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Heat and Thermodynamics question

2019 · 12 Apr · Shift 1 · Q51
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Heat and Thermodynamics question

2019 · 12 Apr · Shift 1 · Q51

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is: JEE Main 2019 (Online) 12th April Morning Slot Physics - Heat and Thermodynamics Question 328 English
  1. A
    120 J
  2. B
    130 J
  3. C
    100 J
  4. D
    140 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:

  • ΔU\Delta UΔU = change in internal energy
  • QQQ = heat absorbed by the gas
  • WWW = work done by the gas
  1. Use the fact that the process is cyclic

The gas goes through cycle a→b→c→aa \to b \to c \to aa→b→c→a. For a complete cycle,

ΔUabca=0\Delta U_{abca} = 0ΔUabca​=0

So,

ΔUabc+ΔUca=0\Delta U_{abc} + \Delta U_{ca} = 0ΔUabc​+ΔUca​=0

Given:

ΔUca=−180 J\Delta U_{ca} = -180\text{ J}ΔUca​=−180 J

Hence,

ΔUabc=180 J\Delta U_{abc} = 180\text{ J}ΔUabc​=180 J
  1. Find total heat absorbed along path abcabcabc

Given:

  • Along ababab, heat absorbed = 250 J250\text{ J}250 J
  • Along bcbcbc, heat absorbed = 60 J60\text{ J}60 J

Therefore,

Qabc=Qab+Qbc=250+60=310 JQ_{abc} = Q_{ab} + Q_{bc} = 250 + 60 = 310\text{ J}Qabc​=Qab​+Qbc​=250+60=310 J
  1. Apply first law on path abcabcabc

For path abcabcabc,

ΔUabc=Qabc−Wabc\Delta U_{abc} = Q_{abc} - W_{abc}ΔUabc​=Qabc​−Wabc​

Substitute values:

180=310−Wabc180 = 310 - W_{abc}180=310−Wabc​

So,

Wabc=310−180=130 JW_{abc} = 310 - 180 = 130\text{ J}Wabc​=310−180=130 J
  1. Match with options
Wabc=130 JW_{abc} = 130\text{ J}Wabc​=130 J

So the correct option is:

B: 130 J

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