I'll start off by saying this is homework, but I ask because I don't understand how the math should work (I don't just want an answer, I'd like an explanation if possible). I understand if this is still against the rules, and I apologize if it is.

I have to calculate the work done by a sample of dry air on the environment in an adiabatic expansion to five times the original volume and an initial temperature of 20˚ C. The answer rounds to 100,000 (it's 99,???).

So far, I have that $c_v dT = -p d\alpha$. I tried integrating, which gives $c_v T = R_d T_0 \ln{5}$ (because the limits of integration give 5 inside the natural log and $R * T_0$ is constant; the IGL gives $p\alpha = RT$).

Plugging numbers in gives $T = \frac{(-287)(293)\ln{5}}{717} = -188.76$. Since the change in internal energy is equal to the negative work done (because $\delta q = 0$ in an adiabatic process), and $dT = 105$, then $\delta w = 105 * 717 = 75,285$.

This answer is incorrect, but I'm not sure where I went wrong.

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## closed as too localized by Qmechanic♦, David Z♦Oct 17 '12 at 17:41

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Hi Steven - we normally don't handle questions where someone is just asking for someone else to check their work. That being said, it's kind of a fine line and this is very close to it, so you shouldn't blame yourself for posting a bad question (because it's not). This is pretty close to being the right way to post a homework question. – David Z Oct 17 '12 at 17:45

Have a look at the Wikipedia article on adiabatic processes. The key equation you need is:

$$PV^\gamma = K$$

where $K$ is a constant and the value of $\gamma$ depends on what type of gas you have: for air $\gamma = 7/5$. As with any expansion the work done is the integral of PdV, so:

$$W = \int_{V_1}^{V_2} PdV = K \int_{V_1}^{V_2} V^{-\gamma}dV$$

So do the integral and you'll get an expression for the work that includes the unknown constant $K$. Then use your initial known values of $P$, $V$ and $T$ to calculate $K$ for your system, substitute this into your equation and you'll get the answer.

I'm happy to do the calculation to check your result, but you don't say what the original pressure is and I need this to get the value of $K$.

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