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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2021 · 25 February · Evening Shift · Physics · Thermodynamics

Thermodynamic process is shown below on a P - V diagram for one...

Thermodynamic process is shown below on a P−VP - V diagram for one mole of an ideal gas. If V2=2V1{V}_{2} = 2{V}_{1}, then the ratio of temperature T2T1\dfrac{{T}_{2}}{{T}_{1}} is :

Thermodynamic process is shown below on a P - V diagram for one mole of an ideal gas. If V _ 2 = 2 V _ 1 , then the

Options

  1. A2\sqrt{2}
  2. B12\dfrac{1}{\sqrt{2}}
  3. C12\dfrac{1}{2}
  4. D22

Correct answer

A. 2\sqrt{2}

Step-by-step solution

Thermodynamic process is shown below on a P - V diagram for one mole of an ideal gas. If V _ 2 = 2 V _ 1 , then the

PV1/2=cP{V}^{1 / 2} = c

nRTVV1/2=c\dfrac{nRT}{V}{V}^{1 / 2} = c

T=c1V1/2T = {c}^{1}{V}^{1 / 2}

T2T1=(V2V1)1/2=(2V1V1)1/2\dfrac{{T}_{2}}{{T}_{1}} = {\left(\dfrac{{V}_{2}}{{V}_{1}}\right)}^{1 / 2} = {\left(\dfrac{2{V}_{1}}{{V}_{1}}\right)}^{1 / 2}

T2T1=2\dfrac{{T}_{2}}{{T}_{1}} = \sqrt{2}

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