Kinetic Theory MCQs with Answers – Part 4 (Class 11 Physics)

Explanation: \( \textbf{Given:} \) \(P=1.0\times10^5\,\text{Pa}\), \(\rho=1.25\,\text{kg m}^{-3}\), and \(M=32\times10^{-3}\,\text{kg mol}^{-1}\). \( \textbf{First relation for rms speed:} \) \[ v_{\text{rms}}=\sqrt{\frac{3P}{\rho}} \] \( \textbf{Substitution:} \) \[ v_{\text{rms}}=\sqrt{\frac{3(1.0\times10^5)}{1.25}} \] \( \textbf{Inside square root:} \) \[ \frac{3.0\times10^5}{1.25}=2.4\times10^5 \] \( \textbf{Speed:} \) \[ v_{\text{rms}}\approx\sqrt{2.4\times10^5}\approx490\,\text{m s}^{-1} \] \( \textbf{Density-temperature relation:} \) \[ \rho=\frac{PM}{RT} \] \( \textbf{Rearrange for } T \textbf{:} \) \[ T=\frac{PM}{\rho R} \] \( \textbf{Substitution:} \) \[ T=\frac{(1.0\times10^5)(32\times10^{-3})}{(1.25)(8.0)} \] \( \textbf{Calculation:} \) \[ T=\frac{3200}{10}=320\,\text{K} \] \( \textbf{Final answer:} \) The closest pair is \(v_{\text{rms}}\approx490\,\text{m s}^{-1}\) and \(T\approx320\,\text{K}\).

Explanation: Gas molecules in a warmer region have higher average kinetic energy than those in a cooler region. Random molecular motion and collisions allow energy to be transferred from one region to another. This microscopic transfer appears macroscopically as thermal conduction. The process does not require molecules to lose their identity or stop moving. Kinetic theory describes conduction in gases as energy transport by moving and colliding molecules.

Explanation: Equipartition is a classical result that assigns \(\frac{1}{2}k_BT\) per active quadratic degree of freedom. It works well for many ideal-gas heat-capacity predictions when the relevant modes are active. However, real substances, low-temperature behaviour, liquids, and complex solids may not follow the simplest prediction exactly. Some modes can be inactive over a given temperature range, and molecular interactions can matter strongly. The model is useful when its assumptions fit the physical situation.

Explanation: \( \textbf{Given pressure:} \) \(P=1.5\times10^5\,\text{Pa}\). \( \textbf{Given volume:} \) \(V=0.020\,\text{m}^3\). \( \textbf{Given temperature:} \) \(T=300\,\text{K}\). \( \textbf{Gas type:} \) Monatomic ideal gas. \( \textbf{First relation:} \) \[ PV=nRT \] \( \textbf{Internal energy relation:} \) \[ U=\frac{3}{2}nRT \] \( \textbf{Eliminate } nRT \textbf{ using } PV=nRT \textbf{:} \) \[ U=\frac{3}{2}PV \] \( \textbf{Substitution:} \) \[ U=\frac{3}{2}(1.5\times10^5)(0.020) \] \( \textbf{Product } PV \textbf{:} \) \[ PV=3.0\times10^3\,\text{J} \] \( \textbf{Internal energy:} \) \[ U=1.5(3.0\times10^3)=4.5\times10^3\,\text{J} \] \( \textbf{Final answer:} \) The internal energy is \(4.5\times10^3\,\text{J}\), and \(R\) is not needed after \(PV=nRT\) is combined with \(U=\frac{3}{2}nRT\).

Explanation: \( \textbf{Initial state:} \) \(P_1=1.0\times10^5\,\text{Pa}\), \(V_1=4.0\times10^{-3}\,\text{m}^3\), and \(T_1=300\,\text{K}\). \( \textbf{Final data:} \) \(P_2=2.0\times10^5\,\text{Pa}\) and \(T_2=600\,\text{K}\). \( \textbf{Condition:} \) Amount of gas is fixed. \( \textbf{Combined gas relation:} \) \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] \( \textbf{Rearrange for } V_2 \textbf{:} \) \[ V_2=V_1\frac{P_1}{P_2}\frac{T_2}{T_1} \] \( \textbf{Substitution:} \) \[ V_2=(4.0\times10^{-3})\left(\frac{1.0\times10^5}{2.0\times10^5}\right)\left(\frac{600}{300}\right) \] \( \textbf{Pressure ratio:} \) \[ \frac{1.0\times10^5}{2.0\times10^5}=\frac{1}{2} \] \( \textbf{Temperature ratio:} \) \[ \frac{600}{300}=2 \] \( \textbf{Combined factor:} \) \[ \frac{1}{2}\times2=1 \] \( \textbf{Final answer:} \) \(V_2=4.0\times10^{-3}\,\text{m}^3\), because doubling pressure and doubling absolute temperature balance each other for fixed \(n\).

