Verified Solution • 2025–26 Session moderate

QJEEM19Ph60be: JEE Main 2019 Physics - Class 11 JEEM19Ph60be Solution: Following figure shows two processes A and B for a gas. If Q A…

Following figure shows two processes A and B for a gas. If Δ\Delta QA and Δ\Delta QB are the amount of heat absorbed by the system in two cases, and Δ\Delta UA and Δ\Delta UB are changes in internal energies, respectively, then : JEE Main 2019 (Online) 9th April Morning Slot Physics - Heat and Thermodynamics Question 314 English

Multiple Choice Options

  • A Δ\Delta QA > Δ\Delta QB ; Δ\Delta UA > Δ\Delta UB
  • B Δ\Delta QA < Δ\Delta QB ; Δ\Delta UA < Δ\Delta UB
  • C Δ\Delta QA > Δ\Delta QB ; Δ\Delta UA = Δ\Delta UB
  • D Δ\Delta QA = Δ\Delta QB ; Δ\Delta UA = Δ\Delta UB

Step-by-Step Verified Solution

Initial and final states for both the processes are same,
∴\therefore Δ\Delta UA = Δ\Delta UB

Work done during process A is greater than in process B. Because area is more By First law of thermodynamics
Δ\Delta Q = Δ\Delta U + W
⇒\Rightarrow Δ\Delta QA > Δ\Delta QB

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