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JEE Main 2021 (25 Feb, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2021 (25 Feb, Shift 1)

Q177JEE Main 2021 · 25 FEB · SHIFT 1Easy

Thermodynamic Processes

A diatomic gas, having Cp=72RC_p = \frac{7}{2}R and Cv=52RC_v = \frac{5}{2}R, is heated at constant pressure. The ratio dU: dQ: dW:

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Answer: (c)

  1. At constant pressure: dU=nCvdTdU = nC_v dT, dQ=nCpdTdQ = nC_p dT, dW=nR dTdW = nR\,dT.
  2. dU:dQ:dW=52R:72R:R=5:7:2dU : dQ : dW = \frac{5}{2}R : \frac{7}{2}R : R = 5 : 7 : 2
  3. So the answer is (c) 5 : 7 : 2.
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