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Heat and Thermodynamics question

2022 · 28 Jun · Shift 2 · Q57
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Heat and Thermodynamics question

2022 · 28 Jun · Shift 2 · Q57

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The workdone by the gas during the part CA is : JEE Main 2022 (Online) 28th June Evening Shift Physics - Heat and Thermodynamics Question 200 English
  1. A
    20 J
  2. B
    30 J
  3. C
    −-− 30 J
  4. D
    −-− 60 J
View written solutionFree

Correct answer: B

  1. Use the first law of thermodynamics

For any process,

ΔU=Q−W\Delta U = Q - WΔU=Q−W

where:

  • QQQ = heat absorbed by the gas
  • WWW = work done by the gas
  1. Given data for each part

  • During ABABAB: heat absorbed =40 J= 40\,\text{J}=40J QAB=+40 JQ_{AB}=+40\,\text{J}QAB​=+40J

  • During BCBCBC: no heat exchange QBC=0Q_{BC}=0QBC​=0

  • During CACACA: gas rejects 60 J60\,\text{J}60J QCA=−60 JQ_{CA}=-60\,\text{J}QCA​=−60J

  • Work of 50 J50\,\text{J}50J is done on the gas during BCBCBC

So work done by the gas during BCBCBC is

WBC=−50 JW_{BC}=-50\,\text{J}WBC​=−50J
  1. Apply cyclic process condition

Since the gas undergoes a cycle ABCAABCAABCA, it returns to the initial state. Therefore,

ΔUcycle=0\Delta U_{\text{cycle}}=0ΔUcycle​=0

Hence over the complete cycle,

Qtotal=WtotalQ_{\text{total}} = W_{\text{total}}Qtotal​=Wtotal​

Calculate total heat absorbed:

Qtotal=QAB+QBC+QCA=40+0−60=−20 JQ_{\text{total}} = Q_{AB}+Q_{BC}+Q_{CA} = 40+0-60 = -20\,\text{J}Qtotal​=QAB​+QBC​+QCA​=40+0−60=−20J

Therefore,

Wtotal=−20 JW_{\text{total}}=-20\,\text{J}Wtotal​=−20J
  1. Relate total work to work in different parts
Wtotal=WAB+WBC+WCAW_{\text{total}} = W_{AB}+W_{BC}+W_{CA}Wtotal​=WAB​+WBC​+WCA​

So,

WAB−50+WCA=−20W_{AB}-50+W_{CA}=-20WAB​−50+WCA​=−20 WAB+WCA=30W_{AB}+W_{CA}=30WAB​+WCA​=30

To find WCAW_{CA}WCA​, we use the figure information: segment ABABAB is vertical on the PVPVPV diagram, so it is an isochoric process.

For an isochoric process,

WAB=0W_{AB}=0WAB​=0

Thus,

WCA=30 JW_{CA}=30\,\text{J}WCA​=30J
  1. Answer

The work done by the gas during CACACA is

30 J\boxed{30\,\text{J}}30J​

So the correct option is B.

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