Chemistry - Thermodynamics - I Question with Solution | TestHub

$\mathrm{n}=5,\mathrm{T}=293\mathrm{K}=$ 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.

$\mathrm{W}=-{\mathrm{P}}_{\text{ext }}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)=-{\mathrm{P}}_{\text{ext }}\left(\frac{\mathrm{nRT}}{{\mathrm{P}}_2}-\frac{\mathrm{nRT}}{{\mathrm{P}}_1}\right)$

or, $\mathrm{W}=-{\mathrm{P}}_{\mathrm{ext}}\mathrm{nRT}{\left(\frac{1}{{\mathrm{P}}_2}-\frac{1}{{\mathrm{P}}_1}\right)}^2$

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

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

$=-15347.7\mathrm{J}$

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

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

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

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

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