Question

$5$ moles of an ideal gas $(C_v=\frac{5}{2}R)$ was compressed adiabatically against a constant pressure of $5atm$, which was initially at $250K$ and $1atm$ pressure. The work done in the process is equal to:

• A. 1450 R
• B. 3562 R
• C. 2500 R
• D. 5000 R

Answer 1

The correct option is B 3562 R

For an irreversible adiabatic process, the work done is given as:
$w=n{C}_v(T_2-T_1)=-p_{ext}\times nR[\frac{T_2}{P_2}-\frac{T_1}{P_1}]$
Substituting the values,
$\Rightarrow 5\times \frac{5}{2}R(T_2-250)=-5\times 5R[\frac{T_2}{5}-\frac{250}{1}]$
$\therefore T_2=535K$

Substituting the value of $T_2$ in the work done equation,
$\therefore w=n{C}_v(T_2-T_1)=5\times \frac{5}{2}R\times (535-250)=3562R$