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MHT CET 2026 (15 May, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from MHT CET 2026 (15 May, Shift 2)

Q15MHT CET 2026 · 15 MAY · SHIFT 2Medium

Thermodynamic Processes

Two moles of a rigid diatomic gas does a work Q4\frac{Q}{4} when the amount of heat supplied is QQ. In this process the total heat capacity of the gas is (RR = gas constant)

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Answer: (d)

  1. First law: ΔU=Q−Q4=3Q4\Delta U = Q - \dfrac{Q}{4} = \dfrac{3Q}{4}.
  2. ΔU=nCvΔT=2×52R ΔT=5R ΔT⇒ΔT=3Q20R\Delta U = nC_v\Delta T = 2 \times \dfrac{5}{2}R\,\Delta T = 5R\,\Delta T \Rightarrow \Delta T = \dfrac{3Q}{20R}
  3. Total heat capacity =QΔT=20R3= \dfrac{Q}{\Delta T} = \dfrac{20R}{3}
  4. So the answer is (d) 20R3\dfrac{20R}{3}.
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