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JEE Main 2006 (AIEEE) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2006 (AIEEE)

Q244JEE Main 2006 · AIEEEMedium

Thermodynamic Processes

The work of 146 kJ146\text{ kJ} is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘C.7^\circ\text{C}. The gas is (R=8.3 J mol−1 K−1)\left(R = 8.3\text{ J mol}^{-1}\text{ K}^{-1}\right)

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Answer: (a)

  1. Adiabatic: Won=ΔU=nCvΔTW_{on} = \Delta U = nC_v\Delta T.
  2. 146×103=103×Cv×7⇒Cv=20.86146\times10^3 = 10^3 \times C_v \times 7 \Rightarrow C_v = 20.86 J mol−1^{-1} K−1^{-1}
  3. CvR=20.868.3≈2.5\frac{C_v}{R} = \frac{20.86}{8.3} \approx 2.5, so Cv=52RC_v = \frac{5}{2}R.
  4. So the answer is (a) diatomic.
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