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(B) For a monoatomic gas,the molar heat capacity $C$ for a process $P = KV$ is calculated as follows:Given $P = KV$,we know $PV = nRT$. Substituting $

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$2RT_0$

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(B) For a monoatomic gas,the molar heat capacity $C$ for a process $P = KV$ is calculated as follows:Given $P = KV$,we know $PV = nRT$. Substituting $P = KV$,we get $KV^2 = nRT$,so $V^2 \propto T$.For a polytropic process $PV^x = 	ext{constant}$,we have $T V^{x-1} = 	ext{constant}$.Since $V^2 \propto T$,we have $T V^{-2} = 	ext{constant}$,which implies $x - 1 = -2$,so $x = -1$.The molar heat capacity $C$ is given by $C = C_v + \frac{R}{1-x} = \frac{3}{2}R + \frac{R}{1-(-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$.The heat supplied is $Q = nC\Delta T = 1 	imes (2R) 	imes (2T_0 - T_0) = 2RT_0$.

$1 \, \text{mole}$ of an ideal monoatomic gas at temperature $T_0$ expands slowly according to the law $P = KV$,where $K$ is a constant. If the final temperature of the gas is $2T_0$,find the heat supplied to the gas.

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