Imagine that 25.0 of ice at -3.0C is placed in a container holding water vapor at temperature of 120.0C . After a certain amount of time, the container has expanded slightly, but holds only water vapor at 110.0C. what is the final volume of the container? you may assume the water vapor is an ideal gas, and no heat is lost to the surroundings. some other information is useful: deltaH fus(H2O) = 6.01KJ/mol

Question

Imagine that 25.0 of ice at -3.0C is placed in a container holding water vapor at temperature of 120.0C . After a certain amount of time, the container has expanded slightly, but holds only water vapor at 110.0C. what is the final volume of the container? you may assume the water vapor is an ideal gas, and no heat is lost to the surroundings. some other information is useful: deltaH fus(H2O) = 6.01KJ/mol
delta H vap(H2O) = 40.66kJ/mol
Cp,m H2O solid =38.02J/mol K
Cp,m H2O liquid = 75.35J/mol K
Cp,m H2O gas = 33.60J/mol K
pressure= 1.000 atm

DrBob222 · Answer

I would calculate the heat required to melt ice, raise T to 100 C, raise T to 110 (in 3 separate steps) and call that q. Then -q = mass water x specific heat water x (Tfinal-Tinitial) to calculate mass water. Convert 25 g ice to moles H2O, convert mass H2O originally in the container to moles, then use PV = nRT to solve for volume @ 110C. Post your work if you get stuck. Check my reasoning.