Thermodynamics

JEE Main level, single correct, medium

115 of 190

Question 115 of 190

A sample of an ideal gas is taken through the cyclic process ABCAABCA as shown in figure. It absorbs, 40J40J of heat during the part ABAB, no heat during BCBC and rejects 60J60J of heat during CACA. A work of 50J50J is done on the gas during the part BCBC. The internal energy of the gas at AA is 1560J1560J. The work done by the gas during the part CACA is

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Answer: (B) 30J30J

In the part A→BA\rightarrow B, the volume remains constant. Thus, the work done by the gas is zero. It is given that the heat absorbed by the gas is 40J40J. From first law of thermodynamics, Q=ΔU+WQ=\Delta U+W
⇒40=ΔU+0⇒ΔU=40J\Rightarrow 40=\Delta U+0\Rightarrow \Delta U=40J
or
ΔU=UB−UA=40J\Delta U={U}_{B}-{U}_{A}=40J
The increase in internal energy from
AA to BB is 40J40J. The internal energy is 1560J1560J at AA, then
UB−1560=40J⇒UB=1600J{U}_{B}-1560=40J\Rightarrow {U}_{B}=1600J
In the part
B→CB\rightarrow C, the work done by the gas is W=50JW=50J and no heat is given to the system. Again using Q=ΔU+WQ=\Delta U+W
0=ΔU−500=\Delta U-50 or ΔU=UC−UB=50J\Delta U={U}_{C}-{U}_{B}=50J
UC−1600=50J⇒UC=1650J{U}_{C}-1600=50J\Rightarrow {U}_{C}=1650J
The change in internal energy from
C→AC\rightarrow A,
∆U=1560−1650=−90J∆U=1560-1650=-90J
The heat given to the system is
Q=−60JQ=-60J
Using
Q=ΔU+WQ=\Delta U+W, we get W=Q−∆U=−60−(−90)=30JW=Q-∆U=-60-\left(-90\right)=30J

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