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(C) For $1 	ext{ mole}$ of an ideal gas,$pV = RT \Rightarrow p = \frac{RT}{V}$.Given the process equation: $p = p_0 \left(1 - \alpha \frac{V^3}{V_0^3

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$\frac{1}{4}$

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(C) For $1 	ext{ mole}$ of an ideal gas,$pV = RT \Rightarrow p = \frac{RT}{V}$.Given the process equation: $p = p_0 \left(1 - \alpha \frac{V^3}{V_0^3}ight)$.Equating the two expressions for $p$: $\frac{RT}{V} = p_0 - \frac{\alpha p_0 V^3}{V_0^3} \Rightarrow T = \frac{p_0 V}{R} - \frac{\alpha p_0 V^4}{R V_0^3}$.For maximum temperature,$\frac{dT}{dV} = 0$: $\frac{dT}{dV} = \frac{p_0}{R} - \frac{4 \alpha p_0 V^3}{R V_0^3} = 0$.This gives $V^3 = \frac{V_0^3}{4 \alpha}$.Substituting $V^3$ back into the expression for $T$: $T_{\max} = \frac{p_0}{R} \left( \frac{V_0}{(4 \alpha)^{1/3}} ight) - \frac{\alpha p_0}{R V_0^3} \left( \frac{V_0^3}{4 \alpha} ight) \left( \frac{V_0}{(4 \alpha)^{1/3}} ight) = \frac{p_0 V_0}{R (4 \alpha)^{1/3}} \left( 1 - \frac{1}{4} ight) = \frac{3}{4} \frac{p_0 V_0}{R (4 \alpha)^{1/3}}$.Given $T_{\max} = \frac{3}{4} \frac{p_0 V_0}{R}$,we have $\frac{3}{4} \frac{p_0 V_0}{R} = \frac{3}{4} \frac{p_0 V_0}{R (4 \alpha)^{1/3}}$.Thus,$(4 \alpha)^{1/3} = 1 \Rightarrow 4 \alpha = 1 \Rightarrow \alpha = \frac{1}{4}$.

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