Heated air below mercury pushes it out of glass tube

In summary: I can't seem to find a function that meets those conditions.In summary, the gas must be heated to a certain temperature in order to expel the mercury from the tube.
  • #1
marcbodea
4
0

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?
p1H9YyR.jpg

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 ##
 
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  • #2
Hello and welcome to PF!
marcbodea said:

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.
 
  • #3
TSny said:
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:

TSny said:
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 ##

TSny said:
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.
 
  • #4
marcbodea said:
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.
 
  • #5
TSny said:
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
 
  • #6
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.
 
  • #7
SteamKing said:
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.
 
  • #8
marcbodea said:
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.
 
  • #9
TSny said:
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..
 
  • #10
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##?
 

1. How does heated air push mercury out of a glass tube?

When air is heated, it expands and becomes less dense. This creates a difference in pressure between the heated air and the surrounding cooler air. As a result, the heated air exerts a force on the mercury, pushing it out of the glass tube.

2. Why does the mercury need to be heated to be pushed out of the glass tube?

Mercury is a liquid metal that is denser than air. In order for it to be pushed out of the glass tube, the force from the heated air must be greater than the force of gravity pulling the mercury down. By heating the air, we decrease its density and increase its ability to exert a force on the mercury.

3. What happens to the mercury after it is pushed out of the glass tube?

The mercury will continue to rise until it reaches a point where the force of gravity pulling it down is equal to the force of the heated air pushing it up. At this point, the mercury will stop rising and remain suspended in the heated air.

4. Is there a limit to how high the mercury can be pushed up by the heated air?

Yes, there is a limit to how high the mercury can be pushed up. The height that the mercury reaches will depend on the temperature and volume of the heated air, as well as the diameter and length of the glass tube.

5. How is this phenomenon used in scientific experiments or applications?

This phenomenon, known as thermal expansion, is used in various scientific experiments and applications. One example is in thermometers, where the expansion of a liquid, such as mercury, is used to measure temperature. It is also used in barometers to measure atmospheric pressure and in some types of engines and turbines to convert thermal energy into mechanical energy.

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