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(A) For an ideal gas,the equation of state is $PV = nRT$.Case $1$: Constant temperature ($T$ is constant).Differentiating $PV = nRT$ with respect to p

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(A) For an ideal gas,the equation of state is $PV = nRT$.Case $1$: Constant temperature ($T$ is constant).Differentiating $PV = nRT$ with respect to pressure,we get $P \Delta V + V \Delta P = 0$,which implies $\Delta V = -\frac{V \Delta P}{P}$.The magnitude of the change in volume is $|\Delta V| = \frac{V |\Delta P|}{P}$.Case $2$: Constant pressure ($P$ is constant).Differentiating $PV = nRT$ with respect to temperature,we get $P \Delta V = nR \Delta T$,which implies $\Delta V = \frac{nR \Delta T}{P}$.The magnitude of the change in volume is $|\Delta V| = \frac{nR |\Delta T|}{P}$.Since the magnitude of the change in volume is the same in both cases:$\frac{V |\Delta P|}{P} = \frac{nR |\Delta T|}{P}$$V |\Delta P| = nR |\Delta T|$$|\Delta T| = \frac{V}{nR} |\Delta P|$From the ideal gas law,$\frac{V}{nR} = \frac{T}{P}$.Substituting this into the equation,we get $|\Delta T| = \frac{T}{P} |\Delta P|$.Given $|\Delta T| = C |\Delta P|$,we have $C = \frac{T}{P}$.Given $T = 300 \ K$ and $P = 2 \ atm$,$C = \frac{300}{2} = 150 \ K/atm$.

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