Explanation: \( \textbf{Given pressure:} \) \(P=2.4\times10^5\,\text{Pa}\). \( \textbf{Given density:} \) \(\rho=1.6\,\text{kg m}^{-3}\). \( \textbf{Required:} \) \(v_{\text{rms}}\). \( \textbf{Kinetic-theory pressure relation:} \) \[ P=\frac{1}{3}\rho v_{\text{rms}}^2 \] \( \textbf{Rearrange for squared rms speed:} \) \[ v_{\text{rms}}^2=\frac{3P}{\rho} \] \( \textbf{Substitution:} \) \[ v_{\text{rms}}^2=\frac{3(2.4\times10^5)}{1.6} \] \( \textbf{Numerator:} \) \[ 3(2.4\times10^5)=7.2\times10^5 \] \( \textbf{Division:} \) \[ v_{\text{rms}}^2=4.5\times10^5\,\text{m}^2\text{s}^{-2} \] \( \textbf{Square root:} \) \[ v_{\text{rms}}=\sqrt{4.5\times10^5}\approx6.7\times10^2\,\text{m s}^{-1} \] \( \textbf{Final answer:} \) The rms speed is approximately \(670\,\text{m s}^{-1}\).

Explanation: \( \textbf{Mass of one molecule:} \) \(m=4.0\times10^{-26}\,\text{kg}\). \( \textbf{Rms speed:} \) \(v_{\text{rms}}=600\,\text{m s}^{-1}\). \( \textbf{Boltzmann constant:} \) \(k_B=1.38\times10^{-23}\,\text{J K}^{-1}\). \( \textbf{Required:} \) Temperature \(T\). \( \textbf{Rms-speed relation:} \) \[ v_{\text{rms}}^2=\frac{3k_BT}{m} \] \( \textbf{Rearrange for } T \textbf{:} \) \[ T=\frac{mv_{\text{rms}}^2}{3k_B} \] \( \textbf{Square of speed:} \) \[ v_{\text{rms}}^2=(600)^2=3.6\times10^5\,\text{m}^2\text{s}^{-2} \] \( \textbf{Substitution:} \) \[ T=\frac{(4.0\times10^{-26})(3.6\times10^5)}{3(1.38\times10^{-23})} \] \( \textbf{Numerator:} \) \[ (4.0\times10^{-26})(3.6\times10^5)=1.44\times10^{-20} \] \( \textbf{Denominator:} \) \[ 3(1.38\times10^{-23})=4.14\times10^{-23} \] \( \textbf{Calculation:} \) \[ T=\frac{1.44\times10^{-20}}{4.14\times10^{-23}}\approx3.5\times10^2\,\text{K} \] \( \textbf{Final answer:} \) The temperature is closest to \(350\,\text{K}\).

Explanation: \( \textbf{Initial volume:} \) \(V_1=6.0\,\text{L}\). \( \textbf{Final volume:} \) \(V_2=2.0\,\text{L}\). \( \textbf{Condition:} \) Isothermal compression with fixed number of molecules. \( \textbf{Pressure relation:} \) \[ P_1V_1=P_2V_2 \] \( \textbf{Pressure ratio:} \) \[ \frac{P_2}{P_1}=\frac{V_1}{V_2}=\frac{6.0}{2.0}=3 \] \( \textbf{Number-density relation:} \) \[ n_v=\frac{N}{V} \] \( \textbf{Density ratio:} \) \[ \frac{n_{v2}}{n_{v1}}=\frac{V_1}{V_2}=3 \] \( \textbf{Mean-free-path dependence:} \) \[ \lambda\propto\frac{1}{n_v} \] \( \textbf{Mean-free-path ratio:} \) \[ \lambda_2=\frac{\lambda_1}{3} \] \( \textbf{Final answer:} \) The final mean free path is \(\frac{\lambda}{3}\) and the final pressure is \(3P_1\).

