A) 400 R In 2
B) 200 R In 2
C) 100 R In 2
D) 300 R In 2
Correct Answer: A
Solution :
[a] Change in internal energy for cyclic process \[\left( \Delta U \right)=0.\] For process \[a\to b\], (P-constant) \[{{W}_{a\to b}}=P\Delta V\] \[=nR\Delta T=-400R\] For process \[b\to c\](T-constant) \[{{W}_{b\to c}}=-2R(300)\,\,\text{ln 2}\] For process \[c\to d\], (P-constant) \[{{W}_{c\to d}}=+\,400R\] For process \[d\to a\], (T-constant) \[{{W}_{d\to a}}=+2R(500)\text{ ln 2 }\] Net work (AW) \[={{W}_{a\to b}}+{{W}_{b\to c}}+{{W}_{c\to d}}+{{W}_{d\to a}}\] \[\Delta W=400R\text{ ln 2}\] \[\therefore dQ=dU+dW\], first law of thermodynamics \[\therefore dQ=400R\text{ ln }2.\]
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