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Heat and Thermodynamics question

2023 · 1 Feb · Shift 1 · Q56
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Heat and Thermodynamics question

2023 · 1 Feb · Shift 1 · Q56

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A sample of gas at temperature TTT is adiabatically expanded to double its volume. The work done by the gas in the process is (given,γ=32)\left(\mathrm{given}, \gamma=\frac{3}{2}\right)(given,γ=23​) :
  1. A
    W=TR[2−2]W=T R[\sqrt{2}-2]W=TR[2​−2]
  2. B
    W=TR[2−2]W=\frac{T}{R}[\sqrt{2}-2]W=RT​[2​−2]
  3. C
    W=RT[2−2]W=\frac{R}{T}[2-\sqrt{2}]W=TR​[2−2​]
  4. D
    W=RT[2−2]W=R T[2-\sqrt{2}]W=RT[2−2​]
View written solutionFree

Correct answer: D

  1. Use the adiabatic relation

For an adiabatic process of an ideal gas, TVγ−1=constant.TV^{\gamma-1}=\text{constant}.TVγ−1=constant.

Given:

  • initial temperature =T=T=T
  • final volume V2=2V1V_2=2V_1V2​=2V1​
  • γ=32\gamma=\dfrac{3}{2}γ=23​

So, T1V1γ−1=T2V2γ−1T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}T1​V1γ−1​=T2​V2γ−1​ TV11/2=T2(2V1)1/2TV_1^{1/2}=T_2(2V_1)^{1/2}TV11/2​=T2​(2V1​)1/2 T=T22T=T_2\sqrt{2}T=T2​2​ T2=T2.T_2=\frac{T}{\sqrt{2}}.T2​=2​T​.

  1. Apply the first law for adiabatic expansion

For an adiabatic process, Q=0.Q=0.Q=0.

Hence, ΔU=Q−W=−W\Delta U=Q-W=-WΔU=Q−W=−W so W=−ΔU.W=-\Delta U.W=−ΔU.

For 1 mole of ideal gas, ΔU=CV(T2−T1).\Delta U=C_V(T_2-T_1).ΔU=CV​(T2​−T1​). Thus, W=CV(T1−T2).W=C_V(T_1-T_2).W=CV​(T1​−T2​).

  1. Find CVC_VCV​ using γ\gammaγ

We know γ=CPCV,CP−CV=R.\gamma=\frac{C_P}{C_V}, \qquad C_P-C_V=R.γ=CV​CP​​,CP​−CV​=R.

Using γ=32,\gamma=\frac{3}{2},γ=23​, we get (32−1)CV=R\left(\frac{3}{2}-1\right)C_V=R(23​−1)CV​=R 12CV=R\frac{1}{2}C_V=R21​CV​=R CV=2R.C_V=2R.CV​=2R.

  1. Calculate the work done

Now, W=CV(T−T2)W=C_V\left(T-\frac{T}{\sqrt{2}}\right)W=CV​(T−2​T​) W=2R(T−T2)W=2R\left(T-\frac{T}{\sqrt{2}}\right)W=2R(T−2​T​) W=2RT(1−12).W=2RT\left(1-\frac{1}{\sqrt{2}}\right).W=2RT(1−2​1​).

Simplify: W=2RT⋅2−12W=2RT\cdot \frac{\sqrt{2}-1}{\sqrt{2}}W=2RT⋅2​2​−1​ W=RT(2−2).W=RT(2-\sqrt{2}).W=RT(2−2​).

  1. Match with the options

Thus, W=RT(2−2)\boxed{W=RT(2-\sqrt{2})}W=RT(2−2​)​ which corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored answer.

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