A sample of 1 mole gas at temperature mathrm{T} is adiabatically expanded to double its volume. If a

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11.Thermodynamics

hard

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

$\mathrm{RT}[2-\sqrt{2}]$

$\frac{\mathrm{R}}{\mathrm{T}}[2-\sqrt{2}]$

$\mathrm{RT}[2+\sqrt{2}]$

$\frac{T}{R}[2+\sqrt{2}]$

(JEE MAIN-2024)

Solution

$\mathrm{TV}^{\gamma-1}=\text { constant }$

$\Rightarrow \mathrm{T}(\mathrm{V})^{\frac{3}{2}-1}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\frac{3}{2}-1}$

$\Rightarrow \mathrm{TV}^{\frac{1}{2}}=\mathrm{T}_{\mathrm{f}}(2)^{\frac{1}{2}}(\mathrm{~V})^{\frac{1}{2}}$

$\Rightarrow \mathrm{T}_{\mathrm{f}}=\left(\frac{\mathrm{T}}{\sqrt{2}}\right)$

$\text { Now, W.D. }=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}=\frac{1 \cdot \mathrm{R}\left[\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right]}{1-\frac{3}{2}}$

$\Rightarrow \text { W.D. }=2 \mathrm{RT}\left[1-\frac{1}{\sqrt{2}}\right]$

$\Rightarrow \text { W.D. }=\mathrm{RT}[2-\sqrt{2}]$

Standard 11

Physics

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easy

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

Solution

Similar Questions

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