Question

The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an initial temperature of $300 \, \text{K}$ is: $\left[\right. \text{R} = 2 \text{cal} / \text{mol} - \text{K} \left]\right.$

Answer 1

To calculate the work done, we neet to know the final temperature.

$T_{1}=300 \, K,\text{ }P_{1}=1 \, atm, \, n=2 \, mole, \, P_{2}=2 \, atm, \, V_{1}=\frac{nRT_{1}}{P_{1}}$

$w=\Delta Y=\left(nC\right)_{v}\left(T_{2} - T_{1}\right)=-P_{2}\left(V_{2} - V_{1}\right)$

For a monatomic gas, $C_{v}=\frac{3}{2}R$

$\Rightarrow \, \, \, \frac{3}{2}nR\left(T_{2} - T_{1}\right)=-P_{2}\left(V_{2} - V_{1}\right)$

$\frac{3}{2}nR\left(T_{2} - T_{1}\right)=-P_{2}\left(nR \frac{T_{2}}{P_{2}} - nR \frac{T_{1}}{P_{1}}\right)$

$\frac{3}{2}\left(T_{2} - T_{1}\right)=-T_{2}+\frac{P_{2}}{P_{1}}\Rightarrow T_{2}=\frac{2}{3}\left(\frac{P_{2}}{P_{1}} + \frac{3}{2}\right)T_{1}$

$T_{2}=0.4\left(2 + 1 .5\right)300=420 \, K$

Work done, $\mathrm{W}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=2 \times \frac{3}{2} \mathrm{R} \times(420-300)$
$\Rightarrow \, W=2\times \frac{3}{2}\times 2\times 120=720 \, cal$