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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2023 · 1 February · Morning Shift · Physics · Thermodynamics

A sample of gas at temperature T is adiabatically expanded to...

A sample of gas at temperature TT is adiabatically expanded to double its volume. The work done by the gas in the process is given, (given γ=32\gamma = \dfrac{3}{2}) :

Options

  1. AW=TR[2−2]W = TR\left[\sqrt{2} - 2\right]
  2. BW=TR[2−2]W = \dfrac{T}{R}\left[\sqrt{2} - 2\right]
  3. CW=RT[2−2]W = \dfrac{R}{T}\left[2 - \sqrt{2}\right]
  4. DW=RT[2−2]W = RT\left[2 - \sqrt{2}\right]

Correct answer

D. W=RT[2−2]W = RT\left[2 - \sqrt{2}\right]

Step-by-step solution

Work done by gas in adiabatic process is:

W=−(P2V2−P1V1)γ−1W = - \dfrac{\left({P}_{2}{V}_{2} - {P}_{1}{V}_{1}\right)}{\gamma - 1}

=−nR(T2−T1)γ−1= - \dfrac{nR\left({T}_{2} - {T}_{1}\right)}{\gamma - 1}

For an adiabatic process, T1V1γ−1=T2V2γ−1{T}_{1}{{V}_{1}}^{\gamma - 1} = {T}_{2}{{V}_{2}}^{\gamma - 1}

⇒TV32−1=T2(2V)32−1⇒TV12=T2(2V)12⇒T2=T2⇒ T{V}^{\dfrac{3}{2} - 1} = {T}_{2}{\left(2V\right)}^{\dfrac{3}{2} - 1} ⇒ TV\dfrac{1}{2} = {T}_{2}\left(2V\right)\dfrac{1}{2} ⇒ {T}_{2} = \dfrac{T}{\sqrt{2}}

Now, work done is W=R(T1−T2)γ−1=R(T−T2)12=RT(2−2)W = \dfrac{R\left({T}_{1} - {T}_{2}\right)}{\gamma - 1} = \dfrac{R\left(T - \dfrac{T}{\sqrt{2}}\right)}{\dfrac{1}{2}} = RT\left(2 - \sqrt{2}\right)

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