A) \[dQ=dU+dW\]
B) \[dQ=dU-dW\]
C) \[dQ=dU\]
D) \[dQ=-dU\]
Correct Answer: C
Solution :
Key Idea: In isochoric process, volume is constant. Work done \[=P\Delta V,\] where P is pressure, \[\Delta V\]the change in volume. For an isochoric process \[\Delta V=0,\] Therefore\[dW=PdV=0\] From first law of thermodynamics \[dQ=dU+dW\] where \[dW=0\] \[\therefore \] \[dQ=dU\]
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