States of Matter and Gaseous StatehardINTEGER

A very long rectangular box is divided into equal compartments with fixed semipermeable membranes (SStates of Matter and Gaseous State Chemistry Question

Question

A very long rectangular box is divided into $N$ equal compartments with $(N-1)$ fixed semipermeable membranes (SPM) numbered from 1 to $(N-1)$ as shown in the figure. The gases are initially present in only first compartment and they can pass through only those SPM whose number is less than or equal to their subscript. For example, gas $A_1$ can pass through only first SPM, gas $A_2$ can pass only through first and second SPM, and so on. If initially all gases have same moles and after a long time, the ratio of partial pressures of gas $A_4$ in third compartment to that of gas $A_{N-1}$ in first compartment is 3, the value of $N$ is

Answer: 0015

💡 Solution & Explanation

Gas $A_k$ can pass through $k$ SPMs, so it occupies $k+1$ compartments evenly. The partial pressure of $A_k$ in any of its accessible compartments is proportional to $1/(k+1)$ (since initially all have same moles $n_0$). Gas $A_4$ occupies 5 compartments, so its partial pressure in the 3rd compartment is $P_0/5$. Gas $A_{N-1}$ occupies $N$ compartments, so its partial pressure in the 1st compartment is $P_0/N$. Ratio is $(P_0/5) / (P_0/N) = N/5$. Given this ratio is 3, $N/5 = 3 \implies N = 15$. Formatted as a four-digit integer, it is 0015. Therefore, correct answer is 0015.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry