Question

Two containers $C_1$ and $C_2$ are kept at a constant temperature T=300 K. Container $C_1$ has a volume of 2 litres and it contains Helium gas at a pressure of 10 atm while container $C_2$ has a volume of 8 litres and that contains Methane $\left(C{H}_4\right)$ gas at a pressure of 5 atm. Assume the effective radius of $C{H}_4$ gas molecule is 4 times that of a He gas molecule.

List-I contains questions related to gases in containers $C_1$ and $C_2$ and List-II contains their answer(s).

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \left(I\right)\text{When both the containers are connected with}& \left(P\right)4\\ \text{a thin tube, the final pressure will be (in atm)}& \\ \left(II\right)\text{The mole fraction of He gas when both}& \left(Q\right)\frac{1}{4}\\ \text{containers are connected with the thin tube}& \\ \left(III\right)\text{The ratio of the mean fee path of}C{H}_4\text{to He}& \left(R\right)\frac{1}{8}\\ \text{before connecting the containers}& \\ \left(IV\right)\text{The ratio of collision frequency of He to}C{H}_4& \left(S\right)\frac{2}{3}\\ \text{before connecting the containers}& \\ & \left(T\right)\frac{1}{3}\\ & \left(U\right)6\end{array}$

Which of the following options has the correct combination considering questions of List-I and answers of List-II

• A. (I), (P) and (III), (Q)
• B. (II), (S) and (I), (P)
• C. (I), (U) and (III), (R)
• D. (IV), (R) and (II), (T)

Answer 1

The correct option is C (I), (U) and (III), (R)

Partial pressure of $He$ after connecting both containers, $=\frac{P_1{V}_1}{V_2}=\frac{10\times 2}{10}=2atm$

Partial pressure of $C{H}_4$ after connecting both containers, $=\frac{P_1{V}_1}{V_2}=\frac{5\times 8}{10}=4atm$

(I) Total pressure after connecting both the containers = 2 + 4 = 6 atm

(II) Mole fraction of $He$, when both the containers are connected, $=\frac{\text{Partial pressure of He}}{Totalpressure}=\frac{2}{6}=\frac{1}{3}$

(III) The ratio of mean free path of $C{H}_4$ to $He$ before connecting the containers,$\frac{{\lambda }_{C{H}_4}}{{\lambda }_{He}}=\frac{\frac{RT}{\sqrt{2}\pi d_{C{H}_4}^2{N}_A{P}_{C{H}_4}}}{\frac{RT}{\sqrt{2}\pi d_{He}^2{N}_A{P}_{He}}}=\frac{10}{4^2\times 5}=\frac{1}{8}$ (d = molecular diameter, $N_A$ = Avogadro no.)

(IV) The ratio of collision frequency of $He$ to $C{H}_4$ before connecting the containers
$\frac{Z_{He}}{Z_{C{H}_4}}=\frac{\frac{V_{av}\left(He\right)}{{\lambda }_{He}}}{\frac{V_{av}\left(C{H}_4\right)}{{\lambda }_{C{H}_4}}}=\frac{{\lambda }_{C{H}_4}}{{\lambda }_{He}}\times \frac{V_{av}\left(He\right)}{V_{av}\left(C{H}_4\right)}=\frac{1}{8}\times \sqrt{\frac{M_{C{H}_4}}{M_{He}}}=\frac{1}{8}\times \sqrt{\frac{16}{4}}=\frac{1}{4}$