Question

Two moles of an ideal gas $(C_v=\frac{5}{2}R)$ was compressed adiabatically against constant pressure of $4atm$. Which was initially at $350K$ and $1atm$ pressure. The work involve in the process is equal to

• A. $1250R$
• B. $1400R$
• C. $1500R$
• D. $1350R$

Answer 1

The correct option is C $1500R$

$\omega =n{C}_v(T_2-T_1)=-p_{ext}\times nR[\frac{T_2}{P_2}-\frac{T_1}{P_1}]$
$\Rightarrow 2\times \frac{5}{2}R(T_2-350)=-4\times 2R[\frac{T_2}{4}-\frac{350}{1}]$
$\therefore T_2=650K$
putting value of $T_2$, we get
$\therefore \omega =n{C}_v(T_2-T_1)=2\times \frac{5}{2}R\times 300=1500R$