# Help to understand fast adiabatic expansion

1. Jul 16, 2008

### juanedilio

Hi,

I am going through a book on thermal physics (specifically on a section on the 2nd Law)

It talks about an extremely fast adiabatic expression (as a gas in one flask separated by a valve to another flask in vacuum and then the valve is opened).

What I fail to understand is that once the valve is opened and the gas is allowed to reach equilibrium then the final temperature will be the same as the initial temperature (even though Q=0). The ideal gas law tells me that for an adiabatic process when volume increases the temperature must decrease.

Thanks

2. Jul 16, 2008

### Hootenanny

Staff Emeritus
You mention that the ideal gas law states that temperature is proportional to volume, which is indeed correct. However, you should also note that temperature is proportional to pressure. So to say that an increase in volume results in a decrease in pressure is not technically correct, this is only the case during an isobaric process. In your thought experiment, the volume does indeed increase, but that pressure also decreased (i.e. the process is no isobaric). An important point to note here is that a rapid adiabatic expansion is irreversible, and as such, whilst we can apply the ideal gas law to the initial and final equilibrium states, we cannot do the same for any intermediate states since these states do not have very well defined state variables.

Noting that for an ideal gas, it's temperature is proportional to it's internal energy, it is perhaps more useful to move away from the ideal gas law and instead consider the first law of thermodynamics:

$$dU = dQ+dW$$

The problem states that the process is adiabatic and therefore the first term is zero, leaving us with:

$$dU = dW$$

So now what you have to ask yourself, is that whether or not the gas does any work on the surroundings?

3. Jul 16, 2008

### juanedilio

Thanks for the response.

I guess I got to caught up in the ideal gas law before checking if the 1st law was met.