Following figure shows two processes A and B for a gas. If Δ QA and Δ QB are the amount of heat absorbed by the system in two cases, and Δ UA and Δ UB are changes in internal energies, respectively, then: (a) Δ QA > Δ QB; Δ UA > Δ UB (b) Δ QA Δ QB; Δ UA = Δ UB (d) Δ QA = Δ QB; Δ UA = Δ UB

Answer: (c). Both processes go from i to f, so Δ UA = Δ UB. Curve A lies above B, so area under A is larger: WA > WB. Q = Δ U + W ⇒ Δ QA > Δ QB So the answer is (c).

Following figure shows two processes A and B for a gas. If Δ QA and Δ… | JEE Main 2019 (9 Apr, Shift 1)

Q60JEE Main 2019 · 9 APR · SHIFT 1Medium

P-V Diagrams & Cyclic Processes

Following figure shows two processes A and B for a gas. If $\Delta$ QA and $\Delta$ QB are the amount of heat absorbed by the system in two cases, and $\Delta$ UA and $\Delta$ UB are changes in internal energies, respectively, then:

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Answer: (c)

Both processes go from $i$ to $f$ , so $\Delta U_A = \Delta U_B$ .
Curve A lies above B, so area under A is larger: $W_A > W_B$ .
$Q = \Delta U + W \Rightarrow \Delta Q_A > \Delta Q_B$
So the answer is (c).

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