0
$\begingroup$

From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure. The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation. If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

(Please correct me if what I know is not true).

$\endgroup$
2
  • 1
    $\begingroup$ Do you mean what the saturation vapor pressure is, or what the dynamic vapor pressure (perhaps as relative humidity) happens to be at a given time? $\endgroup$ – Jon Custer Apr 17 '20 at 18:05
  • $\begingroup$ In such a case, the vapour presure is the pressure of the gas, if all other gases were absent. In river/lake cases, this presure is variable in space and time, not reaching equilibrium. $\endgroup$ – Poutnik Apr 18 '20 at 10:48
1
$\begingroup$

From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure.

This is backwards. The boiling point is when the vapor pressure of the liquid is equal to the atmospheric pressure. So given the average pressure of 1 atmosphere at sea level water would boil at $\pu{100 ^\circ C}$. However on some mountain top water might boil at $\pu{95 ^\circ C}$ since the air pressure is lower.

The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation.

For a particular temperature, in a sealed container, when the rate of condensation equals the rate of evaporation then the system is at equilibrium. The partial pressure of the gaseous form of the liquid is then the equilibrium vapor pressure of the liquid.

Without being at equilibrium the partial pressure could be anything.

If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

If I have an open beaker of benzene on an outside picnic table then the system isn't at equilibrium and benzene has no defined vapor pressure. There just isn't enough benzene to ever reach an equilibrium with the whole atmosphere of the earth.

This wouldn't effect the boiling point of the benzene. Assuming that the air pressure is 1 atmosphere, benzene would boil at the same temperature whether if it was being boiled in the open air or refluxed in a lab.

This bring up another point. If we say that the vapor pressure of water is $\pu{0.0231 atm}$ at $\pu{20 ^\circ C}$, we are referring to "local conditions." In other words, that gravity can be assumed to be a constant within the system. Obviously if I had a cylinder 200 miles high at $\pu{20 ^\circ C}$, the vapor pressure wouldn't be the same at the top and bottom of the cylinder due to the difference in gravity.

$\endgroup$
0
$\begingroup$

For each given pressure $p$ on a liquid there is a temperature at which the liquid will boil and become gaseous.$^1$ This temperature $T_\text{boil}(p)$ is called boiling temperature. The function $T_\text{boil}(p)$ can be calculated by the Clausius–Clapeyron equation.

The pressure on water in a river on earth is $1\ \mathrm{atm}$. The corresponding boiling temperature is $T_\text{boil}(1\ \mathrm{atm})\approx 100\ \mathrm{^\circ C}$.

One can on the other hand invert the relationship and ask for a given temperature $T$ at which pressure the liquid will become gaseous. This pressure is then called vapor pressure.

$^1$ that is assuming that $p$ is smaller or equal than the critical pressure of that fluid.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.