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 $)$
  1. 80.5 kJ
  2. 21.55 kJ
  3. 41.50 kJ
  4. 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 or or or 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 $\Delta U=41.54 \mathrm{~kJ}$

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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