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JEE Main 2026 (5 Apr, Shift 1) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from JEE Main 2026 (5 Apr, Shift 1)

Q2JEE Main 2026 · 5 APR · SHIFT 1Easy

Heat, Work & First Law of Thermodynamics

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R\mathbf{R}

Statement I: Change in internal energy of a system containing nn mole of ideal gas can be written as ΔU=nCv(Tf−Ti)=nRγ−1(Tf−Ti)\Delta U = nC_v\left(T_f - T_i\right) = \frac{nR}{\gamma-1}\left(T_f - T_i\right), where γ=CpCv,Ti=\gamma = \frac{C_p}{C_v}, T_i = initial temperature, Tf=T_f = final temperature. Statement II: Relation between degree of freedom ff and γ (=Cp/Cv)\gamma\,(= C_p/C_v) is (γ=1+2f)\left(\gamma = 1 + \frac{2}{f}\right) Choose the correct answer from the options given below

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Answer: (b)

  1. Statement I: ΔU=nCvΔT\Delta U = nC_v\Delta T for any process of an ideal gas, and Cv=Rγ−1C_v = \frac{R}{\gamma - 1} (from Cp−Cv=RC_p - C_v = R). So I is true.
  2. Statement II: Cv=f2RC_v = \frac{f}{2}R, Cp=(f2+1)R⇒γ=1+2fC_p = \left(\frac{f}{2}+1\right)R \Rightarrow \gamma = 1 + \frac{2}{f}. So II is true.
  3. Statement I follows from Cp−Cv=RC_p - C_v = R, not from the degrees-of-freedom relation, so R does not explain A.
  4. So the answer is (b).
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