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The chemistry problem in my textbook is : Calculate the work one when 2.0 L of methane gas, $\ce{CH4}$ (g) undergoes combustion in excess oxygen at 0ºC and 1.00 bar. Assume the volume of water formed is negligible.

I know that we should use the work equation $\mathrm{Work=-Pressure \cdot \Delta Volume}$, and that the pressure is 1.00 bar and we must convert this into pascals. All i need is the volume and convert that to $\mathrm{m^3}$. I got the equation $\ce{CH4 + 2O2 -> CO2 + 2H2O}$, but what I don't get is that how do we find the change in volume?

The book says let $\ce{O2}$'s volume = $\mathrm{V_{initial}}$, and that the total volume of gases for initial is $\mathrm{V_{i}}$ + 2.0 L. It says the final volume is 2.0 L of $\ce{CO2}$ (which I get because it should be the same amount of liters for carbons), but then it says that the water's volume is $\mathrm{V_{i}}$ - 4.0 L of $\ce{O2}$.

Can someone please explain the water's volume? How did they get that? Why is then the total final volume $\mathrm{V_f}$= $\mathrm{V_i}$ - 2.0 L. I thought the number of moles for initial should be same as the number of moles according to Gay-Lussac's law, so then the number of volume total should be the same for both sides, but this is clearly not correct. Am I applying the Gay-Lussac's law wrong? Why is the final volume not the same as the initial volume even if the initial and final have the same amount of moles on each side?

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I think most of your issues regarding this question will be solved if you actually put the states in your chemical equations, so it becomes: $$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$ It is important to note that all the substances are in the gaseous state except for water which is a liquid. Therefore your application of Gay-Laussac's law is incorrect here as the number of moles of gaseous reactants is not equal to number of moles of gaseous products. Thus the initial and final volume shouldn't be the same. This is also why in the question they have stated to assume that the volume of water is negligible as water is a liquid which has very little volume compared to a gas.

The change in volume can be calculated by noting that 3 moles of gas reacts to form only 1 mole of gas. Since there is 2L of methane, that means that the volume is going to decrease by 4L.

Therefore: $\mathrm{V_f = V_i - 2.0L}$

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