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Question:

The figure shows two processes A and B for a gas. If ΔQA and ΔQB are the amount of heat absorbed by the system in two cases, and ΔUA and ΔUB are changes in internal energies, respectively, then:

ΔQA=ΔQB; ΔUA=ΔUB

ΔQA>ΔQB; ΔUA=ΔUB

ΔQA>ΔQB; ΔUA>ΔUB

ΔQA<ΔQB; ΔUA<ΔUB

Solution:

Correct option is B. ΔQA>ΔQB; ΔUA=ΔUB
Initial and final states for both the processes are the same
∴ ΔUA=ΔUB
Work done during process A is greater than process B
By first law of thermodynamics
ΔQ=ΔU+ΔW ⇒ ΔQA>ΔQB