States of Matter and Gaseous StatehardINTEGER

A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the voluStates of Matter and Gaseous State Chemistry Question

Question

A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at 320 K, the volume of the upper part is 4.0 times greater than that of the lower part. At what temperature (in kelvin) will the ratio of these volumes be equal to 3.0?

Answer: 0450

💡 Solution & Explanation

Let $V$ be the total volume. Initially, $V_u = 4 V_l \implies V_l = 0.2V, V_u = 0.8V$. Pressures: $P_l = \frac{RT}{0.2V}$, $P_u = \frac{RT}{0.8V}$. Piston weight creates pressure difference: $\Delta P = P_l - P_u = \frac{RT}{V}(5 - 1.25) = 3.75 \frac{RT}{V}$. For $T=320\text{ K}$, $\Delta P = 3.75 \frac{320R}{V} = \frac{1200R}{V}$. This weight difference is constant. At new temp $T'$, $V'_u = 3V'_l \implies V'_l = 0.25V, V'_u = 0.75V$. New pressure difference $\Delta P' = P'_l - P'_u = \frac{RT'}{0.25V} - \frac{RT'}{0.75V} = \frac{RT'}{V}(4 - 1.333) = \frac{8}{3} \frac{RT'}{V}$. Equating $\Delta P$: $\frac{8}{3} \frac{RT'}{V} = \frac{1200R}{V} \implies \frac{8T'}{3} = 1200 \implies T' = 450\text{ K}$. Therefore, correct answer is 0450.

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