## Homework Problem 8.93

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Ashley Van Belle 2B
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### Homework Problem 8.93

For 8.93, it says to calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1 mol C6H6 at 25 degrees C and 1 bar. So I figured that I would use the formula w=-P'delta'V and replace it to instead use w=-'delta'nRT. Looking at the solutions, somehow they got 1.5 for the change in moles. How does this work?

Chem_Mod
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### Re: Homework Problem 8.93

The balanced equation for the combustion of one mole of benzene is: $C_{6}H_{6} + \frac{15}{2}O_{2} \rightarrow 6CO_{2} + 3H_{2}O$ where oxygen, carbon dioxide, and water are all in the gas phase while benzene is in the liquid phase. Thus, there are a total of 9 moles of gas on the product side and $\frac{15}{2}$ or 7.5 moles of gas on the reactant side. When you subtract the moles of reactant in the gas phase from the moles of products in the gas phase (9 - 7.5), you get a change of 1.5 moles.