Question #385d0

1 Answer
Stefan V.
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 #"-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 #"Ar"#:

http://encyclopedia.airliquide.com/encyclopedia.asp?LanguageID=11&GasID=3#VaporPressureGraph

The vapor pressure of #"Ar"# is listed as #"1.013 bar"# at #"-185.85"^@"C"#, which is equal to

#"1.013 bar" * ("0.986923267 atm")/("1 bar") = "0.9998 atm"#

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

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