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(C) For an adiabatic process, the relation between pressure and volume is $pV^{\gamma} = 	ext{constant}$.Initially, for part $A$: $p_A = p$, $V_A = 5

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$\frac{10}{3}V$

Vedclass · Answer

(C) For an adiabatic process, the relation between pressure and volume is $pV^{\gamma} = 	ext{constant}$.Initially, for part $A$: $p_A = p$, $V_A = 5V$. For part $B$: $p_B = 8p$, $V_B = V$.When the piston is released, it moves until the pressures in both parts are equal. Let the final pressure be $p_f$ and the final volumes be $V_A'$ and $V_B'$.Since the total volume is constant, $V_A' + V_B' = 5V + V = 6V$.For adiabatic expansion/compression: $p(5V)^{\gamma} = p_f(V_A')^{\gamma}$ and $(8p)(V)^{\gamma} = p_f(V_B')^{\gamma}$.Dividing the two equations: $\frac{p(5V)^{\gamma}}{(8p)V^{\gamma}} = \frac{p_f(V_A')^{\gamma}}{p_f(V_B')^{\gamma}} \implies \frac{5^{\gamma}}{8} = \left(\frac{V_A'}{V_B'}ight)^{\gamma}$.Given $\gamma = 1.5 = \frac{3}{2}$, we have $\frac{5^{3/2}}{8} = \left(\frac{V_A'}{V_B'}ight)^{3/2}$.Taking the power of $2/3$ on both sides: $\left(\frac{5^{3/2}}{8}ight)^{2/3} = \frac{V_A'}{V_B'} \implies \frac{5}{8^{2/3}} = \frac{V_A'}{V_B'} \implies \frac{5}{4} = \frac{V_A'}{V_B'}$.Thus, $V_A' = \frac{5}{4}V_B'$.Substituting into $V_A' + V_B' = 6V$: $\frac{5}{4}V_B' + V_B' = 6V \implies \frac{9}{4}V_B' = 6V \implies V_B' = \frac{24}{9}V = \frac{8}{3}V$.Then $V_A' = 6V - \frac{8}{3}V = \frac{18-8}{3}V = \frac{10}{3}V$.

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