A 50 litre vessel is equally divided into three parts with the help of two stationary (fixed) semi-p — States of Matter and Gaseous State Chemistry Question
Question
A 50 litre vessel is equally divided into three parts with the help of two stationary (fixed) semi-permeable membranes (SPM). The vessel contains 60 g $H_2$ in the left chamber, 160 g $O_2$ in the middle and 140 g $N_2$ in the right chamber. The left SPM allows transfer of only $H_2$ gas while the right SPM allows the transfer of both $H_2$ and $N_2$. If the ratio of final total pressures in the three chambers is $x:y:5$, then the value of $(5 \times x \times y)$ is
💡 Solution & Explanation
Chambers L, M, R have equal volume $V = 50/3$ L. Moles: $n_L(H_2) = 60/2 = 30\text{ mol}$. $n_M(O_2) = 160/32 = 5\text{ mol}$. $n_R(N_2) = 140/28 = 5\text{ mol}$. $H_2$ can pass both SPMs, so it distributes evenly across all 3 chambers: $10\text{ mol}$ per chamber. $N_2$ can pass the right SPM, so it distributes evenly across M and R: $2.5\text{ mol}$ per chamber. $O_2$ cannot pass any SPM, stays in M ($5\text{ mol}$). Final moles: Chamber L = 10 ($H_2$). Chamber M = 10 ($H_2$) + 5 ($O_2$) + 2.5 ($N_2$) = 17.5. Chamber R = 10 ($H_2$) + 2.5 ($N_2$) = 12.5. Pressure ratio is $10 : 17.5 : 12.5$, dividing by 2.5 gives $4 : 7 : 5$. So $x=4, y=7$. Value of $5 \times 4 \times 7 = 140$. Formatted as a four digit integer, it is 0140. Therefore, correct answer is 0140.