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A gas is found to obey the law $P =kV^a$. The initial temperature and volume are $V_0$. If the gas expands to $b V_0$ the sign of the internal energy change depends on which of the following? (Note: $\gamma$ is the adiabatic index)

(A) $a, \gamma$ (B) $b, \gamma$ (C) $a, b$ (D) $a, b, \gamma$

1 Answer

$PV = nRT$ and $P = kV^a$.
$\Rightarrow kV^a V = nRT = kV^{a+1} = nRT$
Differentiating, we get:
$k(a+1)V^adV = nRDT \rightarrow (a+1) P dV = nR dT$
Therefore, the internal energy is dependent on $a$ and $dV$, which in turn is dependent on $b$, so the change in internal energy is dependent on $a, b$.
answered Jan 23, 2014 by balaji.thirumalai

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