Temperature of expanding gas

[mprovod]

Two thermally insulated cylinders, A and B, of equal volume, both equipped with pistons,are connected by a valve. Initially A has its piston fully withdrawn and contains a perfect monatomic gas at temperature Ti, while B has its piston fully inserted, and the valve is closed. The thermal capacity of the cylinders is to be ignored. The valve is fully opened and the gas slowly drawn into B by pulling out the piston B; piston A remains stationary. Show that the final temperature of the gas is Tf = Ti/(2^(2/3))

 

I've been thinking about this one for hours, could someone have a look and let me know if you're successfull?

[insane_alien]

can you show us the working you have already done?

 

seeing as it is a perfect gas you'll want to muck around with the perfect gas equation

 

PV=nRT n and R are constant in this case

[ziggy]

can you show us the working you have already done?

 

seeing as it is a perfect gas you'll want to muck around with the perfect gas equation

 

PV=nRT n and R are constant in this case

 

 

Ive been looking at the same question. So far i can gather that, as its a thermally isolated system, there is no change in heat, but work is done by slowly moving piston B.

So dU = dW = -pdV. And the equation of state is valid for before and after the expansion. So, assuming the valve has negligible volume,

(initial p)*V = Nk(initial T) and (final p)*2V = Nk(final T).

 

So i got 4 variables, and only two equations to work with at this stage. Any ideas?

 

 

[swansont]

The "slowly" is usually code for "reversible process" in questions like this. You can't use PV=nRT because work is being done. dU = dW = -pdV is the right way to go, but you need to do the integral so you need to know the path.

 

 

What you want is this:

http://en.wikipedia.org/wiki/Reversible_adiabatic_process#Isentropic_flow