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 H 2 0=100 o C at 760mmHg (1.00 atm, sea level)

If you reduce the pressure on a liquid, its boiling point decreases BP H 2 0=100 o C at 760mmHg (1.00 atm, sea level) BP H 2 0=95.1 o C at 635mmHg (.836 atm,Denver)

If you reduce the pressure on a liquid, its boiling point decreases BP C 2 H 5 0H=78.4 o C at 760mmHg (1.00 atm, sea level)

If you reduce the pressure on a liquid, its boiling point decreases BP C 2 H 5 0H=78.4 o C at 760mmHg (1.00 atm, sea level) BP C 2 H 5 0H=63.5 o C at 400mmHg (.526 atm,way high)

Consider a phase diagram Pop Quiz—What substance?

Line 1-5 represents increasing the pressure at 0 o C 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

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?

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 H vap )

Ln P = -  H vap /R (1/T) + C P is the pressure (any units) ln P is the natural log of P  H vap 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 -  H vap /R

Clausius-Clayperon graph (c) Ln P 1/T Slope=  H vap /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 H vap 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 17.5 o C 40 mmHg at 43.0 o C Ammonia has the following vapor pressures: 100 mmHg at o C 760 mmHg at o C

What is the normal boiling point? The vapor pressure of Hg is 100 mmHg at 1784 o C. Its heat of vaporization is 259 kJ/mol VP of N 2 =10mmHg at -219 o C,  Hvap=5.58 kJ/mol