# Work done in adiabatic quasistatic compression

1. Oct 15, 2006

### Repetit

Hey!

"Calculate the work done on 1 mole of a perfect gas in an adiabatic quasistatic compression from volume V1 to V2."

The work done on the gas in this compression is:

$$-\int_{V1}^{V2} P dV$$

Because we are talking about an ideal gas the ideal gas law applies so:

$$P=\frac{nRT}{V}$$

Inserting this gives

$$-\int_{V1}^{V2} \frac{nRT}{V} dV = nRT Log[\frac{V1}{V2}]$$

But for some reason this is not the correct result. However, I get the correct result if I use:

$$P V^\gamma = const$$

What's going on? Why can't I use the ideal gas law?

2. Oct 15, 2006

### quasar987

You have assumed that the temperature is constant and you pulled it out of the integral. But the temperature cannot be constant because there is no heat exchange with the environement, and so as work is done on the gaz, its internal energy changes. And for an ideal gaz, internal energy is directly linked to temperature:

$$E(T)=\frac{3}{2}\nu T$$

3. Oct 15, 2006

### Repetit

Hmm, of course... what was i thinking :-) Thanks alot!!