Explanation: \( \textbf{Gas type:} \) Diatomic ideal gas at ordinary temperature. \( \textbf{Active degrees of freedom:} \) \(f=5\). \( \textbf{Given amount and temperature:} \) \(n=2.0\,\text{mol}\), \(T=400\,\text{K}\). \( \textbf{Constant-volume heat capacity:} \) \[ C_V=\frac{f}{2}R=\frac{5}{2}R \] \( \textbf{Internal energy relation:} \) \[ U=\frac{f}{2}nRT \] \( \textbf{Substitution:} \) \[ U=\frac{5}{2}(2.0)(8.3)(400) \] \( \textbf{Coefficient simplification:} \) \[ \frac{5}{2}\times2.0=5 \] \( \textbf{Calculation:} \) \[ U=5(8.3)(400)=16600\,\text{J} \] \( \textbf{Scientific notation:} \) \[ U=1.66\times10^4\,\text{J} \] \( \textbf{Final answer:} \) The internal energy is \(1.66\times10^4\,\text{J}\), and \(C_V=\frac{5}{2}R\).

Explanation: \( \textbf{Molar mass of helium:} \) \(M_{\text{He}}=4\times10^{-3}\,\text{kg mol}^{-1}\). \( \textbf{Molar mass of hydrogen:} \) \(M_{\text{H}_2}=2\times10^{-3}\,\text{kg mol}^{-1}\). \( \textbf{Same-temperature condition:} \) Both gases in the mixture have the same \(T\). \( \textbf{Rms-speed relation:} \) \[ v_{\text{rms}}=\sqrt{\frac{3RT}{M}} \] \( \textbf{Ratio setup:} \) \[ \frac{v_{\text{rms,He}}}{v_{\text{rms,H}_2}}=\sqrt{\frac{M_{\text{H}_2}}{M_{\text{He}}}} \] \( \textbf{Substitution:} \) \[ \frac{v_{\text{rms,He}}}{v_{\text{rms,H}_2}}=\sqrt{\frac{2\times10^{-3}}{4\times10^{-3}}} \] \( \textbf{Simplification:} \) \[ \frac{v_{\text{rms,He}}}{v_{\text{rms,H}_2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \] \( \textbf{Final answer:} \) The ratio is \(1:\sqrt{2}\), so hydrogen has the larger rms speed at the same temperature.

Explanation: \( \textbf{Gas type:} \) Monatomic ideal gas. \( \textbf{Pressure:} \) \(P=1.0\times10^5\,\text{Pa}\). \( \textbf{Initial volume:} \) \(V_1=2.0\times10^{-3}\,\text{m}^3\). \( \textbf{Final volume:} \) \(V_2=5.0\times10^{-3}\,\text{m}^3\). \( \textbf{At constant pressure:} \) \[ \Delta(PV)=P\Delta V \] \( \textbf{For a monatomic ideal gas:} \) \[ U=\frac{3}{2}nRT=\frac{3}{2}PV \] \( \textbf{Therefore:} \) \[ \Delta U=\frac{3}{2}P\Delta V \] \( \textbf{Volume change:} \) \[ \Delta V=5.0\times10^{-3}-2.0\times10^{-3}=3.0\times10^{-3}\,\text{m}^3 \] \( \textbf{Substitution:} \) \[ \Delta U=\frac{3}{2}(1.0\times10^5)(3.0\times10^{-3}) \] \( \textbf{Calculation:} \) \[ \Delta U=1.5(300)=450\,\text{J} \] \( \textbf{Final answer:} \) The change in internal energy is \(450\,\text{J}\).

Explanation: \( \textbf{Starting relation:} \) \[ P=\frac{1}{3}\rho v_{\text{rms}}^2 \] \( \textbf{Initial state:} \) \[ P=\frac{1}{3}\rho v^2 \] \( \textbf{Final state:} \) \[ 4P=\frac{1}{3}\rho v_2^2 \] \( \textbf{Divide final equation by initial equation:} \) \[ \frac{4P}{P}=\frac{v_2^2}{v^2} \] \( \textbf{Simplification:} \) \[ 4=\frac{v_2^2}{v^2} \] \( \textbf{Taking positive square root:} \) \[ v_2=2v \] \( \textbf{Final answer:} \) The new rms speed is \(2v\), because pressure is proportional to \(v_{\text{rms}}^2\) when density is fixed.

