MediumPhysics
+4 / -1
-- 16. In the cyclic process shown in the pressure-volume diagram, an ideal gas undergoes an adiabatic process from B to A. During this process, 50 J of work is done on the gas. When the gas is taken from A to B along the other path, it releases 10 J of heat. Work done by the gas in the process A to B is
A
60 J
B
-40 J
C
40 J
D
-60 J
Explanation
For B to A (Adiabatic): Q=0, W=-50. ΔU = Q - W = 50 J. For A to B: ΔU = -50 J. Q = -10 J. Q = ΔU + W => -10 = -50 + W => W = 40 J? Wait, if 50J work done *on* gas, ΔU = 50. Path A to B, ΔU = -50. Q = -10. W = Q - ΔU = -10 - (-50) = 40 J. Let's re-read: releases 10J heat (Q=-10). ΔU_AB = -50. W = Q - ΔU = -10 - (-50) = 40. However, if path B->A is W=-50, then ΔU = 50. A->B, ΔU = -50. Q = -10. W = Q - ΔU = 40. Re-check options: -60, -40, 40, 60. Process is cyclic? No, just two paths. Let's assume sign convention. Usually releases heat is negative. W done by gas is positive. W = 40J.
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