## 4.7

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

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### 4.7

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25°C and 1.00 bar.

I'm not really sure which work equation to use for this question.

Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

### Re: 4.7

The best one is to use w = -P(deltaV). However, we are not given change in volume, so you must calculate it using the Ideal gas law, in which PV = nRT. You can rearrange the equation to say V = nRT/P, but since we want the change in volume, we can say delta V = deltan(RT/P). I chose change in moles of gas because that is the quickest route (for 1 mol of C6H6 that is burned, what is the net change in moles of gas?). Substitute your unknowns into the work equation, and you will see that P cancels out, leaving you with work = -(delta n)RT.