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JEE Main 2021 (26 Feb, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2021 (26 Feb, Shift 2)

Q196JEE Main 2021 · 26 FEB · SHIFT 2Medium

Thermodynamic Processes

1 mole of rigid diatomic gas performs a work of Q5\frac{Q}{5} when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is xR8\frac{xR}{8}. The value of x is _________. [R = universal gas constant]

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. W=Q5⇒ΔU=4Q5W = \frac{Q}{5} \Rightarrow \Delta U = \frac{4Q}{5}, so Q=54ΔUQ = \frac{5}{4}\Delta U.
  2. C=54CvC = \frac{5}{4}C_v, and rigid diatomic Cv=52RC_v = \frac{5}{2}R.
  3. C=54×52R=25R8C = \frac{5}{4} \times \frac{5}{2}R = \frac{25R}{8}
  4. So x=25x = 25.
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