Question

The work done 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 K is:

[R = 2 cal/mol-K]

• A. 360 cal
• B. 720 cal
• C. 800 cal
• D. 1000 cal

Answer 1

The correct option is B 720 cal

To calculate the work done, We need to know the final temperature.

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

$w=\Delta U=n{C}_v(T_2-T_1)=-P_2(V_2-V_1)$

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

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

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

$\frac{3}{2}(T_2-T_1)=-T_2+\frac{P_2}{P_1}\Rightarrow T_2=\frac{2}{5}(\frac{P_2}{P_1}+\frac{3}{2})T_1$

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

Work done, $W=n{C}_v\Delta T=2\times \frac{3}{2}R\times (420-300)$

$\Rightarrow W=2\times \frac{3}{2}\times 2\times 120=720cal$

Hence, the correct option is $B$