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 a constant external $4.3\mathrm{MPa}$. The heat transferred in this process is $\mathrm{\_\_\_\_\_\_\_\_}{\mathrm{kJmol}}^{-1}.$(Rounded-off the nearest integer).

(Use $\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$)

Answer 1

Answer: 15

$$\begin{gathered} \mathrm{Moles}\left(\mathrm{n}\right)=5 \\ \mathrm{T}=293\mathrm{K} \\ \mathrm{Process}=\mathrm{Iso}\mathrm{T}.\to \mathrm{Irreversible} \\ {\mathrm{P}}^{\mathrm{ini}}=2.1\mathrm{M}\mathrm{Pa} \\ {\mathrm{P}}^{\mathrm{t}}=1.3\mathrm{M}\mathrm{Pa} \\ {\mathrm{P}}^{\mathrm{ext}}=4.3\mathrm{M}\mathrm{Pa} \\ \mathrm{Work}=-{\mathrm{P}}^{\mathrm{ext}}∆\mathrm{v} \\ =-4.3\times \left(\frac{5\times 293\mathrm{R}}{1.3}-\frac{5\times 293\mathrm{R}}{2.1}\right) \\ =-5\times 293\times 8.314\times 43\left(\frac{1}{13}-\frac{1}{21}\right) \\ =-\frac{5\times 293\times 8.314\times 43\times 8}{21\times 13} \\ =-15347.7049\mathrm{J} \\ =-15.34\mathrm{kJ} \end{gathered}$$

$$\begin{gathered} ∆\mathrm{U}=0 \\ \mathrm{W}=-\mathrm{Q} \\ \mathrm{Q}=15.34\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Isothermal process, so $∆\mathrm{U}=0$

$$\begin{gathered} \mathrm{W}=-\mathrm{Q} \\ \mathbf{Q}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{34}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol} \end{gathered}$$

So, the answer is 15.