A closed vessel contains a mixture of two dia | vedclass

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(D) Given: $M_A = 16 M_B$ and $m_A = 2 m_B$.$(i)$ Average kinetic energy per molecule is $\frac{3}{2} k_B T$. Since both gases are in the same vessel

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All are true

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(D) Given: $M_A = 16 M_B$ and $m_A = 2 m_B$.$(i)$ Average kinetic energy per molecule is $\frac{3}{2} k_B T$. Since both gases are in the same vessel at the same temperature $T$,the average kinetic energy per molecule is the same for both. Statement $(i)$ is true.$(ii)$ $RMS$ velocity $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$\frac{v_{rms, B}}{v_{rms, A}} = \sqrt{\frac{M_A}{M_B}} = \sqrt{16} = 4$. So,$v_{rms, B} = 4 v_{rms, A}$. Statement $(ii)$ is true.$(iii)$ Number of moles $n = \frac{m}{M}$. So,$n_A = \frac{m_A}{M_A} = \frac{2 m_B}{16 M_B} = \frac{1}{8} n_B$. Pressure $P = \frac{nRT}{V}$. Since $V$ and $T$ are constant,$P \propto n$. Thus,$P_A = \frac{1}{8} P_B$,or $P_B = 8 P_A$. Statement $(iii)$ is true.$(iv)$ Number of molecules $N = n N_A$. Since $n_B = 8 n_A$,then $N_B = 8 N_A$. Statement $(iv)$ is true.Since all statements are true,the correct option is $(D)$.

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