Antoher Irodov's Problem. Could it be a mistake?

[jaumzaum]

I was trying to solve the following problem from the Irodov book.

"A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at T0 = 300K the volume of the upper part is η = 4.0 times greater than that of the lower part. At what temperature will the ratio of these volumes be equal to η'= 3.0?"

I've first solved the problem the way the book solved it, than I thought this way was wrong and solved another way that gave a close but different result. I would like to know which of them is right, and why is this.

1st way: Be h the altitute of the cylinder. Be S the area. On the first temperature, the altitude of the lower part is x = h/5, and on the second temperature, y = h/4.
PV = nRT
P_lower = P_upper + mass_upper.g
RT₀/Sx = RT₀/4Sx + Mg
RT₀/Sx = 4/3 Mg
And RT/Sy = 3/2 Mg

T = (9/8) (y/x) T₀ = 422K

*In the first way (that's the solution of the book) Irodov considered the pressure of the cylinders to be constant at any height . At the second, I've considered the pressure not constant.

2nd way:

Formulas:
P = P₀e^-Mgh/RT
n = P₀S/Mg (1-e^-Mgh/RT)

Where P is the pressure in the height h. P0 the initial pressure. n is the total number of moles of gas until height h.

This way, at the first temperature:
Be P0 the pressure at the botton of the lower cylinder. Be P1 the pressure at the piston. Be (1-Mg/P₀S) = a

For the lower cylinder.
P₀S/Mg (1-e^-Mgx/RT₀) = 1 mole
e^-Mgx/RT₀ = a
P₁ = P₀e^-Mgx/RT₀

For the upper cylinder.
P₁S/Mg (1-e^-Mgx/RT₀) = 1 mole
P₀e^-Mgx/RT₀S/Mg (1-e^-4Mgx/RT₀) = 1 mole
(1/(1-a)) a (1-a⁴) = 1
a⁴+a³+a²+a=1

By the same reasoning we get b=e^-Mgy/RT and b³+b²+b=1

So T = log_ba y/x T0 = 404K

Which of them are right?
Thanks, John

 

[jaumzaum]

nobody?

 

[ehild]

You completely ignored the weight of the piston. In case of a real cylinder and piston, the weight of the gas is much less than that of piston.

The barometric formula refers to an unbounded column of gas, with zero pressure at the top. Here the volume is closed. The gas is pressed even at the top.

ehild

 

[ehild]

The solution of a⁴+a³+a²+a=1 is about a=0.52. What height does it mean?

ehild

 

[jaumzaum]

You completely ignored the weight of the piston. In case of a real cylinder and piston, the weight of the gas is much less than that of piston.

The barometric formula refers to an unbounded column of gas, with zero pressure at the top. Here the volume is closed. The gas is pressed even at the top.

ehild

I assumed the mass of the piston to be negligible, as the problem does not say anything about that (even knowing that a piston mass would never be negligible if compared to a gas mass, it doesn't matter how heavy it is).

About the barometric formula, I don't think it can only be applied with columns of gases that have a zero pressure at the top. The P₀ of the formula already does this job. I know that the gas is the vessel can be in a very big (or very low) compression, in this case, P0 would be greater/lesser, and the pressure at the given point would change.

Look at the proof I posted here:
https://www.physicsforums.com/showthread.php?t=652529

The solution of a⁴+a³+a²+a=1 is about a=0.52. What height does it mean?

ehild

The a stands for e^-Mgx/RT₀ = (1-Mg/P0S)

I've only considered this variable to simplify the calculation.
Applying ln from the both sides we get
ln_a = -Mgx/RT₀

The same for b
When we divide both, we get the final result

 

[ehild]

If you calculate how long the cylinder must be to ensure that the upper part is 4 times as long as the bottom part, while the amount of gas is the same, you will find it a couple of km-s. Is it realistic?

It is the illustration what can be an arrangement when the weight of the gas column counts so you need to apply the barometric formula and assuming zero weight of the piston.

The problem does not specify the mass of the piston, but it has to be taken into account. For realistic lengths of the cylinder, the mass of the gas is negligible to that of the piston. The solution of the book is right.

ehild

 

[jaumzaum]

Now I've got what you mean. I calculated the height and I've got 5.5km for the vessel (that's a little bit high ;D )

If we consider the difference of pression to be caused by the piston mass (and not the gas mass), we get the book result (I was confised because I thought that M stood for molar mass instead of piston mass).

Thanks ehild, it helped a lot
[]

 

[jaumzaum]

Now I've got what you mean. I calculated the height and I've got 5.5km for the vessel (that's a little bit high ;D )

If we consider the difference of pression to be caused by the piston mass (and not the gas mass), we get the book result (I was confised because I thought that M stood for molar mass instead of piston mass).

Thanks ehild, it helped a lot
[]'s
John