Question

Six moles of an ideal gas performs a cycle shown in the figure. If the temperature is $T_{A}=600 \, K$ , $T_{B}=800 \, K$ , $T_{C}=2200 \, K$ and $T_{D}=1200 \, K$ , the work done per cycle is

• A. $20 \, kJ$
• B. $30 \, kJ$
• C. $40 \, kJ$
• D. $60 \, kJ$

Answer 1

Answer: (C)

$A \rightarrow B$ and $C \rightarrow D$
$V=$ constant $W_{A B}=W_{C D}=0$
$W_{B C}=P\Delta V$
$W_{B C}=nR\left(\Delta T\right)_{B C}$
$W_{D A}=nR\left(\Delta T\right)_{D A}$
Total work done in a complete cycle
$W_{t o t a l}=W_{A B}+W_{B C}+W_{C D}+W_{D A}$
$W_{t o t a l}=0+W_{B C}+0+W_{D A}$
$W_{t o t a l}=0+nR\left(\Delta T\right)_{B C}+0+nR\left(\Delta T\right)_{D A}$
$W_{t o t a l}=nR\left(T_{C} - T_{B} + T_{A} - T_{D}\right)$
$W_{t o t a l}=40 \, kJ$