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A gas is found to obey the law $P^3V =$ constant. The initial temperature and volume are $T_0$ and $V_0$. If the gas expands to $2 V_0$ what happens to the internal energy?

(A) increases (B) decreases (C) remains the same (D) insufficient data to resolve answer

1 Answer

$PV = nRT$
$P^3 V = k $ (a constant) $ \rightarrow P = (kV^{-1})^{\large\frac{1}{3}}$
$\Rightarrow (kV^{-1})^{\large\frac{1}{3}} V = nRT$
$\Rightarrow k^{\large\frac{1}{3}} V^{\large\frac{2}{3}} = nRT$
Differentiating, we get, $ k^{\large\frac{1}{3}} \large\frac{2}{3}$$ V^{\large\frac{1}{3}} dV = nRdT$
$\Rightarrow \large\frac{2}{3}$$P(\Delta V) = nR \Delta T$
Since $\Delta V$ is positive, $\Delta T$ is positive, and hence internal energy increases (i.e, it has a positive $\Delta$).
answered Jan 23, 2014 by balaji.thirumalai

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