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5 moles of Hydrogen $\left(\gamma=\frac{7}{5}\right)$ initially at S.T.P. are compressed adiabatically so that its temperature becomes $400^{\circ} \mathrm{C}$. The increase in the internal energy of the gas in kilo-joules is
$$
\left(R=8.30 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)
$$
Options:
$$
\left(R=8.30 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)
$$
Solution:
2900 Upvotes
Verified Answer
The correct answer is:
$41.55$
Initial temperature $T_1=0^{\circ} \mathrm{C}$
Final temperature $T_2=400^{\circ} \mathrm{C}$
Work done
$$
\begin{aligned}
& W=\frac{\mu R}{\gamma-1} \times \Delta t \\
= & \frac{5 \times 8.31}{\frac{7}{5}-1} \times(400-0) \\
= & \frac{5 \times 8.31 \times 400}{2 / 5} \mathrm{~J} \\
= & \frac{5 \times 5 \times 8.31 \times 400}{2} \mathrm{~kJ} \\
= & 41.55 \mathrm{~kJ}
\end{aligned}
$$
For adiabatic process,
Increase in internal energy $=$ work done $=41.55 \mathrm{~kJ}$
Final temperature $T_2=400^{\circ} \mathrm{C}$
Work done
$$
\begin{aligned}
& W=\frac{\mu R}{\gamma-1} \times \Delta t \\
= & \frac{5 \times 8.31}{\frac{7}{5}-1} \times(400-0) \\
= & \frac{5 \times 8.31 \times 400}{2 / 5} \mathrm{~J} \\
= & \frac{5 \times 5 \times 8.31 \times 400}{2} \mathrm{~kJ} \\
= & 41.55 \mathrm{~kJ}
\end{aligned}
$$
For adiabatic process,
Increase in internal energy $=$ work done $=41.55 \mathrm{~kJ}$
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