m}}$.
Thus drag $=4 \rho A \sqrt{\frac{k T}{m}} v_{0}$.
