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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2019 · 9 April · Morning Shift · Physics · Thermodynamics

Following figure shows two processes A and B for a gas. If ∆ Q _ A...

Following figure shows two processes AA and BB for a gas. If ∆QA∆ {Q}_{A} and ∆QB∆ {Q}_{B} are the amount of heat absorbed by the system in two cases, and ∆UA∆ {U}_{A} and ∆UB∆ {U}_{B} are changes in internal energies, respectively, then:
Following figure shows two processes A and B for a gas. If ∆ Q _ A and ∆ Q _ B are the amount of heat absorbed by the

Options

  1. A∆QA<∆QB;∆UA<∆UB∆ {Q}_{A} < ∆ {Q}_{B}; ∆ {U}_{A} < ∆ {U}_{B}
  2. B∆QA=∆QB;∆UA=∆UB∆ {Q}_{A} = ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}
  3. C∆QA>∆QB;∆UA=∆UB∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}
  4. D∆QA>∆QB;∆UA>∆UB∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} > ∆ {U}_{B}

Correct answer

C. ∆QA>∆QB;∆UA=∆UB∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}

Step-by-step solution

⇒∆UA=∆UB⇒ ∆ {U}_{A} = ∆ {U}_{B} as change in internal energy depends on initial and final state and not the path.
⇒⇒ By comparing area under the curve,
WA>WB{W}_{A} > {W}_{B}
From Q=U+WQ = U + W
We can say, QPA>QPB{Q}_{{P}_{A}} > {Q}_{{P}_{B}}
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