Answer:
a) Pf = 689.4 bar
b) P = 226.6 bar
c) T = 269.99 K
Explanation:
a)
The molar volume of ice is equal to:
Vi = m/p = (18.02x10^-3 kg H2O/1 mol H2O)*(1 m^3/920 kg) = 1.96x10^-5 m^3 mol^-1
The molar volume of liquid water is equal to:
Vl = (18.02x10^-3 kg/1 mol)*(1 m^3/997 kg) = 1.8x10^-5 m^3 mol^-1
The change in volume is equal to:
ΔVchange = Vi-Vl = 1.96x10^-5 - 1.8x10^-5 = 1.5x10^-6 m^3 mol^-1
using the Clapeyron equation:
Pf = Pi + ((ΔHf*ΔT)/(ΔVf*Ti)) = 1.013x10^5 Pa + ((6010 J mol^-1 * 4.7 K)/(1.5x10^-6 * 273.15 K)) = 6.89x10^7 Pa = 689.4 bar
b)
For the pressure we will use the equation:
P = (m*g)/A, where m is the mass, g is the acceleration of gravity and A is the area. Replacing values:
P = (79 kg * 9.81 m s^-2)/(1.9x10^-4 m * 0.18 m) = 2.26x10^7 Pa = 226.6 bar
c)
From Clapeyron´s expression we need to clear ΔT:
ΔT = ((Pf-Pi)*ΔV*Ti)/ΔHf = ((2.26x10^7 - 6.89x10^7)*1.5x10^-6*273.15)/6010 = -3.16 K
you can evaluate the new melting point of ice:
T = Ti + ΔT = 273.15 K - 3.16 K = 269.99 K
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