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MHT CET 2026 (19 Apr, Shift 2) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from MHT CET 2026 (19 Apr, Shift 2)

Q4MHT CET 2026 · 19 APR · SHIFT 2Medium

Heat, Work & First Law of Thermodynamics

An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which dW=0dW = 0 and dQ<0dQ < 0. Then, for the gas

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Answer: (c)

  1. dW=0⇒dW = 0 \Rightarrow volume constant (isochoric).
  2. First law: dU=dQ−dW=dQ<0dU = dQ - dW = dQ < 0, so internal energy falls.
  3. For an ideal gas U∝TU \propto T, so TT decreases (and P∝TP \propto T decreases, not 'may increase').
  4. So the answer is (c) T will decrease.
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