[LHM]

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 don't need help with getting the right number or anything, I just want to know how come the work would be positive, as opposed to negative? Because isn't work being done by the system on the atmosphere? Am I misunderstanding anything?

Thanks!

[Trave11er]

So you get a positive value for work that system does on the atmosphere, don't you? There is nothing wrong with it. The work would be negative if we the system was compressed.

[LHM]

Forgive me if I'm stubbornly missing the point or something, but isn't work being done by the system negative, and work done on the system positive?

Like I just googled it, and here it says:
"If you view this from the gas' point of view, work done BY the gas has a negative sign as it lowers the internal energy of the gas and work done ON the gas has a positive sign since it increases the internal energy."

http://answers.yahoo.com/question/index?qid=20090130100829AAGwOXI
I know that that isn't always a source that should be trusted, but that's what I always thought too, did I get it backwards or something?

[Twigg]

I know what you mean. These kinds of questions are, frankly, annoying. The trick is that you're supposed to calculate the work done "against the atmosphere" and not by the gas; in other words, the work done with respect to the atmosphere. The atmosphere gains energy, so the work is positive.