Question

Two moles of diatomic gas is taken through cyclic process ABCA as shown in figure. Then,

• A. heat absorbed in process AB is $2.5P_0{V}_0$.
• B. heat absorbed in process BC is $7P_0{V}_0$.
• C. heat rejected in process CA is $9P_0{V}_0$.
• D. heat rejected in process CA is $11P_0{V}_0$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A heat absorbed in process AB is $2.5P_0{V}_0$.
B heat absorbed in process BC is $7P_0{V}_0$.
C heat rejected in process CA is $9P_0{V}_0$.

$\Delta Q_{AB}=n{C}_v\Delta T=\frac{5}{2}nR(2T_0-T_0)=\frac{5P_0{V}_0}{2}\Delta Q_{BC}=n{C}_p\Delta T=\frac{7}{2}nR(4T_0-2T_0)=7P_0{V}_0\Delta Q_{CA}=\Delta W_{CA}+\Delta U_{CA}=-\frac{1}{2}[P_0+2P_0]V_0-\frac{5}{2}nR(T_c-T_a)=-\frac{1}{2}[P_0+2P_0]V_0-\frac{5}{2}nR(\frac{4P_0{V}_0}{nR}-\frac{P_0{V}_0}{nR})=-9P_0{V}_0$