What Is the Equilibrium Pressure Inside Two Connected Containers of Ideal Gas?

[Aestus]

Homework Statement

Two rigid containers contain the same type of ideal gas and are connected by a thin tube with a valve. Container two is 3 times the volume of container 1. The initial pressures inside the containers are different, as are the temperatures. When the valve is opened, the temperatures inside the containers are maintained. Find an expression for the equilibrium pressure inside the containers.

Homework Equations

PV = nRT[/B]

The Attempt at a Solution

P₁V₁=n₁RT₁ and P₂V₂=n₂RT₂
But, V₂=3V₁
Total volume = V₁ + V₂, but V₂ = 3V₁
ΣV=(n₁RT₁)/P₁+(n₂RT₂)/3P₂
I decided to take the molar volume, as I can't calculate the number of moles, But I'm not sure if i can do this
ΣV_,m=(RT₁)/P₁+(RT₂)/3P₂

It's this step that really stumps me, because the temperature inside the two tanks are maintained at different values, so for this last step I took the average temperature.

P₃ΣV_,m=RT₃ where P₃ is the equilibrium temp and T₃ is the average temperature of the two tanks.
P₃((RT₁)/P₁+(RT₂)/3P₂)=RT₃

After some rearranging i come to the answer P₃=T₃/((T₁/P₁)+(T₂/3P₂))

My real questions are, was I correct in taking the molar volume, and am I incorrect in taking the average temperature of the two tanks?

Sorry for the hard to read formatting, it's my first time posting on here.

 

[Chestermiller]

Homework Statement

Two rigid containers contain the same type of ideal gas and are connected by a thin tube with a valve. Container two is 3 times the volume of container 1. The initial pressures inside the containers are different, as are the temperatures. When the valve is opened, the temperatures inside the containers are maintained. Find an expression for the equilibrium pressure inside the containers.

Homework Equations

PV = nRT[/B]

The Attempt at a Solution

P₁V₁=n₁RT₁ and P₂V₂=n₂RT₂
But, V₂=3V₁
Total volume = V₁ + V₂, but V₂ = 3V₁
ΣV=(n₁RT₁)/P₁+(n₂RT₂)/3P₂

This equation looks incorrect to me. Please reconsider.
If V1=V and V2 = 3V, please express the number of moles of each container in terms of V, T1, P1, T2, and P2.
What is the total number of moles in terms of these parameters?

 

[Aestus]

Ah, I think I see.
n₁=P1V1/RT1
n₂=P2V2/RT2
n₁+n₂=P₁V₁/RT₁ + P₂V₂/RT₂

And then I can sub V₂ for 3V₁

n₁+n₂=P₁V₁/RT₁ + P₂3V₁/RT₂

Bringing out the V₁

n₁+n₂=V₁(P₁/RT₁ + 3P₂/RT₂)

Divide left side by (P₁/RT₁ + 3P₂/RT₂) and right side by (n₁+n₂)

V_,m=RT₁/P₁ + RT₂/3P₂

Which after looking up above gives the same result, so I'm not sure what I've done.

 

[Chestermiller]

Ah, I think I see.
n₁=P1V1/RT1
n₂=P2V2/RT2
n₁+n₂=P₁V₁/RT₁ + P₂V₂/RT₂

And then I can sub V₂ for 3V₁

n₁+n₂=P₁V₁/RT₁ + P₂3V₁/RT₂

Bringing out the V₁

n₁+n₂=V₁(P₁/RT₁ + 3P₂/RT₂)

Let P be the final pressure in both tanks. In terms of P, V1, and T1, what is the final number of moles in tank 1? In terms of P, V1, and T2, what is the final number of moles in tank 2? In terms of these parameters, what is the final sum of the number of moles in the two tanks? Has the total number of moles in the two tanks changed from the initial to the final state? What is the final pressure?

 

[Aestus]

Tank 1, n₁=PV₁/RT₁
Tank 2, n₂=PV₂/RT₂

n₁+n₂=PV₁/RT₁ + PV₂/RT₂

moles of gas doesn't change so P_fV₁/RT₁ + P_fV₂/RT₂ = P₁V₁/RT₁ + P₂V₂/RT₂

The question wants me to work out the pressure without any values of volume given, so can I equate the molar volumes?

RT₁/P₁ + RT₂/3P₂ = RT₁/P_f + RT₂/3P_f

P₁/T₁ + 3P₂/T₂ = P_f/T₁ + 3P_f/T₂

P_f = ((P₁/T₁ + 3P₂/T₂)/(1/T₁ + 3/T₂))

 

[Aestus]

Still felt uneasy with what I worked out above. In particular the molar volume. If the molar volume is (total volume/n₁+n₂) then, it must be,
(V₁+3V₁/n₁+n₂)

So, in the case of

n₁+n₂ = PV₁/RT₁ + P3V₁/RT₂

Dividing by total volume, 4V₁

n₁+n₂/4V₁=P₁V₁/4V₁RT₁+ P₂3V₁/4V₁RT₂

n₁+n₂/4V₁=P₁/4RT₁+ 3P₂/4RT₂

So, 4V₁/(n₁+n₂)=4RT₁/P₁+ 4RT₂/3P₂

V_,m = 4RT₁/P₁+ 4RT₂/3P₂ = 4RT₁/P+ 4RT₂/3P

After solving, it comes out the same as above, Pf = ((P1/T1 + 3P2/T2)/(1/T1 + 3/T2))

I think this proof was a bit clearer, hopefully it's right.

 

[Chestermiller]

Tank 1, n₁=PV₁/RT₁
Tank 2, n₂=PV₂/RT₂

n₁+n₂=PV₁/RT₁ + PV₂/RT₂

moles of gas doesn't change so P_fV₁/RT₁ + P_fV₂/RT₂ = P₁V₁/RT₁ + P₂V₂/RT₂

The question wants me to work out the pressure without any values of volume given, so can I equate the molar volumes?

No. In your last equation above, you simply substitute V₂=3V₁. This gives you:

P_f = ((P₁/T₁ + 3P₂/T₂)/(1/T₁ + 3/T₂))

Then, to make the equation look more presentable, you multiply number and denominator by T₁ T₂ to obtain:

$$P_f=\frac{p_1T_2+3P_2T_1}{T_2+3T_1}$$

 

[Aestus]

Thank you for you help, greatly appreciated. :)