Consider a long, vertical tube of water, as shown below:

long tube of water

Where $P_\mathrm{atm}$ is the atmospheric pressure and $P_\mathrm{bottom}$ is the pressure at the bottom of the tube.

The tube is long enough such that $P_\mathrm{atm}$ is much less than $P_\mathrm{bottom}$.

It is known that water at the surface (top of the tube) will boil at a temperature such that the water's vapor pressure equals $P_\mathrm{atm}$. Let's call this temperature $T$. For example, $T=100~^\circ\mathrm{C}$ when $P_\mathrm{atm}$ is about $1\mathrm~{atm}.

Question: At what temperature will the water at the bottom boil? Will it be higher than $T$ or equal to $T$? Why?

References, if available, would also be greatly appreciated.

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    $\begingroup$ Your setup, if heated from below, is actually a pretty good working model of a geyser. Water at the bottom will reach 100∘C and above, and still remain liquid. But as soon as the bubbles start to form, and some of the liquid spills over the top, the pressure decreases, and suddenly the whole bulk of water would start boiling explosively, throwing everything out. $\endgroup$ – Ivan Neretin Oct 8 '15 at 12:57

Here's a cool example that should help illustrate the correct answer to your question. As for why it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles. Is it harder or easier to do that when the ambient pressure is high?

As for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described here. You will need some values for $P_0$, $T_0$ and $\Delta H_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.

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  • $\begingroup$ Thanks for the cool example - 464 °C! Wow. It is now clear from the 2-D phase diagram that the boiling temperature is higher. $\endgroup$ – boxofchalk1 Jan 10 '15 at 3:59
  • $\begingroup$ Just to followup, I read that "A liquid boils at a temperature at which its vapor pressure is equal to the pressure of the gas above it." This only applies to the water on top surface right? Because the water on the bottom boils at a much higher temperature at which the vapor pressure is much higher than $P_{atm}$, which is the "pressure of the gas above it". $\endgroup$ – boxofchalk1 Jan 13 '15 at 0:41
  • $\begingroup$ I think a better statement would be "A liquid boils at a temperature at which its vapor pressure is equal to the external pressure." FYI the original statement is here (chem.purdue.edu/gchelp/liquids/boil.html). (For some reason that site automatically downloaded a useless file to my computer). $\endgroup$ – boxofchalk1 Jan 13 '15 at 0:44
  • $\begingroup$ Look up "black smokers", undersea volcanoes spouting water at 200 C or higher that does not boil because of the immense pressure at depth. At high enough temperature and pressure, though, it can become a supercritical fluid, q.v. $\endgroup$ – DrMoishe Pippik Jan 13 '15 at 17:12

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