A piece of ice at 0.0 degrees C weighing 2.50 g is placed in a 5.00 L container

Question

A piece of ice at 0.0 degrees C weighing 2.50 g is placed in a 5.00 L container filled with water vapor at 125.0 degrees C and a total pressure of 500.0 torr. The system is allowed to come to equilibrium. At equilibrium, what is the temperature, and how many grams of water are present in each phase? Neglect the effect of pressure on the boiling point of water.

Data for water: Delta H fusion= 333 J/g, Delta H vap= 2256 J/g, Specific Heat: solid= 2.06 J/K g, liquid= 4.184 J/K g, gas= 2.10 J/K g

Find: Final Temp., g(ice), g(liq), g(vap)

Explanation / Answer

find number of moles of water vapor

PV = nRT {P = 500.0 torr = 0.658 atm, R = 0.082 L atm / K mol , V = 5.0 L, T = 398 K}

n = PV / RT = { (0.658 atm x 5.0 L) / (0.082 x 398 K) } = 0.1 mole = (0.1 x 18 g/mol) = 1.8 g

q = m Cp T

m1 Cp  T = m2 Cp  T ( Cp = specific heat capacity)

2.50 g x 2.06 J/K g x (T + 273 K) = 1.8 g x 2.10 J/K g x ( 398 K - T)

5.15 ( T - 273 K) = 3.78 (398 K - T)

5.15 T - 1405.95 = 1504.44 - 3.78 T

T = 325.911 K

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