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JEE Main 2022 (24 Jun, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (24 Jun, Shift 2)

Q171JEE Main 2022 · 24 JUN · SHIFT 2Medium

Thermodynamic Processes

A monoatomic gas performs a work of Q4\frac{Q}{4} where Q is the heat supplied to it. The molar heat capacity of the gas will be ______________ R during this transformation. Where R is the gas constant.

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

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Answer: 2

  1. W=Q4⇒ΔU=Q−W=3Q4W = \frac{Q}{4} \Rightarrow \Delta U = Q - W = \frac{3Q}{4}, so Q=43ΔUQ = \frac{4}{3}\Delta U.
  2. C=QnΔT=43⋅ΔUnΔT=43CvC = \frac{Q}{n\Delta T} = \frac{4}{3} \cdot \frac{\Delta U}{n\Delta T} = \frac{4}{3}C_v
  3. Monatomic: Cv=32RC_v = \frac{3}{2}R, so C=43×32R=2RC = \frac{4}{3} \times \frac{3}{2}R = 2R.
  4. Answer: 2
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