Supercooled Steam Homework: Solving for Tf, Condensation Fraction & ΔS

AI Thread Summary
The discussion revolves around a thermodynamic problem involving 0.50 mol of supercooled steam at 95˚C, which partially condenses in a thermally-insulated vessel at constant pressure. The final equilibrium temperature is determined to be 100˚C, the boiling point of water, as the steam must reach this temperature to avoid being supercooled. The condensation fraction is calculated by equating the heat gained by the condensing water to the heat lost by the steam, using the specific heat capacity and temperature change. The change in entropy (ΔS) can be derived from the final temperature and the initial conditions. The heat of vaporization of water at 100˚C is also relevant for determining the energy changes during the phase transition.
Ariel Jo
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Homework Statement


You have 0.50 mol of steam which has been supercooled to 95˚C at 1 atm. Since the steam is below the boiling/condensation point, it proceeds to partially condense into liquid water. This happens in a thermally-insulated vessel at constant pressure.
(a) When the system re-attains equilibrium, what will the final temperature be?
(b) What fraction of the water will condense?
(c) Calculate ΔS for this process.

Homework Equations


ideal gas equation (the gas is assumed to behave ideally)
deltaS=n*Cp*ln(Tf/Ti)

The Attempt at a Solution


From PV=nRT, obtained an initial volume of steam of 15.1L but unsure how to proceed... Vf and Tf are both unknowns but should not be equal to Vi and Ti (from common sense - steam at 95degC will both condense and cool spontaneously.) Can easily find change in entropy from knowing final temperature... Would U=3/2RT be useful in this situation?
 
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Some hints:

Lacking other data, assume the water condenses until it reaches equilibrium.

Since the pressure is held constant, assume the volume is variable.

This seems to give a final temperature of 100º C. (The boiling point of water.)

So the basic question seems to be, how much water has to condense to heat the vapor back to 100º C.?
 
q(gained by water)=q(lost by steam), and at constant pressure this would be q=n*Cp*deltaT for both substances. Setting them equal to each other would give moles of each, but there are no calculations to get 100degC as the equilibrium temp, just the (seemingly sound, but I'm not a physics specialist) logic that for steam to not be supercooled it must be at 100degC or higher (when the pressure is a constant 1atm), if I understand you correctly.
Thanks for the help!
 
How much heat is given off (per mole) when steam condenses to liquid water at a constant temperature of 100 C? What is the heat of vaporization of water at 100 C?

Chet
 
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