Question

A 50 litre vessel is equally divided into three parts with the help of two stationary semi permeable membrane(SPM). The vessel contains 60g $H_2$ gas in the left chamber, 160g $O_2$ in the middle and 140g $N_2$ in the right one. The left SPM allows transfer of only $H_2$ gas while the right one allows the transfer of both $H_2$ and$N_2$. The final ratio of pressure in the three chambers:

• A. 4:7:5
• B. 5:4:7
• C. 7:4:5
• D. 5:7:4

Answer 1

The correct option is A 4:7:5

SPM will permit the flow of asses till (concentration became equal in each chambers. (of respective gas)

At equilibrium: Moles of $H_2$ in each member $=\frac{30}{2}=10$ moles.

Moles. of $N_2$ in right and middle chamber $=\frac{5}{2}=2.5moles$ So, total number of moles in left chamber $=10$.

total number of moles in middle chamber $=\frac{(2\cdot 5+10)}{+5}=17\cdot 5$. total number of moles in right chamber $=(2.5+10)=12.5.$

Now, here volume and temperature of each chamber remain constant

So, $p\alpha n$

$\begin{array}{c}\text{Thus,}{P}_L:P_m:P_R=n_L:n_m:n_R=10:17.5:12.5\\ =4:7:5\end{array}$

So, option (A) is correct.