[Boxxxed]

Am I on the right track with this question? This is not homework, but mere speculation. Hence I don't have a real question.

Ice will sublime if temperature is increased enough and the surrounding air is sufficiently undersaturated (low vapour pressure).

How would I figure out how much of a certain mass of ice will sublime in a contained system given the temperature change and volume of container?

H2O(s)<---->H2O(g)

G could be calculated from enthalpy and entropy values at final and initial temperature giving two Gibbs energy values using G=H-TS.

From this, two K values can be calculated for each temperature and equilibrium concentrations of H2O(g) can be calculated for for each temperature.

K1 at T1
K2 at T2

K=H2O(g)

Solving for H2O at T1 would give concentration of H2O gas and therefore initial equilibrium vapour pressure using P=nRT/V

K2=H2O(g)+[H2O(g) at K1]; Since the initial vapour will inhibit farther sublimation after the temperature increase.

Since reaction is endothermic K will increase with temperature allowing additional sublimation to continue in addition to the initial sublimation at T1. We can solve for secondary sublimation with H2O(g) = K2 -[H2O(g) at T1]

 By adding H2O(g) concentrations from both T1 and T2 I could find the mass of H2O sublimed? From this equilibrium vapour pressure at T2 could be calculated.

[Boxxxed]

No replies at all?

[fledarmus]

Wouldn't it be easier just to find the vapor pressure of water vapor over the solid at each temperature? Then, if your starting atmosphere is absolutely dry, you add ice and allow it to equilibrate, the partial pressure of the water vapor and the size of the container should give you the mass of ice that has evaporated.

A table of vapor pressures and the equations for calculating them can be found here:

http://en.wikipedia.org/wiki/Phase_diagram_of_water#Phase_diagram

[Boxxxed]

n=PV/RT using equilibrium vapour pressure of water right? That would give me moles and thus mass sublimed at each temperature.


How would I calculate equilibrium vapour pressure of water at each temperature if it is not given? Since the ice is also decreasing in volume the volume occupied by the gas increases. How do I take this into account?

[fledarmus]

You can take the decrease in volume of the solid into account by calculated how much mass has been lost, and using the density of the ice to determine how much volume was lost. In a container large enough to get substantial sublimation, I think you will find that the change in volume of the solid is small enough to be negligible, but it doesn't hurt to do the calculation.

I'm not sure what you mean by:

How would I calculate equilibrium vapour pressure of water at each temperature if it is not given?

The table at the bottom of the page gives the formulas it used for calculation:

Formulas
 
The table values for −100 °C to 100 °C were computed by the following formulas, where T is in kelvins and vapor pressures, Pw and Pi, are in pascals.
 
Over liquid water
 loge(Pw)  =  –6094.4642 T−1 + 21.1249952 – 2.724552×10−2 T + 1.6853396×10−5 T2 + 2.4575506 loge(T)
For temperature range: 173.15 K to 373.15 K or equivalently −100 °C to 100 °C
 
Over ice
 loge(Pi)  =  –5504.4088 T−1 – 3.5704628 – 1.7337458×10−2 T + 6.5204209×10−6 T2 + 6.1295027 loge(T)
For temperature range: 173.15 K to 273.15 K or equivalently −100 °C to 0 °C

Are you asking how the equations were derived?