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JEE Main 2018 (15 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2018 (15 Apr, Shift 1)

Q220JEE Main 2018 · 15 APR · SHIFT 1Easy

Thermodynamic Processes

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27∘C.27^\circ\text{C}. The work done on the gas will be:

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Answer: (c)

  1. Isothermal: Won=nRTln⁡P2P1W_{on} = nRT\ln\frac{P_2}{P_1}.
  2. n=1n = 1, T=300T = 300 K, P2P1=2\frac{P_2}{P_1} = 2
  3. Won=300Rln⁡2W_{on} = 300R\ln 2
  4. So the answer is (c) 300Rln⁡2300R\ln 2.
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