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JEE Main 2022 (30 Jun, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (30 Jun, Shift 1)

Q167JEE Main 2022 · 30 JUN · SHIFT 1Medium

Thermodynamic Processes

A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm3 to 150 cm3 adiabatically. In this process, the value of workdone on the gas will be:

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Answer: (b)

  1. Monoatomic: γ=53\gamma = \frac{5}{3}. P2=P1(V1V2)γ=75×85/3=75×32=2400P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 75 \times 8^{5/3} = 75 \times 32 = 2400 kPa.
  2. Work done on gas W=P2V2−P1V1γ−1W = \frac{P_2V_2 - P_1V_1}{\gamma - 1}.
  3. P2V2=2400×103×150×10−6=360P_2V_2 = 2400\times10^3 \times 150\times10^{-6} = 360 J, P1V1=75×103×1200×10−6=90P_1V_1 = 75\times10^3\times1200\times10^{-6} = 90 J.
  4. W=360−902/3=405W = \frac{360 - 90}{2/3} = 405 J
  5. So the answer is (b) 405 J.
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