question_answer

A diatomic ideal gas undergoes a thermodynamic change according to the \[P\]-\[V\] diagram shown in figure. The total heat given to the gas is nearly

A) \[2.5{{P}_{0}}{{V}_{0}}\]

B) \[1.4{{P}_{0}}{{V}_{0}}\]

C) \[3.9{{P}_{0}}{{V}_{0}}\]     

D) \[1.1{{P}_{0}}{{V}_{0}}\]

Correct Answer: C

Solution :

[c] \[{{Q}_{AB}}=\Delta {{U}_{AB}}+{{W}_{AB}}\] \[{{W}_{AB}}\]=0 \[\Delta {{U}_{AB}}=\frac{f}{2}nR\Delta T\] \[\frac{f}{2}(\Delta PV)\Delta {{U}_{AB}}=\frac{5}{2}(\Delta PV)\] \[{{Q}_{AB}}=2.5{{P}_{0}}{{V}_{0}}\] Process BC, \[{{Q}_{BC}}=\Delta {{U}_{BC}}+{{W}_{BC}}=0+2{{P}_{0}}{{V}_{0}}\log 2=1.4{{P}_{0}}{{V}_{0}}\] \[{{Q}_{net}}={{Q}_{AB}}+{{Q}_{BC}}=3.9{{P}_{0}}{{V}_{0}}\]