CBSE · NCERT · Class 11 Physics · Chapter 12

NCERT Solutions: Class 11 Physics Chapter 12 - Kinetic Theory

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Chapter-wise NCERT intext questions and exercise answers for Kinetic Theory, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 9
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1Exercises9 questions
Q.12.1Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.v
Solution

For one mole, molecular volume is $N_A\frac43\pi r^3$, with $r=1.5\times10^{-10}\,\text{m}$. Thus molecular volume $=6.02\times10^{23}\times\frac43\pi(1.5\times10^{-10})^3=8.51\times10^{-6}\,\text{m}^3$. Actual molar volume at STP is $22.4\,\text{L}=2.24\times10^{-2}\,\text{m}^3$. Fraction $=8.51\times10^{-6}/2.24\times10^{-2}=3.8\times10^{-4}$.

Answer:

$3.8\times10^{-4}$ approximately.

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Q.12.2Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.v
Solution

For one mole of an ideal gas, $V=RT/P$. At STP, $T=273.15\,\text{K}$ and $P=1.013\times10^5\,\text{Pa}$. Hence $V=8.31(273.15)/(1.013\times10^5)=2.24\times10^{-2}\,\text{m}^3=22.4\,\text{L}$.

Answer:

$22.4\,\text{L}$.

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Q.12.4An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u).v
Solution

Use absolute pressures: initially $P_1=16\,\text{atm}$ and finally $P_2=12\,\text{atm}$. With $V=0.030\,\text{m}^3$, $T_1=300\,\text{K}$ and $T_2=290\,\text{K}$, $n_1=P_1V/(RT_1)=19.50\,\text{mol}$ and $n_2=P_2V/(RT_2)=15.13\,\text{mol}$. Moles removed $=4.37$, so mass removed $=4.37(0.032)=0.140\,\text{kg}$.

Answer:

$0.14\,\text{kg}$ approximately.

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Q.12.5An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?v
Solution

At the bottom, $P_1=P_0+\rho gh=1.013\times10^5+1000(9.8)(40)=4.93\times10^5\,\text{Pa}$. At the surface, $P_2=P_0=1.013\times10^5\,\text{Pa}$. Temperatures are $T_1=285\,\text{K}$ and $T_2=308\,\text{K}$. From $PV/T=\text{constant}$, $V_2=V_1(P_1/P_2)(T_2/T_1)=1.0(4.87)(1.08)=5.26\,\text{cm}^3$.

Answer:

$5.3\,\text{cm}^3$ approximately.

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Q.12.6Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.v
Solution

Using $PV=Nk_BT$, $N=PV/(k_BT)=\frac{(1.013\times10^5)(25.0)}{(1.38\times10^{-23})(300)}=6.1\times10^{26}$ molecules.

Answer:

$6.1\times10^{26}$ molecules.

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Q.12.7Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).v
Solution

For a monatomic atom, average translational thermal energy is $\frac32k_BT$. At $300\,\text{K}$, this is $1.5(1.38\times10^{-23})(300)=6.2\times10^{-21}\,\text{J}$. At $6000\,\text{K}$ it is $1.24\times10^{-19}\,\text{J}$. At $10^7\,\text{K}$ it is $2.07\times10^{-16}\,\text{J}$.

Answer:

(i) $6.2\times10^{-21}\,\text{J}$. (ii) $1.24\times10^{-19}\,\text{J}$. (iii) $2.07\times10^{-16}\,\text{J}$.

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Q.12.8Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ?v
Solution

Equal $P$, $V$ and $T$ imply equal $N$ from $PV=Nk_BT$. The rms speed is $v_{rms}=\sqrt{3k_BT/m}$, so at the same temperature it is larger for smaller molecular mass. Neon is lighter than chlorine and uranium hexafluoride, so neon has the largest rms speed.

Answer:

Yes, they contain equal numbers of molecules. No, their rms speeds are not the same; $v_{rms}$ is largest for neon.

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Q.12.9At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).v
Solution

Equal rms speeds require $T_{Ar}/M_{Ar}=T_{He}/M_{He}$. The helium temperature is $253\,\text{K}$. Thus $T_{Ar}=253(39.9/4.0)=2.53\times10^3\,\text{K}$.

Answer:

$2.5\times10^3\,\text{K}$ approximately.

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Q.12.10Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).v
Solution

At $P=2.0\,\text{atm}$ and $T=290\,\text{K}$, number density $n=P/(k_BT)=5.06\times10^{25}\,\text{m}^{-3}$. Molecular diameter $d=2.0\times10^{-10}\,\text{m}$. Mean free path $l=1/(\sqrt2\pi d^2n)=1.11\times10^{-7}\,\text{m}$. The rms speed is $v=\sqrt{3RT/M}=\sqrt{3(8.31)(290)/0.028}=508\,\text{m s}^{-1}$. Collision frequency $=v/l=4.6\times10^9\,\text{s}^{-1}$ and free time $=l/v=2.2\times10^{-10}\,\text{s}$. A collision time of order $d/v=4\times10^{-13}\,\text{s}$ is far smaller.

Answer:

Mean free path $\approx1.1\times10^{-7}\,\text{m}$; collision frequency $\approx4.6\times10^9\,\text{s}^{-1}$. Collision duration is about $4\times10^{-13}\,\text{s}$, much smaller than the free-flight time $2.2\times10^{-10}\,\text{s}$.

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