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JEE Main 2022 (28 Jun, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (28 Jun, Shift 1)

Q162JEE Main 2022 · 28 JUN · SHIFT 1Medium

Thermodynamic Processes

Given below are two statements:

Statement I: When μ\mu amount of an ideal gas undergoes adiabatic change from state (P1, V1, T1) to state (P2, V2, T2), then work done is W=μR(T2−T1)1−γW = \frac{\mu R(T_2 - T_1)}{1-\gamma}, where γ=CpCv\gamma = \frac{C_p}{C_v} and R = universal gas constant. Statement II: In the above case, when work is done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

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Answer: (a)

  1. Adiabatic work by gas W=μR(T1−T2)γ−1=μR(T2−T1)1−γW = \dfrac{\mu R(T_1 - T_2)}{\gamma - 1} = \dfrac{\mu R(T_2 - T_1)}{1 - \gamma}, so Statement I is true.
  2. Work done on the gas with ΔQ=0\Delta Q = 0 goes into ΔU>0\Delta U > 0, so temperature rises; Statement II is true.
  3. So the answer is (a).
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