For a National Board Exam Review:

To what height will a barometer column rise if the atmospheric conditions are 13.9psia and 68F and the barometer fluid is mercury?

Answer 28.3in

I have encountered questions with irrelevant variables; I assume temperature is irrelevant here; so I try;

$${ P = \rho gh }$$ $${ 13.9\frac{lb}{in^2} = (13.6)(62.4 \frac{lb}{ft^3})(\frac{1ft}{12in})^3 (32.2\frac{ft}{s^2})(\frac{12in}{ft})h }$$

Im getting h = 0.073...

Why isn't my solution with the right answer?

  • 1
    $\begingroup$ does unit cancelation work out? $\endgroup$ – ratchet freak Aug 21 '15 at 14:25
  • $\begingroup$ yes. its in inches. but it does not work $\endgroup$ – james Aug 21 '15 at 14:26
  • 4
    $\begingroup$ Did you make the distinction between pounds-force and pounds-mass? psi pounds is pounds-force and density is pounds-mass, there is a factor of 32 difference. $\endgroup$ – ratchet freak Aug 21 '15 at 14:44
  • $\begingroup$ Temperature is not entirely irrelevant here, density is a function of temperature. It's not a huge variance for small temperature differences, and most standard density numbers are for around room temperature, so your density figure is okay, but it's not something to completely ignore. $\endgroup$ – Trevor Archibald Aug 21 '15 at 15:39
  • $\begingroup$ @ratchetfreak That's accounted for though. The 'g' term changes the mass in the density into a force, canceled out in the pressure term. When you use metric units, you don't need an extra extra term to account for the Newtons used in pressure and the kilograms used in density; that's what the gravity term is there for. $\endgroup$ – Trevor Archibald Aug 21 '15 at 16:10

You are making this too complicated. You look up the density of mercury at 68°F (20°C) and find that it's 0.49057 pounds-mass/in³.

In 1 g, (13.9 pounds/in²)/(0.49057 pounds/in3) = 28.3 inches. Yes, it's really that simple.

Temperature is slightly relevant in that the density of mercury changes slightly with temperature.

Note that pounds-force and pounds-mass come out to the same thing under 1 g acceleration, such as when sitting on the surface of the earth.

  • $\begingroup$ He's not really overcomplicating, he's just mistaking there being a difference in the types of pounds being used in the units. Contrary to ratchet freak's comment, the pounds are both lbm; pressure is not calculated using lbf. $\endgroup$ – Trevor Archibald Aug 21 '15 at 16:23
  • $\begingroup$ @TrevorArchibald pressure is equal to force over area. Who in there right mind would mass units for that? $\endgroup$ – ratchet freak Aug 21 '15 at 18:59
  • 2
    $\begingroup$ @ratchetfreak I've gotten all turned around at this point. Olin's method yields the correct answer, but a dimensional analysis of the pressure number shows it using lbf. I barely understand how to go between lbf and lbm mass anyway, the only consistent way I get the right answer is by converting everything to metric and then back. Thus I conclude that Imperial units are stupid. $\endgroup$ – Trevor Archibald Aug 21 '15 at 19:25
  • $\begingroup$ @Trevor: In this case since we are working at 1 g, as I stated, lbf and lbm can be considered equivalent. It does depend on 1 g. This barometer itself depends on gravity, and would read differently under different gravity but with the same pressure. As for imperial units, in this case the problem was stated in those units and the answer was expected in those units, so it made sense to do the computations in that space. You can, of course, convert back and forth as you want. $\endgroup$ – Olin Lathrop Aug 21 '15 at 22:32

You were using the wrong unit for density. It must be in slugs/ft3, not in lbm/ft3. By the way, 32.2 lbm = 1 slug.


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