Heated air below mercury pushes it out of glass tube

[marcbodea]

Homework Statement

vertical glass tube with height of 2H. H is 760mm.
the lower half is full of an unknown gas. the upper half is full of mercury.
The gas gets heated so it pushes the mercury out of the glass tube.
What temperature must the gas be heated to?

here is the photo if, for some reason, it doesn't show up http://i.imgur.com/p1H9YyR.jpg

Homework Equations

$p_1 = -p_0 - \rho h g$
$p_2 = -p_0$ because tube doesn't have mercury
$h = H = 760 mm$
$\rho h g = 10^5 Pa$

The Attempt at a Solution

$\frac{p*V}{T} = ct$
$\frac{p_1*V_1}{T_1}=\frac{p_2*V_2}{T_2}$
$\frac{p_1*V_1}{p_2*V_2}=\frac{T_1}{T_2}$
$\frac{(-p_0 - \rho h g) S H}{(-p_0) S 2 H} = \frac{T_1}{T_2}$
$\frac{-p_0 - \rho h g}{-2p_0} = \frac{T_1}{T_2}$
$\frac{p_0+ \rho h g}{2p_0} = \frac{T_1}{T_2}$
but $\rho h g = p_0$
so $\frac{T_1}{T_2} = 1$
and $T_1 = T_2$

 

[TSny]

Hello and welcome to PF!

Homework Equations

$p_1 = -p_0 - \rho h g$
$p_2 = -p_0$ because tube doesn't have mercury

Why are the pressures negative?

3. The Attempt at a Solution

$\frac{p*V}{T} = ct$
$\frac{p_1*V_1}{T_1}=\frac{p_2*V_2}{T_2}$
$\frac{p_1*V_1}{p_2*V_2}=\frac{T_1}{T_2}$
$\frac{(-p_0 - \rho h g) S H}{(-p_0) S 2 H} = \frac{T_1}{T_2}$
$\frac{-p_0 - \rho h g}{-2p_0} = \frac{T_1}{T_2}$
$\frac{p_0+ \rho h g}{2p_0} = \frac{T_1}{T_2}$
but $\rho h g = p_0$
so $\frac{T_1}{T_2} = 1$
and $T_1 = T_2$

The negative signs that you had in the pressures cancel out, so your answer for $T_2$ is correct (when the last bit of mercury is expelled).
But, you need to consider the temperatures required at intermediate positions of the mercury column.

 

[marcbodea]

Hello and welcome to PF!
Why are the pressures negative?

The negative signs that you had in the pressures cancel out, so your answer for $T_2$ is correct (when the last bit of mercury is expelled).
But, you need to consider the temperatures required at intermediate positions of the mercury column.

Thank you. Here is the explanation:

Why are the pressures negative?

im sorry, i got it wrong. it was something like this $p_1-p_0-h\rho g = 0$ , so $p_1=p_0 + \rho h g$

The negative signs that you had in the pressures cancel out, so your answer for T2T_2 is correct (when the last bit of mercury is expelled).
But, you need to consider the temperatures required at intermediate positions of the mercury column.

I'm thinking that the max temperature and pressure are at the initial point, when the gas has to push the biggest amount of mercury.

 

[TSny]

I'm thinking that the max temperature and pressure are at the initial point, when the gas has to push the biggest amount of mercury.

Initially, the temperature will need to be increased in order to push some of the mercury out of the tube. So, the max temperature is not the initial temperature.

 

[marcbodea]

Initially, the temperature will need to be increased in order to push some of the mercury out of the tube. So, the max temperature is not the initial temperature.

yes, sorry, I'm really tired. That's what i meant to say, that the max temperature and pressure are the temperature and pressure at the point that the gas starts pushing the mercury

 

[SteamKing]

I'm not sure what the point of this exercise is. Mercury is a liquid at room temperature and freezes at temperatures below -39° C. Unless the initial temperature of the tube is such that the mercury has solidified, the situation with liquid mercury over gas is not stable. The liquid mercury is going to wind up in the bottom of the tube, with the gas on top.

 

[TSny]

I'm not sure what the point of this exercise is. Mercury is a liquid at room temperature and freezes at temperatures below -39° C. Unless the initial temperature of the tube is such that the mercury has solidified, the situation with liquid mercury over gas is not stable. The liquid mercury is going to wind up in the bottom of the tube, with the gas on top.

Yes. I guess it's just an academic exercise. They should have included a thin, massless piston at the bottom of the mercury.

Anyway, I thought it was kind of interesting to work through.

 

[TSny]

That's what i meant to say, that the max temperature and pressure are the temperature and pressure at the point that the gas starts pushing the mercury

I think you will find that the max temperature occurs at a point where quite a bit of mercury has been pushed out.

 

[marcbodea]

I think you will find that the max temperature occurs at a point where quite a bit of mercury has been pushed out.

Thank you for all the help, but I can't tell when the temperature will reach it's max value. I've tried to do something like $T=max <=> p V =max$ , but I didn't get anywhere..

 

[TSny]

Suppose that some of the mercury has been pushed out so that the height of the mercury left inside the tube is $h$. Can you find an expression for the temperature of the gas in terms of $h$?