# Adiabatic process btwn 2 containers

1. Sep 16, 2006

### Kenny Lee

Two cylinders of equal volume are connected by a valve. We are given the pressure for one of them, and then told to find the final equilibrium pressure after the valve is opened. Adiabatic process may be assumed.

I know the solution, but not sure why its that way... initially, I tried PiV1^gamma = Pf(V1+V2)^gamma. But the correct method is: PiV1^gamma =PfV1^gamma+PfV2^gamma.

Someone please tell me how the latter is valid. It seems to me that it violates the basic formula PV^gamma = constant.

2. Sep 16, 2006

### Chi Meson

PV^gamma is constant for an adiabatic process if n is constant. Opening up the chambers changes the number of moles of gas in the first cylinder. That which stays constant for the two chambers together is total internal energy.

Edit: hmm, I'm assuming that this was an example of "adiabatic free expansion." Was there total vacuum in the second cylinder? IF so, then in this situation, no work is done when the gas expands (you see, there is no piston involved). That's why internal energy remains constant.

Last edited: Sep 16, 2006
3. Sep 16, 2006

### Kenny Lee

thanks, I get it now. Hey, youve helped me out quite a few times already. Just so you know, I genuinely appreciate it

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