A cylinder containing gas at 27,^{circ}C is d | vedclass

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(A) Initial state: $T_1 = 27 + 273 = 300\,K$,$V_1 = V_2 = 100\,cm^3$,$P_1 = P_2 = P$. Final state: One part is at $T_1' = 300\,K$,other at $T_2' = 100

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(A) Initial state: $T_1 = 27 + 273 = 300\,K$,$V_1 = V_2 = 100\,cm^3$,$P_1 = P_2 = P$. Final state: One part is at $T_1' = 300\,K$,other at $T_2' = 100 + 273 = 373\,K$. Let the piston move by distance $x$. The new volumes are $V_1' = 100 - 10.85x$ and $V_2' = 100 + 10.85x$. Since the piston is in equilibrium,the final pressures are equal: $P_1' = P_2' = P'$. Using the ideal gas law $PV = nRT$,and noting that the number of moles $n$ remains constant: $P_1 V_1 / T_1 = P_1' V_1' / T_1' \implies P(100) / 300 = P'(100 - 10.85x) / 300 \implies P' = P \cdot 100 / (100 - 10.85x)$. $P_2 V_2 / T_2 = P_2' V_2' / T_2' \implies P(100) / 300 = P'(100 + 10.85x) / 373$. Equating $P'$ from both: $P \cdot 100 / (100 - 10.85x) = P \cdot 100 \cdot 373 / (300(100 + 10.85x))$. $300(100 + 10.85x) = 373(100 - 10.85x)$. $30000 + 3255x = 37300 - 4047.05x$. $7302.05x = 7300 \implies x \approx 1\,cm$.

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