A) \[2.5{{P}_{0}}{{V}_{0}}\]
B) \[1.4{{P}_{0}}{{V}_{0}}\]
C) \[1.1{{P}_{0}}{{V}_{0}}\]
D) \[3.9{{P}_{0}}{{V}_{0}}\]
Correct Answer: D
Solution :
[d] \[{{Q}_{AB}}=\Delta {{U}_{AB}}+{{W}_{AB}}\] \[{{W}_{AB}}=0\] \[\Delta {{U}_{AB}}=\frac{f}{2}nR\Delta T\Rightarrow \frac{f}{2}\left( \Delta PV \right)\] \[\Delta {{U}_{AB}}=\frac{5}{2}\left( \Delta PV \right)\Rightarrow {{Q}_{AB}}=2.5{{P}_{0}}{{V}_{0}}\] Process BC: \[{{Q}_{AB}}=\Delta {{U}_{BC}}+{{W}_{BC}}=2{{P}_{0}}{{V}_{0}}\text{ ln 2=1}\text{.4 }{{P}_{0}}{{V}_{0}}\] \[{{Q}_{net}}={{Q}_{AB}}+{{Q}_{BC}}=3.9\text{ }{{P}_{0}}{{V}_{0}}\]
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