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Q.

Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in fig. The temperature of the gas at (A) and (B) are 300K and 500K respectively. A total of 1200 J heat is withdrawn from the sample in the process. Find the work done by the gas in part BC. Take R = 8.3 J/mol -K

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a

–3000 J

b

+3000 J

c

– 4520 J

d

+4500 J

answer is C.

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Detailed Solution

$Q_{n e t} = W_{A B} + W_{B C} + W_{C A}$
$- 1200 = (2 X 8.3 X 200) + W_{B C} + 0$
$W_{B C} = - 4520 J$

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