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Physics Question on Thermodynamics

A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32\gamma=\frac{3}{2} , then the work done by the gas in the process is:

A

RT[2−2]RT[2-\sqrt2]

B

RT[2−2]\frac{R}{T}[2-\sqrt2]

C

RT[2+2]RT[2+\sqrt2]

D

TR[2−2]\frac{T}{R}[2-\sqrt2]

Answer

RT[2−2]RT[2-\sqrt2]

Explanation

Solution

Step 1: Adiabatic condition The adiabatic condition is given by:

TVγ−1=constant.TV^{\gamma-1} = \text{constant.}

For the initial and final states:

T(V)32−1=Tf(2V)32−1.T(V)^{\frac{3}{2}-1} = T_f(2V)^{\frac{3}{2}-1}.

Simplify:

TV12=Tf(2V)12,TV^{\frac{1}{2}} = T_f(2V)^{\frac{1}{2}},

TVV=Tf2V.TV\sqrt{V} = T_f\sqrt{2V}.

Cancel V\sqrt{V}:

T=Tf2.T = T_f\sqrt{2}.

Solve for TfT_f:

Tf=T2.T_f = \frac{T}{\sqrt{2}}.

Step 2: Work done in adiabatic expansion The work done in an adiabatic process is given by:

W.D.=nR1−γ[Tf−T].W.D. = \frac{nR}{1-\gamma}\left[T_f - T\right].

Substitute Tf=T2,γ=32,T_f = \frac{T}{\sqrt{2}}, \gamma = \frac{3}{2}, and n=1n = 1:

W.D.=R1−32[T2−T].W.D. = \frac{R}{1-\frac{3}{2}}\left[\frac{T}{\sqrt{2}} - T\right].

Simplify:

W.D.=R−12[T2−T],W.D. = \frac{R}{-\frac{1}{2}}\left[\frac{T}{\sqrt{2}} - T\right],

W.D.=−2R[T2−T].W.D. = -2R\left[\frac{T}{\sqrt{2}} - T\right].

Factorize:

W.D.=2RT[1−12].W.D. = 2RT\left[1 - \frac{1}{\sqrt{2}}\right].

Simplify further:

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

Final Answer: RT[2−2].RT\left[2 - \sqrt{2}\right].