The heat of fusion of ice is 80 cal/g at 0 °C and 1 atm, and...

The heat of fusion of ice is 80 cal/g at 0 °C and 1 atm, and the ratio of the specific volume of

water to that of ice is 1.000:1.091. The saturated vapor pressure and the heat of vaporization of

water at 0 °C are 6.026 × 10-3 atm and 600 cal/g, respectively. Estimate the triple point of water

using these data

Solutions

Expert Solution

Clausius-Clapeyron Equation:

For vapouration curve: lnP_v = -H_v/RT +K₁

Sublimation curve: lnPs = -H_s/RT +K₂

At triple point: Vapour pressure of solid = Vapour pressure of liquid = Vapour pressure of gas

At triple point, sublimation curve and vapourisation intersects

-H_v/RT +K₁ = -H_s/RT +K₂

(Here T is the temperature of triple point)

H_s = H_v + H_f = 80 + 600 = 680 cal/g

We need to find out values of constant K₁ and K₂

lnP_v = -H_v/RT +K₁

ln0.006026 = -600 / 2 x 273 +K1

K1 = -5.11 + 600 / 2 x 273

K₁ = - 4.01

lnPs = -H_s/RT +K₂

ln1 = 680/2 x 273 +K2

0-680/2 x 273 = 680/2 x 273

K₂ = -1.245

-H_v/RT +K₁ = -H_s/RT +K₂

-600/T- 4.01 =- 680/T -1.245

T = 28.9K

queen_honey_blossom answered 1 year ago

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