Question

What is the amount of work done when $0.5$moles of methane, ${\mathrm{CH}}_4\left(\mathrm{g}\right)$, is subjected to combustion at $300\mathrm{K}$? (given, $\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$)

Answer 1

Step 1: Chemical reaction of combustion of methane

• The following is the balanced equation of combustion of methane
• ${\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Finding $∆\mathrm{n}$:

• when one mole of methane undergoes combustion than $∆\mathrm{n}$ (mole difference) is
• $$\begin{gathered} \mathrm{\Delta n}=\mathrm{mole}\mathrm{of}\mathrm{methane}-(\mathrm{mole}\mathrm{of}{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2) \\ \mathrm{\Delta n}=1-(1+2)=-2 \end{gathered}$$
• when half a mole of methane undergoes combustion than $∆\mathrm{n}$ is
• $\mathrm{\Delta n}=-2\times 0.5=-1$

Step 3: Finding $∆\mathrm{V}$:

• $∆\mathrm{V}$ is the change in volume
• $$\begin{gathered} \mathrm{\Delta V}=\mathrm{One}\mathrm{mole}\mathrm{of}\mathrm{an}\mathrm{ideal}\mathrm{gas}\mathrm{at}\mathrm{standard}\mathrm{temperature}\mathrm{and}\mathrm{pressure}\times \mathrm{mole}\mathrm{difference}\times \frac{\mathrm{final}\mathrm{temperature}}{\mathrm{initial}\mathrm{temperature}} \\ \mathrm{\Delta V}=22.4\mathrm{L}\times \mathrm{\Delta n}\times \frac{300\mathrm{K}}{273\mathrm{K}} \\ \mathrm{\Delta V}=-24.6\mathrm{L} \end{gathered}$$

Step 4: Finding work done:

• Here P is pressure
• $∆\mathrm{V}$ is the change in volume
• $$\begin{gathered} \mathrm{W}=-\mathrm{P\Delta V} \\ \mathrm{W}=-1\mathrm{atm}\times (-24.6\mathrm{L}) \\ \mathrm{W}=+24.6\mathrm{Latm} \\ 1\mathrm{Latm}=-101.33\mathrm{J} \\ \mathrm{W}=-24.6\mathrm{Latm}\times -101.33\mathrm{J}{\mathrm{L}}^{-1}{\mathrm{atm}}^{-1} \\ \mathrm{W}==+2494\mathrm{J} \end{gathered}$$

Therefore, the amount of work done = $+2494\mathrm{J}$.