Question

The normal boiling point of methanol is 64.70 °C. At 21.20 °C, the vapour pressure of methanol is 0.1315 atm. What is the vapour pressure of methanol when the temperature is 31.00 °C? Please show all work and reasoning, thanks.

Answer 1

1st find HVap:

T1 = 64.7 oC

=(64.7 + 273)K

= 337.7 K

T2 = 21.2 oC

=(21.2 + 273)K

= 294.2 K

P1 = 1 atm

P2 = 0.1315 atm

we have below equation to be used:

ln(P2/P1) = (Hvap/R)*(1/T1 - 1/T2)

ln(0.1315/1) = ( Hvap/8.314)*(1/337.7 - 1/294.2)

-2.0287 = (Hvap/8.314)*(-4.378*10^-4)

Hvap = 38523 J/mol

Now calculate:

T1 = 64.7 oC

=(64.7 + 273)K

= 337.7 K

T3 = 31.0 oC

=(31.0 + 273)K

= 304.0 K

P1 = 1 atm

Hvap = 38523 J/mol

we have below equation to be used:

ln(P3/P1) = (Hvap/R)*(1/T1 - 1/T3)

ln(P3/1) = (38523.0/8.314)*(1/337.7 - 1/304.0)

ln(P3/1) = 4634*(-3.283*10^-4)

P2 = 0.2185 atm

Answer: 0.2185 atm