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(D) For $1$ mole of an ideal gas,$pV = RT$ ... $(i)$.At constant pressure,$P dV = R dT$ ... (ii).From equations $(i)$ and (ii),we get $\frac{dV}{V} =

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All of the above.

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(D) For $1$ mole of an ideal gas,$pV = RT$ ... $(i)$.At constant pressure,$P dV = R dT$ ... (ii).From equations $(i)$ and (ii),we get $\frac{dV}{V} = \frac{dT}{T}$.The coefficient of volume expansion at constant pressure is defined as $\gamma = \frac{1}{V} \frac{dV}{dT} = \frac{1}{T}$. Since this depends only on $T$,it is the same for all ideal gases.Mean free path $\lambda$ is given by $\lambda = \frac{kT}{\sqrt{2} \pi d^2 P}$. As pressure $P$ decreases,$\lambda$ increases.The average translational kinetic energy of a molecule is $\langle K \rangle = \frac{3}{2} kT$. This value depends only on the temperature $T$ and is independent of the mass or nature of the gas molecules. Thus,in a mixture at thermal equilibrium,all components have the same average translational kinetic energy.Therefore,all the given statements are correct.

From the following statements,concerning an ideal gas at any given temperature $T$,select the correct one$(s)$:

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