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Question
Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in following figure. The temperature of the gas at A and B are 300 K and 500 K respectively. A total of 1200 J heat is withdrawn from the sample in this process. Find the work done by the gas in part BC. (R = 8.3 J/mol K)
Numerical
Solution
Q = ΔU + W
W = PΔV = nRΔT
Q = ΔU + W
Q = WABCD ...(As internal energy ΔU = 0 )
Q = WAB + WBC + WCA
As, the volume remains constant during path CA,
WCA = 0
-1200 = WAB + WBC + 0
WAB = nRΔT
= nR(TB - TA)
= 2.0 x 8.3(500 - 300)
WAB = 3320J
From equation (i)
3320 + WBC = -1200
WBC = -4520 J
Negative sign indicates that the work is done on the gas.
Work done by the gas in part BC = + 4520 J
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First Law of Thermodynamics
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