Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673…
Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 \mathrm{~K}$. The increase in internal energy of the gas is $\left(R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \gamma=1.4\right.$ for diatomic gas $)$
80.5 kJ
21.55 kJ
41.50 kJ
65.55 kJ
Solution
Given, temperature, $T_2=673 \mathrm{~K}$
initial temperature $T_1=273 \mathrm{~K}$
rate of heat flow, $R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
$\gamma=1.4$ for diatomic gas, moles of hydrogen gas,
$n=5$
Now, change in internal energy due to adiabatic process is
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or
$\Delta U=n C_v \Delta T$
$\Delta U=5 \times \frac{R}{\gamma-1}\left(T_2-T_1\right)$
$\left(\because\right.$ Specific heat, $\left.C_v=\frac{R}{\gamma-1}\right)$
$=5 \times \frac{8.3}{1.4-1}(673-273)$
$=5 \times \frac{8.3}{0.4} \times 400$
$\Delta U=41.5 \mathrm{~kJ}$
So, the increase in internal energy of the gas is $\Delta U=41.54 \mathrm{~kJ}$