# Five moles of an ideal gas

Question:

Five moles of an ideal gas at $293 \mathrm{~K}$ is expanded isothermally from an initial pressure of $2.1 \mathrm{MPa}$ to $1.3 \mathrm{MPa}$ against at constant external pressure 4.3 MPa. The heat transferred in this process is_______________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Rounded-off to the nearest integer) [Use $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]

Solution:

$\mathrm{n}=5, \mathrm{~T}=293 \mathrm{~K}=\mathrm{const}, \Delta \mathrm{U}=0$

$\mathrm{P}_{1}=2.1 \mathrm{MPa}, \mathrm{P}_{2}=1.3 \mathrm{MPa}$

$\mathrm{P}_{\text {ext }}=4.3 \mathrm{MPa}=$ const.

$\mathrm{W}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=-\mathrm{P}_{\mathrm{ext}}\left(\frac{\mathrm{nRT}}{\mathrm{P}_{2}}-\frac{\mathrm{nRT}}{\mathrm{P}_{1}}\right)$

or, $\mathrm{W}=-\mathrm{P}_{\mathrm{ext}} \mathrm{nRT}\left(\frac{1}{\mathrm{P}_{2}}-\frac{1}{\mathrm{P}_{1}}\right)$

$=-4.3 \times 5 \times 8.314 \times 293\left(\frac{1}{1.3}-\frac{1}{2.1}\right)$

$=-4.3 \times 5 \times 8.314 \times 293\left(\frac{2.1-1.3}{1.3 \times 2.1}\right)$

$=-15347.7 \mathrm{~J}$

or, $W=-15.35 \mathrm{~kJ}$

$\Delta \mathrm{U}^{0}=\mathrm{q}+\mathrm{W}$

$\therefore \quad \mathrm{q}=-\mathrm{W}$

or, $\mathrm{q}=15.35 \mathrm{~kJ}$ (for 5 moles)

$\therefore \mathrm{q} / \mathrm{mole}=\frac{15.35}{5}=3 \mathrm{~kJ} \mathrm{~mol}^{-1}$