$20 g m$ of $N_{2}$ at $300 K$ is compressed reversibly and adiabatically from $20 {d m}^{3}$ to $10 {d m}^{3}$ . Change in internal energy for the process is :

284.8 J

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142.46 J

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1424.69 J

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3462.89 J

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Solution

The correct option is D $284.8 J$
$20$ gm of $N_{2}$ of $300 k$ is composed reventively and adiabatically.
Mole of $N_{2} = \frac{20}{14} = 1.42$
$V_{1} = 20 d m^{3}$
$V_{2} = 10 d m^{3}$
Process is
$T = 300 k$
$\begin{matrix} w = - n R T ln (v_{2} / v_{1}) = - 1.42 \times 0.0821 \times ln \frac{10}{20} = 284.8 J q = 0 Δ U = q + w = 284.8 J \end{matrix}$
Hence, option $A$ is correct answer.

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