A sample of gas at temperature T is adiabatically expanded to dou… — Solution | JEE Physics

MCQMediumJEE 2023Isothermal & Adiabatic Processes

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

A
$W = TR [\sqrt{2} - 2]$
B
$W = \frac{T}{R} [\sqrt{2} - 2]$
C
$W = RT [2 - \sqrt{2}]$
D
$W = RT [2 - \sqrt{2}]$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Initial temperature is $T_1 = T$ , the gas expands adiabatically to double its volume so $V_2 = 2V_1$ , and $\gamma = \frac{3}{2}$ .

Find: The work done by the gas.

For an adiabatic process,

T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}

Using temperature relation and work formula

Since $\gamma - 1 = \frac{1}{2}$ ,

T_1 V_1^{1/2} = T_2 V_2^{1/2}

Using $V_2 = 2V_1$ ,

T V^{1/2} = T_2 (2V)^{1/2}

So,

T_2 = \frac{T}{\sqrt{2}}

Direct work expression

Now use the adiabatic work formula shown in the solution,

W = \frac{R(T_1 - T_2)}{\gamma - 1}

Substituting $T_1 = T$ , $T_2 = \frac{T}{\sqrt{2}}$ and $\gamma - 1 = \frac{1}{2}$ ,

W = \frac{R\left(T - \frac{T}{\sqrt{2}}\right)}{1/2}

W = RT(2 - \sqrt{2})

Therefore, the correct option is C. Note that the answer key says option $4$ , but the solution explicitly concludes option C and the expression matches option C.

Common mistakes

Using the isothermal relation instead of the adiabatic relation is incorrect because temperature changes during adiabatic expansion. Use $T V^{\gamma-1} = \text{constant}$ , not $TV = \text{constant}$ .
Substituting $\gamma = \frac{3}{2}$ but forgetting that $\gamma - 1 = \frac{1}{2}$ gives the wrong temperature dependence. Compute $\gamma - 1$ first before applying the exponent.
Taking the final temperature as $T\sqrt{2}$ instead of $\frac{T}{\sqrt{2}}$ is wrong because volume increases in adiabatic expansion, so temperature must decrease. Use $V_2 = 2V_1$ carefully in the adiabatic relation.

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Answer

Step-by-step solution

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