Explanation: \( \textbf{Number of molecules:} \) \(N=5.0\times10^{23}\). \( \textbf{Volume:} \) \(V=0.025\,\text{m}^3\). \( \textbf{Temperature:} \) \(T=320\,\text{K}\). \( \textbf{Boltzmann constant:} \) \(k_B=1.38\times10^{-23}\,\text{J K}^{-1}\). \( \textbf{Molecule-scale ideal-gas equation:} \) \[ PV=Nk_BT \] \( \textbf{Rearrange for pressure:} \) \[ P=\frac{Nk_BT}{V} \] \( \textbf{Substitution:} \) \[ P=\frac{(5.0\times10^{23})(1.38\times10^{-23})(320)}{0.025} \] \( \textbf{First product:} \) \[ (5.0\times10^{23})(1.38\times10^{-23})=6.9 \] \( \textbf{Numerator:} \) \[ 6.9(320)=2208\,\text{J} \] \( \textbf{Pressure:} \) \[ P=\frac{2208}{0.025}=88320\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure is closest to \(8.8\times10^4\,\text{Pa}\).

Explanation: \( \textbf{Given temperature:} \) \(T=300\,\text{K}\). \( \textbf{Given pressure:} \) \(P=1.38\times10^5\,\text{Pa}\). \( \textbf{Molecular diameter:} \) \(d=3.0\times10^{-10}\,\text{m}\). \( \textbf{Number density relation:} \) \[ n_v=\frac{P}{k_BT} \] \( \textbf{Substitution for number density:} \) \[ n_v=\frac{1.38\times10^5}{(1.38\times10^{-23})(300)} \] \( \textbf{Simplification:} \) \[ n_v=\frac{10^5}{300\times10^{-23}}=\frac{1}{3}\times10^{26}\approx3.33\times10^{25}\,\text{m}^{-3} \] \( \textbf{Mean-free-path relation:} \) \[ \lambda=\frac{1}{\sqrt{2}\pi d^2n_v} \] \( \textbf{Square of diameter:} \) \[ d^2=(3.0\times10^{-10})^2=9.0\times10^{-20}\,\text{m}^2 \] \( \textbf{Denominator estimate:} \) \[ (4.4)(9.0\times10^{-20})(3.33\times10^{25})\approx1.32\times10^7\,\text{m}^{-1} \] \( \textbf{Reciprocal:} \) \[ \lambda\approx7.6\times10^{-8}\,\text{m} \] \( \textbf{Final answer:} \) The mean free path is closest to \(1.0\times10^{-7}\,\text{m}\).

Explanation: \( \textbf{Amount of each gas:} \) \(n=3.0\,\text{mol}\). \( \textbf{Temperature rise:} \) \(\Delta T=80\,\text{K}\). \( \textbf{Monatomic constant-volume heat capacity:} \) \[ C_{V,\text{mono}}=\frac{3}{2}R \] \( \textbf{Diatomic constant-volume heat capacity:} \) \[ C_{V,\text{dia}}=\frac{5}{2}R \] \( \textbf{Difference in heat at constant volume:} \) \[ \Delta Q=n(C_{V,\text{dia}}-C_{V,\text{mono}})\Delta T \] \( \textbf{Heat-capacity difference:} \) \[ C_{V,\text{dia}}-C_{V,\text{mono}}=\frac{5}{2}R-\frac{3}{2}R=R \] \( \textbf{Substitution:} \) \[ \Delta Q=(3.0)(8.3)(80) \] \( \textbf{Calculation:} \) \[ \Delta Q=1992\,\text{J} \] \( \textbf{Final answer:} \) The diatomic gas requires about \(1.99\times10^3\,\text{J}\) more heat at constant volume.

Explanation: \( \textbf{Molecule-scale ideal-gas equation:} \) \[ PV=Nk_BT \] \( \textbf{Initial pressure:} \) \[ P_1=\frac{N_1k_BT_1}{V} \] \( \textbf{Changed molecule count:} \) \[ N_2=2N_1 \] \( \textbf{Changed temperature:} \) \[ T_2=\frac{T_1}{2} \] \( \textbf{Final pressure:} \) \[ P_2=\frac{N_2k_BT_2}{V} \] \( \textbf{Substitution:} \) \[ P_2=\frac{(2N_1)k_B(T_1/2)}{V} \] \( \textbf{Simplification:} \) \[ P_2=\frac{N_1k_BT_1}{V}=P_1 \] \( \textbf{Final answer:} \) The pressure remains \(P\), because the doubled molecule count is exactly balanced by halving absolute temperature.

