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 \[dW=0\] \[\therefore \] \[dQ=dU\]You need to login to perform this action.
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