Question

Physical chemistry

The vapor pressure of water is 760 torr at 100 °C and 100 torr at 51.6 °C. Estimate the molar heat of vaporization (delta Hvap,m) at the midpoint of this temperature range, and delta Hvap,m

at room temperature assuming delta Cp,m =    - 41.5 J/K.mol

Answer 1

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

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

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

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

Change negative signs

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

ln(100/760) = Hvap/8.314*(1/(100+273) - 1/(51.6+273))

Hvap = 8.314 * ln(100/760) / (1/(100+273) - 1/(51.6+273))

Havp = 42181.46 J/mol

HVap = 42.18 kJ/mol