Question
Answer: Option A
:
A
Along AB heat was input. Along the line AB, the quantity pV increased, so NkT increased so ΔU=NkΔT was positive. ΔU=Q+W, and W = 0 since pΔV=0. So ΔU=Q is positive.
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:
A
Along AB heat was input. Along the line AB, the quantity pV increased, so NkT increased so ΔU=NkΔT was positive. ΔU=Q+W, and W = 0 since pΔV=0. So ΔU=Q is positive.
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