Question

9) Using the Clausius-Clapeyron equation determine the vapor pressure of water at 50.0 C. The molar heat of

vaporization of water is 40.7 kJ/mol. (Review Sample Exercise: Example 13-4)

ln (

R = 8.314 J

) =

∆Hvap

P2

P1

R

mol∙K

(

1

−

T1

; T = Kelvin; P = Consistent units

10) Referring to the figure below, describe the phase changes (and the temperatures at which they occur) when CO2 is

heated from −80 °C to 0 °C at:

(a) a constant pressure of 1 atm

(b) a constant pressure of 10 atm

11) As the intermolecular attractive forces between molecules increase in magnitude, do you expect each of the

following to increase or decrease in magnitude?

(a) heat of vaporization

(b) vapor pressure

(c) surface tension

(d) critical temperature

Answer 1

Q10).

Solution :-

At normal boiling point that is 373 K vapor pressure of water is 760 torr

Vapor pressure of water at 50 C + 273 = 323 K =

Delta H vap = 40.7 kJ/mol * 1000 J / 1 kJ = 40700 J per mol

Ln(P2/P1) = delta H/R [(1/T1)-(1/T2)]

Ln[760/P1] = 40700 J per mol / 8.314 J per mol K * [(1/323)-(1/373)]

Ln[760/P1] = 2.0316

760/P1 = antiln 2.0316

760/P1 = 7.626

760/7.626 = P1

99.7 torr = P1

So the vapor pressure of the water at 50 C is 99.7 torr

Q11)

when CO2 is

heated from −80 °C to 0 °C at:

(a) a constant pressure of 1 atm

Solution :- The solid CO2 wil change to the gas form

b). a constant pressure of 10 atm

Solution :- the solid CO2 will change to liquid and gas equilibrium

Q12)

When the intermolecular forces of the attraction between the molecules increases then following change occure in the magnitude of each property

a). Heat of vaporization increases because the boiling point increases

b) vapor pressure decreases because vapor pressure is having inverse relation ship with intermolecular forces

c)surface tension increases

d)critical temperature increases.