Question

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

• A. 250 R
• B. 300 R
• C. 380 R
• D. 500 R

Answer 1

The correct option is C 380 R

For irreversible adiabatic process,

$W=-P_{\text{ext}}[\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1}]$

$P_2=P_{\text{ext}}=2\text{atm}$

$P_1=1\text{atm}{T}_1=300K$

$W=-\left(2\text{atm}\right)[\frac{2\left(R\right)T_2}{2\text{atm}}-\frac{2R\left(350\right)}{1\text{atm}}]$

$\text{and}W=2C_v(T_2-350)=2\times \frac{5}{2}R(T_2-350)$

$5R(T_2-350)=(2\times 750R-2R{T}_2)$

$5T_2-1750=1400-2T_2$

$7T_2=3150T_2=450K$

$W=2\times C_v(450-350)$

$=2\frac{5}{2}R\times \left(100\right)=500R$