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JEE Main 2020 (2 Sep, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2020 (2 Sep, Shift 1)

Q207JEE Main 2020 · 2 SEP · SHIFT 1Medium

Thermodynamic Processes

An engine takes in 5 moles of air at 20oC and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is________.

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

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Answer: 46

  1. Rigid diatomic: γ=1.4\gamma = 1.4, Cv=52RC_v = \frac{5}{2}R. TVγ−1=TV^{\gamma-1} = const.
  2. T2=293×100.4=293×2.512≈736T_2 = 293 \times 10^{0.4} = 293 \times 2.512 \approx 736 K, so ΔT≈443\Delta T \approx 443 K.
  3. ΔU=nCvΔT=5×2.5×8.314×443≈46×103\Delta U = nC_v\Delta T = 5 \times 2.5 \times 8.314 \times 443 \approx 46\times10^3 J
  4. X=46X = 46 kJ
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