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JEE Main 2021 (25 Feb, Shift 2) · P-V Diagrams & Cyclic Processes

P-V Diagrams & Cyclic Processes question from JEE Main 2021 (25 Feb, Shift 2)

Q49JEE Main 2021 · 25 FEB · SHIFT 2Medium

P-V Diagrams & Cyclic Processes

Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is:

Figure

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Answer: (a)

  1. PV1/2=PV^{1/2} = constant and P=nRT/VP = nRT/V give TV−1/2=TV^{-1/2} = constant, i.e. T∝VT \propto \sqrt{V}.
  2. T2T1=V2V1=2\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}} = \sqrt{2}
  3. So the answer is (a) 2\sqrt{2}.
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