A system goes from state a to state c along paths adc and abc on the P–V diagram. Initially Uₐ = 10 J. Along adc, the heat absorbed is Q₁ = 50 J and work obtained is W₁ = 20 J. Along abc, the heat absorbed is Q₂ = 36 J. The work obtained along abc is:
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Question

A system goes from state a to state c along paths adc and abc on the P–V diagram. Initially Uₐ = 10 J. Along adc, the heat absorbed is Q₁ = 50 J and work obtained is W₁ = 20 J. Along abc, the heat absorbed is Q₂ = 36 J. The work obtained along abc is:

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G1 Physics · Accepted Answer

Step 1: The first law is ΔU = Q − W, where W is work done BY the gas. Step 2: Along path a→d→c: Q = 50 J and W = 20 J. Step 3: Thus change of internal energy from a to c: ΔU = 50 − 20 = +30 J. Step 4: Internal energy is a state function; therefore change from a to c is also +30 J along path a→b→c. Step 5: Along path a→b→c, Q = 36 J. Thus 30 = 36 − W. Step 6: Required work: W = 36 − 30 = 6 J. Step 7: Starting at U = 10 J in state a, final U in state c is 10 + 30 = 40 J (consistency check). Final…