Question

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 $U = aV^{3}$ , where $a$ is a constant. The work done during this process is

• A. $\frac{3RT}{2}$
• B. $\frac{5RT}{2}$
• C. $\frac{5RT}{3}$
• D. $\frac{7RT}{3}$

Answer 1

Answer: (D)

In the process given, internal energy is
U$=aV^{3} $
$\Rightarrow \frac{fnRT}{2}=aV^{3}$
where, $f = $degree of freedom $= 3$ (for idealgas)
and $n =$ number of moles $= 1.$
$\Rightarrow \frac{3}{2}RT=aV^{3}$
$\frac{\Rightarrow 3}{2}pV =aV^{3} \left[\because pV =RT\right] $
So, pressure in this process is given by
$p=\frac{2a}{3}.V^{2}$
Now, work done during the process is
$W =\int\limits_{V}^{2V} pdV =\int\limits_{V}^{2V} \frac{2a}{3}V^{2}dV$
$=\frac{2a}{3}\times\frac{1}{3}\left(\left(2V\right)^{3} -V^{3}\right)$
$=\frac{2a}{9}\left(7V^{3}\right) $
$=\frac{2}{9}\times7\times\frac{3}{2}RT\left[\because aV^{3}=\frac{3}{2}R\right]$
$=\frac{7}{3}RT$