# Work Done by an Adiabatic Expansion

I am given the information that a parcel of air expands adiabatically (no exchange of heat between parcel and its surroundings) to five times its original volume, and its initial temperature is 20° C. Using this information, how can I determine the amount of work done by the parcel on its surroundings?

I know that $dq = 0$, and that $du + dW = dq = 0$, but I don't know what to do with this information. $dW = pdV$, which seems like it should be helpful, but I don't know what to do for the pressure.

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I just wanted to comment because some people are bound to think this violates our homework policy - but personally I think you just make it into the domain of acceptable homework-like questions because you narrowed the problem down to the underlying concept, which is that you don't see a way to get the pressure. –  David Z Oct 9 '12 at 17:58
If I use $pV = RT$, then $p = \frac{RT}{V}$. Using $\delta w = pdV$, then $w = RT \int{\frac{1}{V}} dV$. Solving this gives me $w = RT\ln(5)$, which, after plugging numbers in, gives me $w = (287)(293)\ln5 = 1.35339 * 10^5$. And I was just sent an email saying that the book's answer is wrong and this one is correct. Thank you so much for your help, I really appreciate it (and I do completely get it now)! –  Vaindil Oct 9 '12 at 18:42
Hmm, actually that's wrong. The temperature is a function of volume as well, so you can't just assume the pressure is constant. The way to do it is to note that for a reversible adiabatic expansion $PV^\gamma$ is a constant, where $\gamma$ is a constant related to the type of gas. –  John Rennie Oct 9 '12 at 19:31