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(B) Given the process equation: $p V^2 = C$.For an ideal gas,we know that $p V = n R T$,which implies $p = \frac{n R T}{V}$.Substituting this into the

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if $V_2 > V_1$ then $T_2 < T_1$

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(B) Given the process equation: $p V^2 = C$.For an ideal gas,we know that $p V = n R T$,which implies $p = \frac{n R T}{V}$.Substituting this into the process equation:$\left(\frac{n R T}{V}ight) V^2 = C$$n R T V = C$Since $n, R,$ and $C$ are constants,we have $T V = 	ext{constant}$,or $T \propto \frac{1}{V}$.If $V_2 > V_1$,then according to the relation $T \propto \frac{1}{V}$,the temperature $T_2$ must be less than $T_1$ $(T_2 Therefore,option $(b)$ is correct.

An ideal gas follows a process described by $p V^2 = C$ from $(p_1, V_1, T_1)$ to $(p_2, V_2, T_2)$,where $C$ is a constant. Then,

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