Use the two point form of the Clausius-Clapeyron equation to calculate the vapor pressure of carbon disulfide at 15 degrees C. This temperature in kelvin would be 288. The normal boiling point of CS2 is 46 degrees C. This temperature in kelvin.......?

Would it not be lower?

#P_2 = "260 torr"#

Well, you'd need to write the equation...

#ln (P_2/P_1) = -(DeltaH_(vap))/R [1/T_2 - 1/T_1]#

where:

• Each #P# are the vapor pressure at some atmospheric temperature #T# in #"K"#.
• #R = "8.314472 J/mol"cdot"K"# is the universal gas constant.
• #DeltaH_(vap) = "26.8 kJ/mol"# is the enthalpy of vaporization of #"CS"_2#.

As usual, the units must work out. The left-hand side has no units, so the right-hand side must not either.

#("J/mol")/("J/mol"cdot"K") cdot 1/"K" = "no units"# #color(blue)(sqrt"")#

So now we can proceed... once we recognize that NORMAL pressure is #"760 torr"#.

#P_2/P_1 = "exp"(-(DeltaH_(vap))/R [1/T_2 - 1/T_1])#

The second vapor pressure is:

#color(blue)(P_2) = P_1"exp"(-(DeltaH_(vap))/R [1/T_2 - 1/T_1])#

#= "760 torr" cdot "exp"(-("26800 J/mol")/("8.314472 J/mol"cdot"K") [1/(15 + "273.15 K") - 1/(46+"273.15 K")])#

#= "760 torr" cdot "exp"(-"3223.3 K" cdot 3.3709 xx 10^(-4) "K"^(-1))#

#= "760 torr" cdot 0.3374#

#=# #color(blue)ul("256.4 torr")#

but you only get two sig figs, so #"260 torr"# I suppose... Does this pressure make sense? Why should the vapor pressure be lower at a lower temperature?