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One mole of an ideal gas at initial temperature $T$, undergoes a quasi-static process during which the volume $V$ is doubled. During the process, the internal energy $U$ obeys the equation $U=a V^3$, where $a$ is a constant. The work done during this process is
$\frac{3 R T}{2}$
$\frac{5 R T}{2}$
$\frac{5 R T}{3}$
$\frac{7 R T}{3}$
Solution
(d)
In the process given, internal energy is
$U=a V^3$
$\Rightarrow \quad \frac{f n R T}{2} =a V^3$
where, $f=$ degree of freedom $=3$ (for ideal gas) and $n=$ number of moles $=1$.
$\Rightarrow \quad \frac{3}{2} R T=a V^3$
$\Rightarrow \quad \frac{3}{2} p V=a V^3 \quad[\because p V=R T]$
So, pressure in this process is given by
$p=\frac{2 a}{3} V^2$
Now, work done during the process is
$W=\int \limits_V^{2 V} p d V=\int \limits_V^{2 V} \frac{2 a}{3} V^2 d V$
$=\frac{2 a}{3} \times \frac{1}{3}\left((2 V)^3-V^3\right)$
$=\frac{2 a}{9}\left(7 V^3\right)$
$=\frac{2}{9} \times 7 \times \frac{3}{2} R T \quad\left[\because a V^3=\frac{3}{2} R^2 T\right]$
$=\frac{7}{3} R T$
