Kinetic theory of gases practice · Mixed difficulty

Option B is correct. Both the assertion and reason are true, but the reason doesn't explain why the speeds are of the same order.

Option A (${E_1} = {E_2}$) is correct because at the same temperature, the average kinetic energy depends only on the temperature and degrees of freedom, which are the same for both diatomic gases.

Option A ($1.1 V_0, 1.1 T_0$) is correct because the final volume and temperature are calculated using the heat transferred at constant pressure.

Option D (523 m/s) is correct because the RMS speed is calculated using the formula Vrms = $\sqrt{{3kT/m}}$.

Option C (Only the translational part of the energy, because during collisions with the wall, pressure related to change in linear momentum) is correct because the pressure exerted on the walls of the container is due to the change in linear momentum of the molecules.

Option C ($\frac{1}{\sqrt{2} n \pi d^2}$) is correct because this is the standard formula for the mean free path of a molecule in a gas.

Option C is correct. Carbon dioxide (CO2) molecule has 6 degrees of freedom, while a water molecule in water vapor has more than 6 degrees of freedom due to vibrational modes at high temperatures.

Option B (3) is correct because a diatomic gas, like any gas, can move freely in three-dimensional space, corresponding to three translational degrees of freedom.

Option A is correct because the temperature of the gas will increase due to the increase in kinetic energy of the gas molecules as the train moves with high speed.

Option D ($\frac{1}{{{\mathrm{V}_A}}} < \frac{1}{{{\mathrm{V}_B}}} < \frac{1}{{{\mathrm{V}_C}}}$) is correct because the root mean square speed is inversely proportional to the square root of the mass.

Option A (01:01) is correct because the average rotational kinetic energy depends only on temperature and degrees of freedom, which are the same for both gases.

Option D is correct because, according to Maxwell's distribution, the most probable velocity is less than the root mean square velocity.

Option C ($\frac{{1}}{{2}} k_B T$) is correct because, according to the equipartition theorem, each degree of freedom of a molecule contributes $\frac{{1}}{{2}} k_B T$ to its average energy.

Option D ($d^{-2}$) is correct because the mean free path is inversely proportional to the square of the diameter of the gas molecules.

Option A (120 R) is correct because the work done is calculated by integrating pdV, using the given relation between V and T.

Option D ($0$) is correct because at resonance in an LCR series circuit, the inductive and capacitive reactances cancel each other out, resulting in a phase difference of zero between voltage and current.

Option C (17.4 J / mol-K) is correct because it is calculated by using the molar specific heats of helium and hydrogen and their respective mole fractions in the mixture.

Option B (3) is correct because a diatomic gas, like any gas, can move freely in three-dimensional space, thus having 3 translational degrees of freedom.

Option B is correct. The assertion is true because balloons stop rising at a certain height, but the reason is false because atmospheric pressure doesn't become zero at maximum height; rather, the density of the balloon equals the density of the surrounding air.

Option C ($24.9 \times {10^5}Pa$) is correct because it is calculated using Dalton's law of partial pressures and the ideal gas law.

Option B ($v_{\text{rms}} > v_{av} > v_{{mp }}$) is correct because the root mean square velocity is always greater than the average velocity, which is greater than the most probable velocity.

Option B is correct because the mean free path is inversely proportional to the square of the diameter of the gas molecules, as shown in the formula.

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.

Option C (5) is correct because a diatomic molecule has 3 translational and 2 rotational degrees of freedom, totaling 5.

Option A is correct because the ratio of change in internal energy to work done is 3/2, derived from the constant pressure process.

Option A (${10^4}\mathrm{ }K$) is correct because it is the temperature at which the root mean square velocity equals the escape velocity, calculated using the given constants.

Option B is correct because the mean kinetic energy depends only on temperature, and since both gases are at the same temperature, their mean kinetic energies are equal.

Option A (Kinetic theory of matter) is correct because Brownian motion is the random movement of particles suspended in a fluid, providing evidence for the continuous motion of molecules as described by the kinetic theory.

Option A (Kinetic energy) is correct because at 0 K, all molecular motion ceases, resulting in zero kinetic energy.

Option A ($\frac{{400}}{{\sqrt{{3}}}}$) is correct because it is calculated using the relationship between RMS velocity and temperature.