How is boiling point related to pressure?

…it’s more than you think!

• If you reduce the pressure on a liquid, its boiling point decreases

• BP H20=100oC

at 760mmHg (1.00 atm, sea level)

• If you reduce the pressure on a liquid, its boiling point decreases

• BP H20=100oC

at 760mmHg (1.00 atm, sea level)

• BP H20=95.1oC

at 635mmHg (.836 atm,Denver)

• If you reduce the pressure on a liquid, its boiling point decreases

• BP C2H50H=78.4oC

at 760mmHg (1.00 atm, sea level)

• If you reduce the pressure on a liquid, its boiling point decreases

• BP C2H50H=78.4oC

at 760mmHg (1.00 atm, sea level)

• BP C2H50H=63.5oC

at 400mmHg (.526 atm,way high)

Consider a phase diagram

• Pop Quiz—What substance?

• Line 1-5 represents increasing the pressure at 0oC

• Line 6-9 represents warming at 1.00 atm.

• The slope of the left hand line shows that the substance is water (that and the normal melting point)

• The item of interest today is the slope of the right hand line—

How does the boiling point respond to pressure?

• Compare

• Compare

• Yes, I know

the MP and BP

are different.

Look at the slopes

• Water has a steeper liquid/gas line

• Carbon dioxide has a shallower liquid/gas line

• Water has a steeper liquid/gas line

--its boiling point changes (↔) only a little with changes in pressure (↨)

• Carbon dioxide has a shallower liquid/gas line

• --its boiling point changes more with changes in pressure

Why?

Why?

• Because water has a large heat of vaporization

The Clausius-Clayperon equation

• --relates heat of vaporization to changes in vapor pressure at different temperatures

• --If a small change in T makes a big change in vapor pressure, the substance is easy to boil (low Hvap)

• Ln P = -Hvap/R (1/T) + C

• P is the pressure (any units)

• ln P is the natural log of P

• Hvap is the heat of vaporization in J/mol

• R is the ideal gas constant 8.31 J/mol k

• T is the absolute temperature

Please notice:

• Don’t use P

• Don’t use T

Please notice:

• Don’t use P, use ln P

• Don’t use T, use 1/T

Please notice:

• Don’t use P, use ln P

• Don’t use T, use 1/T

• This is a linear relationship

• The slope of the line is -Hvap/R

Clausius-Clayperon graph

(c)

Ln P

1/T

Slope=Hvap/R

• There are two ways this equation is used

• --to find the heat of vaporization of a substance

• You will need two sets of (T,VP) data• Convert to (1/T,lnP), graph, find the slope, and

solve OR• Calculate by difference

• There are two ways this equation is used

• --to find the normal boiling point• You will need one set of (T,VP) data, and Hvap

• Convert to (1/T,lnP), graph, calculate the slope, and extend the line to ln(1atm) OR

• Calculate C and use with ln(1atm) to find 1/T, solve for T

What is the heat of vaporization?

• Acetic Acid has the following vapor pressures: 10 mmHg at 17.5oC

40 mmHg at 43.0oC

• Ammonia has the following vapor pressures: 100 mmHg at -68.4oC

760 mmHg at -33.6oC

What is the normal boiling point?

• The vapor pressure of Hg is 100 mmHg at 1784oC. Its heat of vaporization is 259 kJ/mol

• VP of N2=10mmHg at -219oC,

Hvap=5.58 kJ/mol