## 8.93 Expansion?

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

Hannah Chew 2A
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### 8.93 Expansion?

For part 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.

Why is this considered expansion, and why is work being done against the atmosphere?
The equation I had was: C6H6 (l) +15/2 O2 (g)--> 6CO2 (g) + 3H2O (l)
This would result in a negative change in gaseous moles, so why is the gas expanding?
The answer in the solution manual was work = 3.72 kJ, which is positive and implies compression?

Tim Foster 2A
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### Re: 8.93 Expansion?

If you write your equation with water as a gas instead of a liquid, you will find the change in moles to be +3 and you will get -3.71 KJ as the change in energy of the system due to work. If a system is losing energy due to work, that means it is performing work. Thus the work done against the atmosphere is a positive 3.71 KJ

Hannah Chew 2A
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### Re: 8.93 Expansion?

^ but 1) combustion involves liquid water, not gas and 2) the solution manual says that work is +3.72 kj/mol, so I am still confused.

Tim Foster 2A
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### Re: 8.93 Expansion?

1) I'm not sure about this either, I know that combustion can produce water vapor when it is carried out at temperatures above 100 C, but this reaction appears to be at 25 C. 2) However, I do know that if you consider water vapor as a product instead of liquid water, you will find that the system loses 3.71 KJ of energy as work (or 3.72). When a system loses energy due to work, that means it is performing work on its surroundings. This problem is asking us for the amount of work done BY the system, thus giving us our positive value for w.

Jana Sun 1I
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### Re: 8.93 Expansion?

2) I'm still a little confused about this. I previously thought that if work is done by the system, work would be negative. If work is done on the system, work would be positive. In this case, work is done BY the system, so shouldn't our calculated w be negative instead of positive?

But if I'm understanding Tim correctly, since they're just asking for work done by the system, we'd just provide the work that the system loses (so a positive value). But if they ask for work in general, would we provide a negative value? In other words, when would we ever write work as a negative value then?