In a cyclic process shown in the figure an ideal gas is adiabatically taken from B and A, the work done on the gas during the process B → A is 30 J, when the gas is taken from A → B the heat absorbed by the gas is 20 J. The change in internal energy of the gas in the process A → B is:
Text Solution
Verified by ExpertsThe correct answer is:
B
B → A
Δ Q = 0
0 = – 30 + Δ U AB
Δ U BA = 30 J
∴ Δ U AB = – Δ U BA = – 30 J
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