Explanation: \( \textbf{Given density:} \) \(\rho=0.80\,\text{kg m}^{-3}\). \( \textbf{Molar mass:} \) \(M=20\times10^{-3}\,\text{kg mol}^{-1}\). \( \textbf{Temperature:} \) \(T=500\,\text{K}\). \( \textbf{Gas constant:} \) \(R=8.0\,\text{J mol}^{-1}\text{K}^{-1}\). \( \textbf{Density relation:} \) \[ \rho=\frac{PM}{RT} \] \( \textbf{Rearrange for pressure:} \) \[ P=\frac{\rho RT}{M} \] \( \textbf{Substitution:} \) \[ P=\frac{(0.80)(8.0)(500)}{20\times10^{-3}} \] \( \textbf{Numerator:} \) \[ (0.80)(8.0)(500)=3200 \] \( \textbf{Division by molar mass:} \) \[ P=\frac{3200}{0.020}=1.6\times10^5\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure is \(1.6\times10^5\,\text{Pa}\).

Explanation: \( \textbf{Initial rms speed:} \) \(v_1=500\,\text{m s}^{-1}\). \( \textbf{Mean free path:} \) \(\lambda=2.5\times10^{-7}\,\text{m}\). \( \textbf{Temperature change:} \) \[ T_2=4T_1 \] \( \textbf{Rms-speed proportionality for same gas:} \) \[ v_{\text{rms}}\propto\sqrt{T} \] \( \textbf{New speed:} \) \[ v_2=v_1\sqrt{4}=2v_1=1000\,\text{m s}^{-1} \] \( \textbf{Why } \lambda \textbf{ is unchanged:} \) Number density and molecular diameter are unchanged, so \(\lambda=\frac{1}{\sqrt{2}\pi d^2n_v}\) is unchanged. \( \textbf{Collision frequency:} \) \[ z=\frac{v_{\text{rms}}}{\lambda} \] \( \textbf{Substitution:} \) \[ z_2=\frac{1000}{2.5\times10^{-7}} \] \( \textbf{Calculation:} \) \[ z_2=4.0\times10^9\,\text{s}^{-1} \] \( \textbf{Final answer:} \) The new collision frequency is \(4.0\times10^9\,\text{s}^{-1}\).

Explanation: \( \textbf{Monatomic amount:} \) \(n_m=2.0\,\text{mol}\). \( \textbf{Diatomic amount:} \) \(n_d=1.0\,\text{mol}\). \( \textbf{Temperature:} \) \(T=300\,\text{K}\). \( \textbf{Monatomic internal energy:} \) \[ U_m=\frac{3}{2}n_mRT \] \( \textbf{Diatomic internal energy at ordinary temperature:} \) \[ U_d=\frac{5}{2}n_dRT \] \( \textbf{Total internal energy:} \) \[ U=U_m+U_d \] \( \textbf{Substitution:} \) \[ U=\frac{3}{2}(2.0)RT+\frac{5}{2}(1.0)RT \] \( \textbf{Coefficient total:} \) \[ \frac{3}{2}(2.0)+\frac{5}{2}(1.0)=3.0+2.5=5.5 \] \( \textbf{Numerical calculation:} \) \[ U=5.5(8.3)(300) \] \( \textbf{Result:} \) \[ U=13695\,\text{J}\approx1.37\times10^4\,\text{J} \] \( \textbf{Final answer:} \) The total internal energy is closest to \(1.37\times10^4\,\text{J}\).

Explanation: \( \textbf{Pressure:} \) \(P=1.0\times10^5\,\text{Pa}\). \( \textbf{Volume:} \) \(V=0.030\,\text{m}^3\). \( \textbf{Molecule count:} \) \(N=7.5\times10^{23}\). \( \textbf{Boltzmann constant:} \) \(k_B=1.38\times10^{-23}\,\text{J K}^{-1}\). \( \textbf{Molecule-scale ideal-gas equation:} \) \[ PV=Nk_BT \] \( \textbf{Rearrange for temperature:} \) \[ T=\frac{PV}{Nk_B} \] \( \textbf{Numerator:} \) \[ PV=(1.0\times10^5)(0.030)=3.0\times10^3\,\text{J} \] \( \textbf{Denominator:} \) \[ Nk_B=(7.5\times10^{23})(1.38\times10^{-23})=10.35\,\text{J K}^{-1} \] \( \textbf{Calculation:} \) \[ T=\frac{3000}{10.35}\approx290\,\text{K} \] \( \textbf{Final answer:} \) The temperature is closest to \(290\,\text{K}\).