Question

A handbook lists the normal haline pint of an organic liquid is enthalpy of vaporization as 59.8 kJ/mol. Calculate the vapor pressure in tort of this compound at 138.0°C 4. Hydrazine, N2H4, has a normal boiling point of 113.5°C and a critical point at 380.0°C and 145.4 atm. Estimate the vapor pressure of hydrazine at 40.0°C. (Treat the critical point as last point on the vapor pressure curve.)

Answer 1

Finding $\Delta$H_vap :

State-1(boiling point):

At boiling point, pressure, P = 1.00 atm. Hence

P1 = 1.00 atm, T1 = 113.5 oC + 273 = 386.5 K

State-2:

P2 = 145.4 atm, T2 = 380.0 oC + 273 = 653 K

$//img.homeworklib.com/questions/0c8472e0-ab7e-11ea-9dff-bf44f12249ef.png?x-oss-process=image/resize,w_560$_vapH = ??

Applying Clasious Claperon equation:

ln(P2/P1) = ($//img.homeworklib.com/questions/0d0d8050-ab7e-11ea-a3be-91017f954406.png?x-oss-process=image/resize,w_560$H_vap / R) * (1/T1 - 1/T2)

=> ln(145.4 atm / 1 atm) = ($//img.homeworklib.com/questions/0c8472e0-ab7e-11ea-9dff-bf44f12249ef.png?x-oss-process=image/resize,w_560$_vapH / 8.314 J.mol^-1.K^-1) * (1/386.5 K - 1/653K)

=> 4.98 = ($//img.homeworklib.com/questions/0c8472e0-ab7e-11ea-9dff-bf44f12249ef.png?x-oss-process=image/resize,w_560$_vapH / 8.314 J.mol^-1.K^-1) * 0.001056

=> $//img.homeworklib.com/questions/0c8472e0-ab7e-11ea-9dff-bf44f12249ef.png?x-oss-process=image/resize,w_560$_vapH = (4.98/0.001056)*8.314 J/mol

=> $//img.homeworklib.com/questions/0c8472e0-ab7e-11ea-9dff-bf44f12249ef.png?x-oss-process=image/resize,w_560$_vapH = 39206.7 J/mol or 32.068 kJ/mol

# Finding vapor pressure at 40.0 oC:

State-1:

vapor pressure, P1 = ??, T1 = 40.0 oC + 273 = 313 K

State-2:

P2 = 145.4 atm, T2 = 380.0 oC + 273 = 653 K

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

=> ln(145.4 atm / P1) = (39206.7 J.mol^-1 / 8.314 J.mol^-1.K^-1) * (1/313 - 1/653)

=> ln(145.4 atm / P1) = 7.8446

=> 145.4 atm / P1 = (exp)^7.8446 = 2551.95

=> P1 = 0.057 atm (Answer)