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The ratio of specific heats of a gas is $\gamma$. The change in internal energy of one mole of the gas, when the volume changes from $V$ to $2 V$ at constant pressure $p$ is
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The correct answer is:
$\frac{p V}{\gamma-1}$
Change in internal energy,
$d U=d Q-d W$
At constant pressure,
$\begin{aligned}
d U & =C_p d T-p d V \\
& =C_p d T-R d T \\
& =\left(C_p-R\right) d T=C_V d T \\
& =\frac{R}{\gamma-1} d T \\
& =\frac{R}{\gamma-1} \times \frac{p V}{R} \\
& =\frac{p V}{\gamma-1}
\end{aligned}$
$d U=d Q-d W$
At constant pressure,
$\begin{aligned}
d U & =C_p d T-p d V \\
& =C_p d T-R d T \\
& =\left(C_p-R\right) d T=C_V d T \\
& =\frac{R}{\gamma-1} d T \\
& =\frac{R}{\gamma-1} \times \frac{p V}{R} \\
& =\frac{p V}{\gamma-1}
\end{aligned}$
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