Two moles of an ideal gas are confined in a piston-cylinder arrangement. Initially, the temperature is 300 K and the pressure is 1 bar. If the gas is compressed isothermally to 5 bar, how much work is done on the gas?

Question

Accepted Answer

Since pressure is not constant in this process, we must put it in terms of Volume
The Ideal Gas constant we will use for this problem is

[latex]0.08206 \frac{\mathrm{L} \,\mathrm{atm}}{\mathrm{mol} \,\mathrm{K}}\left(\frac{1.01325 \mathrm{bar}}{1 \mathrm{~atm}} ight)=0.08315 \frac{\mathrm{L\,bar}}{\mathrm{mol} \,\mathrm{K}}[/latex]

as well as [latex]8.314 \frac{\mathrm{J}}{\mathrm{mol} \mathrm{K}}[/latex]

[latex]\begin{aligned}& \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}} \& \mathrm{W}_{\mathrm{EC}}=-\int_{ ext {final }} \mathrm{PdV} \& \mathrm{W}_{\mathrm{EC}}=-\int_{ ext {initial }} \frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{dV}=-\mathrm{nRT} \int_{ ext {initial }}^{ ext {final }} \frac{1}{\mathrm{~V}} \mathrm{dV}=-\mathrm{nRT}\left(\ln \frac{\mathrm{V}_{ ext {final }}}{V_{ ext {initial }}} ight) \& \mathrm{V}_{ ext {initial }}=\frac{(2 \mathrm{~mol})\left(0.08315 \frac{\mathrm{L} ext { bar }}{\mathrm{mol} \mathrm{~K}} ight)(300 \mathrm{~K})}{1 \mathrm{bar}}=49.89 \mathrm{~L} \& \mathrm{~V}_{ ext {final }}=\frac{(2 \mathrm{~mol})\left(0.08315 \frac{\mathrm{L} ext { bar }}{\mathrm{mol} \mathrm{~K}} ight)(300 \mathrm{~K})}{5 \mathrm{bar}}=9.98 \mathrm{~L} \& -\mathrm{nRT}\left(\ln \frac{\mathrm{V}_{ ext {final }}}{\mathrm{V}_{ ext {initial }}} ight)=-(2 \mathrm{~mol})\left(8.314 \frac{\mathrm{J}}{\mathrm{mol} \mathrm{~K}} ight)(300 \mathrm{~K})\left(\ln \frac{9.98 \mathrm{~L}}{49.89 \mathrm{~L}} ight)=\mathbf{8 0 2 8} \mathbf{~ J}\end{aligned}[/latex]

Question 2.19: Two moles of an ideal gas are confined in a piston-cylinder ...

Related Answered Questions