# Column of Mercury at Sea Level

1. Sep 3, 2008

### Andrusko

1. The problem statement, all variables and given/known data

At sea level, the pressure of the standard terrestrial atmosphere will support a column of mercury of density 1.359 x 10^4 kg/m^3 and of height 0.7600m at a place where the acceleration due to gravity is 9.80665 m/s^2. Show that 1 atmosphere = 760 mm Hg = 1013hPa.

I really just want to know what "support" means.

Does it mean that at this pressure the mercury can only expand so far? So if you had a 1m long tube and you put the mercury in it it would rise 0.76m up the tube. Is it the case that the expansion of the mercury is independent of its area?

If you had a receptacle of cross sectional area A and one of cross sectional area B and you put the same amount of mercury into both would they rise the same distance? Because this doesn't seem right at all.

2. Sep 3, 2008

### LowlyPion

http://www.idahoop.org/education/lesson8b.htm

It means the air pushing toward the vacuum can only push 760 mm of Hg.

3. Sep 6, 2008

### Andrusko

Thanks, that's cleared most of the problem up but I'm still having issues.

By definition atmospheric pressure can be measured in mm/Hg by how far up a tube mercury is pushed by it. The question specifies that the mercury is pushed up 760mm. How does one go about mathematically proving that? Or do I just state exactly what I've said? I got my answer in hPa like so:

I set the zero point to be the mercury-air interface and did a calculation with this formula:

$$P = P_{0} + \rho(y_{0}-y_{1})g$$

$$P_{0} = 0.76*\rho*g$$

where $$P_{0}$$ is atmospheric pressure. and $$P$$ is the vacuum at the top of the tube, and thus zero. After calculating, I get

$$P_{0} = 101287 Pa$$ which when divided by 100 to get it in hPa is 1013 hPa. Fair enough, but how can I prove the mm/Hg bit?

4. Sep 7, 2008

### LowlyPion

I'm at a loss to say it any other way than that there is minimal vapor pressure in the near vacuum, .01mm or something like that, and it offers little additional offset to help the mercury on the one side to the weight of the air on the other. So ...

$$P_{0} = \rho*g*h$$

There is a temperature dependence to the vapor pressure, but I think that it is much tinier per degree C.

5. Sep 7, 2008

### Andrusko

Fair enough, that makes sense.

Thanks for the help, LowlyPion.