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: $[\text{R}=2\text{cal}/\text{mol}-\text{K}]$

Answer 1

Answer: 720

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

$T_1=300K,P_1=1atm,n=2mole,P_2=2atm,V_1=\frac{nR{T}_1}{P_1}$

$w=\Delta Y={\left(nC\right)}_v\left(T_2-T_1\right)=-P_2\left(V_2-V_1\right)$

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

$\Rightarrow \frac{3}{2}nR\left(T_2-T_1\right)=-P_2\left(V_2-V_1\right)$

$\frac{3}{2}nR\left(T_2-T_1\right)=-P_2\left(nR\frac{T_2}{P_2}-nR\frac{T_1}{P_1}\right)$

$\frac{3}{2}\left(T_2-T_1\right)=-T_2+\frac{P_2}{P_1}\Rightarrow T_2=\frac{2}{3}\left(\frac{P_2}{P_1}+\frac{3}{2}\right)T_1$

$T_2=0.4\left(2+1.5\right)300=420K$

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