## Final 2011 Question 1

$\Delta U=q+w$

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### Final 2011 Question 1

I have a question on both parts a and b of question 1 on the 2011 final. On #1 when it asks how much methane needs to be combusted to get 25 degree celsius liquid water to boil, why does the answer only take into account the heat to raise the water to 100 degrees and not also include the phase change it would need to undergo? On part b it says to calculate the work associated with the irreversible combustion of 1mole methane at 25 degrees. The answer uses the formula w=-deltanRt\T but why couldn't you just say deltaU is zero becaue of constant temperature so -q=w and just use the heat of combustion of methane provided in part a?

Chem_Mod
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### Re: Final 2011 Question 1

The wording of part a implies that we are only interested in the energy to get the water to begin boiling, not vaporize completely.

For part b, we can't use the expression from part a because $\Delta T$ is zero (T is constant). We must calculate the work of the molecules of gas created pushing against the atmosphere.