# Question #385d0

Jan 21, 2015

Vapor pressure and boiling point for a substance depend on external pressure, which more often than not is the atmospheric pressure.

Argon's normal boiling point of $\text{-186"^@"C}$ implies an atmospheric pressure of 1 atm. Any liquid will begin to boil when its vapor pressure equals the atmospheric pressure, which, in this case, implies that Argon's vapor pressure at its boiling point will be approximately 1 atm.

Here's the vapor pressure graph for $\text{Ar}$:

The vapor pressure of $\text{Ar}$ is listed as $\text{1.013 bar}$ at $\text{-185.85"^@"C}$, which is equal to

$\text{1.013 bar" * ("0.986923267 atm")/("1 bar") = "0.9998 atm}$

This value is, for all intended purposes, equal to 1 atm.