Cylinder,piston termodynamics problem

[flyerpower]

Homework Statement

A closed vertical cylinder contains a mobile piston that splits it in two parts of volumes V1 and V2 with a ratio V1/V2=3. Both parts contain the same mass of an ideal gas at the same temperature T.
Both chambers are heated to a temperature of 4T/3, find the ratio of the new volumes.

Here's a little diagram to make it more clear.

The Attempt at a Solution

Now I'm somehow confused because they don't specify if the mass of the piston is negligible.

So, if the piston is in equilibrium and its mass is negligible then the resultant force that apply to the piston must be 0 => p1=p2, which is false because p1/p2=1/3. (where p=pressure)

If the piston's mass is not negligible it has a mass m.

p2=p1+mg/S
p2'=p1'+mg/S (where g=gravitational constant, S=surface area of the piston)

p2'-p1'=p2-p1
4nRT1/3 (1/V2' - 1/V1') = nRT1(1/V2-1/V1)

4/3 ( 1/V2' - 1/V1') = 1/V2 - 1/V1, from this i should conclude the V1'/V2' ratio but something might be wrong here, the problem statement is kind of misleading.

 

[gsal]

Well, I am no thermodynamics kind of guy so take my comments with a grain of salt

First of all, this is some kind of idealized textbook problem, in other words, don't worry too much about mass vs mass-less or even gravity...' cause then, you would need to worry about friction or where is the rest of that piston which look pretty much like just a disc with no mechanism to be moved or anything...anyway...

If the initial condition was an equilibrium condition, I would say, yes, you need to find out how much pressure the piston is helping with to the weakest of the chambers...then, you keep that piston help constant and heat up both chambers and see what the final volumes need to be so the pressure still balance.

On the other hand, if the initial condition has nothing to do with anything and was not even steady and the piston is massless and is not exerting any force, then, the final position of the piston should be right in the middle... :-)

 

[Andrew Mason]

You know that there is a net downward force of the piston from the pressure difference. If it is frictionless, you could calculate the mass of the piston if you knew the area, but you don't need that. You just need to know that there is a net downward force that is constant (ie. the same force applies on the lower section after the contents are heated).

AM

 

[flyerpower]

You know that there is a net downward force of the piston from the pressure difference. If it is frictionless, you could calculate the mass of the piston if you knew the area, but you don't need that. You just need to know that there is a net downward force that is constant (ie. the same force applies on the lower section after the contents are heated).

AM

The piston is frictionless.

And that's pretty much what i said in the second part: p2=p1+mg/S where mg/S is due to the piston's mass. But i cannot find the ratio V1'/V2' from my calculations. I would really apprecied if you checked them.
Thanks.

 

[gsal]

I think you just need to come up with enough equations to find the ratio between the volumes after heating things up to 4T/3

For example, initially, you know that
(1) V₁/V₂ = 3
which can also be expressed as
(2) V₁ - 3V₂ = 0

Because both chambers have the same amount of mass and are at the same temperature, you know that
P₁V₁ = nRT = P₂V₂
and, thus
(3) P₁V₁ = P₂V₂
(4) P₁/P₂ = V₂/V₁
(5) P₂/P₁ = V₁/V₂
(6) P₂/P₁ = 3 [ using (1) ]
(7) P₂ = 3P₁
(8) P₂ = P₁ + 2P₁

Also, because of the initial equilibrium condition, you know that
(9) P₂ = P₁ + P_piston

See the similarity between (8) and (9) ? We now know the pressure that comes from the piston without having to know its mass or anything, we got it in terms of P1!

So, you need to continue with the problem and keep Ppiston constant.

After heating things up, you know have

(10) P₁' V₁' = 4nRT/3 = P₂' V₂'

And you also now have:

(11) P₂' = P₁' + P_piston
(12) P₂' = P₁' + 2P₁

And of course
(13) V₁ + V₂ = V_total
(14) V₁' + V₂' = V_total
and so
(15) V₁ + V₂ = V₁' + V₂'

etc., etc.