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An ideal gas undergoes a cyclic process consisting of one isochoric, one isothermal, and one adiabatic process. During the isothermal process, 40 J of heat is released by the gas, and during the isochoric process, 80 J of heat is absorbed by the gas. If W1 is the work done by the gas during the adiabatic process and W2 is the work done during the isothermal process, find W1/W2.

1. -2
2. 2
3. 1
4. -1/2

Correct answer: -2

Solution

For the full cycle: delta U_total = 0. Sum of Q = sum of W. Total Q = 80 - 40 = 40 J. Total W = W1 + W2 + 0 = 40 J. W2 = -40 J (isothermal release means work done BY gas = -40 J). So W1 = 40 - (-40) = 80 J. W1/W2 = 80/(-40) = -2.

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