Posted by S on Monday, December 27, 2010 at 12:43am.
The temperature will not change since both phases (ice and liquid) still coexist. The vapor pressure is a function of temperature and that will not change either.
So I would pick answer D, however:
The melting of some ice will decrease total liquid + solid volume. This will open up more slightly more gas volume in the container, for air and water vapor.
If the container is sealed and originaly had some air, the total pressure will decrease slightly as a partial vacuum is created. This will have a very small effect upon the temperature of the two-phase mixture. I don't think they expect you to worry about that.
The equilibrium vapor pressure of ice and water at 273K is slightly different, but not significantly. So I agree D. See http://apollo.lsc.vsc.edu/classes/met130/notes/chapter7/graphics/esw_esi.gif
Agreed, D is the only sensible answer, given the information supplied.
The temperature is constant, so the vapour pressure for the mixture is constant.
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