Thermodyn.: ideal gases in 2 two chambers separated by a sliding barrier

[sunrah]

Homework Statement

Two different ideal gases are separated by a sliding barrier that can move vertically. The gas in the upper chamber 1 has n moles of material, whilst the gas in the lower chamber 2 has 3n moles of material. At T₀ the weight of the barrier is such that the volumes of the two chambers are equal. What is the ratio of V₂/V₁at T₁.

Homework Equations

P = F/A
W = mg
PV = nRT

The Attempt at a Solution

Equilibrium at T₀: F₂ = F₁ + W
where the F_i are the forces from the i gas on the barrier.

therefore W = F₂ - F₁ = A(P₂ - P₁)
\frac{3AnRT_{0}}{V_{2}} - \frac{AnRT_{0}}{V_{1}} = \frac{2AnRT_{0}}{V_{0}}, because V_{0} = V_{1}(T_{0}) = V_{2}(T_{0})
further W = \frac{4AnRT_{0}}{V}, because V_{0} = \frac{V}{2} where V is the total volume of the two chambers!

Now I want to check equilibrium forces at T₁ but using F₂ = F₁ + W for T₁ and substituting for constant weight will not allow me to calculate the ratio V₂/V₁!

 

[TSny]

Hi, sunrah.

Can you use the fact that V_total = V₁ + V₂ to eliminate V₂, say, from your equation F₂=F₁+W at T₁? Then maybe you can solve for V₁ in terms of T₁, T₀, and V_total. Once you have V₁ you should be able to get an expression for V₂ and then construct the ratio of the final volumes.