Problem on heat of sublimation

Using the vapor pressure data provided below, estimate

i) the heat of sublimation of ice,

ii) the heat of vaporization of water,

iii) the heat of fusion of ice and compare your estimate with the published value of 6010 J/moL

iv) the triple point of water (pressure and temperature). Solve the nonlinear.

T(°C)                             Vapor Pressure (mm Hg)
Ice                                       -4                                            3.280
-2                                            3.880
Water                                   +2                                            5.294
+4                                            6.101

E

Expert

##### Verified

At 1 atm pressure, ice melts at 0°C

density of ice = 920 kg/m3
density of liquid water = 997 kg/m3

Now using the Clapeyron equation

dP/dT = LP/RT2  [ where Vg>>>Vl]

On integration:

Ln[P1/P2] = -L/R[ 1/T1 -1/T2]

Ln(3.880/3.280)= -L/R [ 1/271-1/269]

So Ice  L = ΔH sublimation = - 50911.5 J/mol =-50.9115KJ/mole

Similarly for water L = ΔH vap =- 44929.9 J/Mole =- 44.929 KJ/mole

Now At triple point Ice /Water /Saturate water Vap co-exists Now using the Thermo concept at triple point

ΔH sublimation = ΔH vap + ?H fusion

Similarly  at triple vap pressure of water = vap pressure of ice

LnP = ln(3.880)-50911.5/8.3145[1/T – 1/271] = ln(5.294)-44929.9/8.3145[1/T-1/275]

So Solving for T we get T =273.1297K

Now Solving for P we get

LnP = ln(3.880)-50911.5/8.3145[1/273.129 – 1/271]
LnP = 1.532014

So P = 3.589546 mmHg

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