# An engine takes in 5 mole of air

Question:

An engine takes in 5 mole of air at $20^{\circ} \mathrm{C}$ and 1 atm, and compresses it adiabaticaly to $1 / 10^{\text {th }}$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be $X \mathrm{~kJ}$. The value of $X$ to the nearest integer is_______

Solution:

$(46)$

For adiabatic process, $T V^{\gamma-1}=$ constant

or, $T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$

$T_{1}=20^{\circ} \mathrm{C}+273=293 \mathrm{~K}, V_{2}=\frac{V_{1}}{10}$ and $\gamma=\frac{7}{5}$

$T_{1}\left(V_{1}\right)^{\gamma-1}=T_{2}\left(\frac{V_{1}}{10}\right)^{\gamma-1}$

$\Rightarrow 293=T_{2}\left(\frac{1}{10}\right)^{2 / 5} \Rightarrow T_{2}=293(10)^{2 / 5} \simeq 736 \mathrm{~K}$

$\Delta T=736-293=443 \mathrm{~K}$

During the process, change in internal energy

$\Delta U=N C_{V} \Delta T=5 \times \frac{5}{2} \times 8.3 \times 443 \simeq 46 \times 10^{3} \mathrm{~J}=X \mathrm{~kJ}$

$\therefore X=46$