Grade 11 : Physics - Kinetic Theory (Set 3)

Explanation: The RMS speed of gas molecules is given by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, which shows that $v_{\text{rms}} \propto \sqrt{T}$. If the absolute temperature quadruples ($T_2 = 4 T_1$), the new RMS speed is $v_{\text{rms},2} = \sqrt{4 T_1} \propto 2 \sqrt{T_1}$. Thus, the ratio is $2 : 1$.

Explanation: A rigid diatomic gas has $f = 5$ degrees of freedom (3 translational + 2 rotational). The internal energy per mole is $U_m = \frac{f}{2} R T = \frac{5}{2} R T$. For $n = 2\text{ moles}$, total internal energy is $U = n \times \left(\frac{5}{2} R T\right) = 2 \times \frac{5}{2} R T = 5 R T$.

Explanation: Mean free path is given by $\lambda = \frac{1}{\sqrt{2}\pi n d^2}$, where $n = \frac{P}{k_B T}$ is the number density. Substituting $n$, we get $\lambda = \frac{k_B T}{\sqrt{2}\pi P d^2}$. At constant temperature, $\lambda \propto \frac{1}{P}$. Doubling the pressure $P$ reduces the mean free path to half its original value.

Explanation: The average translational kinetic energy per molecule depends solely on absolute temperature: $E = \frac{3}{2} k_B T$. Since $E \propto T$, doubling the absolute temperature from 300 K to 600 K doubles the average kinetic energy to $2E$.

Explanation: According to kinetic theory, pressure is related to density and RMS speed by $P = \frac{1}{3} \rho v_{\text{rms}}^2$. Substituting values: $P = \frac{1}{3} \times 1.2 \times (500)^2 = 0.4 \times 250,000 = 100,000\text{ Pa} = 10^5\text{ Pa}$.

Explanation: The characteristic speeds are $v_{\text{mp}} = \sqrt{\frac{2RT}{M}} \approx 1.414 \sqrt{\frac{RT}{M}}$, $v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \approx 1.596 \sqrt{\frac{RT}{M}}$, and $v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \approx 1.732 \sqrt{\frac{RT}{M}}$. Therefore, $v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$.

Explanation: For Helium ($n_1=1, C_{v1}=1.5R$) and Oxygen ($n_2=1, C_{v2}=2.5R$), $C_{v,\text{mix}} = \frac{1(1.5R) + 1(2.5R)}{1+1} = 2R$. Since $C_p = C_v + R$, $C_{p,\text{mix}} = 2R + R = 3R$. Thus $\gamma = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{3R}{2R} = 1.50$.

Explanation: The law of equipartition of energy states that for any dynamical system in thermal equilibrium, the total energy is equally distributed among all degrees of freedom, with each degree of freedom contributing $\frac{1}{2} k_B T$ per molecule.

Explanation: Using the ideal gas equation $P V = n R T$: $n = \frac{P V}{R T} = \frac{3 \times 10^5 \times 0.0831}{8.31 \times 300} = \frac{24930}{2493} = 10\text{ moles}$.

Explanation: Most probable speed is $v_{\text{mp}} = \sqrt{\frac{2RT}{M}}$. Since temperature is constant, $v_{\text{mp}} \propto \frac{1}{\sqrt{M}}$. Therefore, $\frac{v_{\text{mp}, H_2}}{v_{\text{mp}, O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$. The ratio is $4 : 1$.

Explanation: A non-rigid diatomic molecule has 3 translational, 2 rotational, and 2 vibrational degrees of freedom (1 kinetic + 1 potential), giving $f = 7$. Thus $C_v = \frac{7}{2} R$ and $C_p = C_v + R = \frac{9}{2} R$. The ratio $\gamma = \frac{C_p}{C_v} = \frac{9}{7}$.

Explanation: Number of moles of Hydrogen: $n_{H_2} = \frac{2}{2} = 1\text{ mol}$. Number of moles of Oxygen: $n_{O_2} = \frac{16}{32} = 0.5\text{ mol}$. Total moles: $n_{\text{total}} = 1.5\text{ mol}$. By Dalton's law, $\frac{P_{H_2}}{P_{\text{total}}} = \frac{n_{H_2}}{n_{\text{total}}} = \frac{1}{1.5} = \frac{2}{3}$.

Explanation: Mean free path is $\lambda = \frac{1}{\sqrt{2}\pi n d^2}$, where $n = \frac{N}{V}$ is the number density. Because volume $V$ and particle number $N$ are fixed in a rigid container, $n$ remains constant. Therefore, heating the gas at constant volume leaves $\lambda$ unchanged.

Explanation: Collision frequency is given by $f_c = \frac{v_{\text{avg}}}{\lambda}$. At constant volume, mean free path $\lambda$ is constant. Average speed $v_{\text{avg}} \propto \sqrt{T}$. Consequently, $f_c \propto v_{\text{avg}} \propto \sqrt{T}$.

Explanation: Translational kinetic energy for $n$ moles of any ideal gas is $E_K = n \left(\frac{3}{2} R T\right)$. Moles of $H_2$: $n_1 = 2/2 = 1$. Moles of He: $n_2 = 8/4 = 2$. At the same $T$, $\frac{E_{K1}}{E_{K2}} = \frac{n_1}{n_2} = \frac{1}{2}$. Ratio is $1 : 2$.

Explanation: At high temperatures, high kinetic energy minimizes intermolecular attractive forces. At low pressure, volume per molecule is large, making intermolecular volume negligible compared to container volume. Thus, real gases behave ideally at high temperature and low pressure.

Explanation: RMS speed is proportional to square root of temperature: $v_{\text{rms}} \propto \sqrt{T} \implies T \propto v_{\text{rms}}^2$. If speed triples to $3v$, the required temperature becomes $T_2 = 3^2 \times T_1 = 9 \times 300\text{ K} = 2700\text{ K}$.

Explanation: A non-linear triatomic molecule has 6 degrees of freedom (3 translational + 3 rotational). Therefore, $C_v = \frac{f}{2} R = \frac{6}{2} R = 3 R$.

Explanation: RMS speed is inversely proportional to the square root of molar mass at constant $T$: $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$. Ratio is $\frac{1}{\sqrt{4}} : \frac{1}{\sqrt{16}} : \frac{1}{\sqrt{64}} = \frac{1}{2} : \frac{1}{4} : \frac{1}{8}$. Multiplying by 8 gives $4 : 2 : 1$.

Explanation: Pressure of an ideal gas is related to translational kinetic energy per unit volume by $P = \frac{2}{3} E_V$. Substituting $E_V = 1.5 \times 10^5\text{ J/m}^3$: $P = \frac{2}{3} \times (1.5 \times 10^5) = 1.0 \times 10^5\text{ Pa}$.