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2 moles of an ideal monoatomic gas is carried from a state $\left(p_{0}, V_{0}\right)$ to state $\left(2 p_{0}, 2 V_{0}\right)$ along a straight line path in a $p-V$ diagram. The amount of heat absorbed by the gas in the process is given by
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The correct answer is:
$6 p_{0} V_{0}$
The internal energy $\Delta U=n C_{r} \Delta T$
$C_{v}=$ Specific heat of gas at constant volume
$$
\begin{aligned}
\Rightarrow \quad \Delta U &=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}}{n R}-\frac{p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}-p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2} \cdot \frac{3 p_{0} V_{0}}{n R} \\
&=\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
Work done by the gas $W=\left(2 p_{0}+p_{0}\right) \frac{V_{0}}{2}=\frac{3 p_{0} V_{0}}{2}$
From first law of thermodynamics,
$$
\begin{aligned}
\Delta Q &=d W+d U \\
&=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
[from Eqs. (i) and (ii)]
$$
\begin{array}{l}
=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0} \\
=\frac{12 p_{0} V_{0}}{2}=6 p_{0} V_{0}
\end{array}
$$
$C_{v}=$ Specific heat of gas at constant volume
$$
\begin{aligned}
\Rightarrow \quad \Delta U &=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}}{n R}-\frac{p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}-p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2} \cdot \frac{3 p_{0} V_{0}}{n R} \\
&=\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
Work done by the gas $W=\left(2 p_{0}+p_{0}\right) \frac{V_{0}}{2}=\frac{3 p_{0} V_{0}}{2}$
From first law of thermodynamics,
$$
\begin{aligned}
\Delta Q &=d W+d U \\
&=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
[from Eqs. (i) and (ii)]
$$
\begin{array}{l}
=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0} \\
=\frac{12 p_{0} V_{0}}{2}=6 p_{0} V_{0}
\end{array}
$$
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