Finding Mole Fractions in a Two-Phase Equilibrium System

In summary, The final mole fraction of the vapor phase is found to be almost entirely due to the variation in pressure.
  • #1
FranciscoSili
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Homework type problem posted in wrong forum, so no template
Hello everyone, I'm doing revision for a final exam in Thermodynamics and i found this exercise i can't solve:

A particular material has a latent heat of vaporization Δh, constant along the coexistence curve. One mole of this material exists in two.phase (liquid-vapor) equilibrium in a container of fixed volume V0, at an initial temperature T0 and a pressure P0. The system is heated at constant volume increasing its pressure to 2P0. The vapor phase can be treated as an monoatomic ideal gas, and the molar volume of the liquid can be neglected relativa to that of the gas. Find the initial and final mole fractions of the vapor phase. ( x≡Ng/(Ng+Nl) ).

I started using the Clapeyron equation to find the final temperature (Tf) of the gas (and so the complete system i think). I integrated from T0 to Tf in one side and from P0 to Pf in the other. The only unknown variable was Tf. Then using the ideal gas equation ( vg(molar volume of gas)=NRTf ) and dividing that by N = Ng + Nl I could leave the final mole fraction of the gas as a function of Pf, Tf and final vg, which i don't know.

So, how can i find the value of the final molar volume? This problem is in a book called Herbert Callen - "Thermodynamics and an Introduction to Thermostatistics" 2nd Edition. page 233 problem 9·3-7

Thank you and sorry i don't use formulas, i don't know how to write them here.
 
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  • #2
You're told the volume of the container, that its value is fixed, and that the volume of liquid is negligible. . . .
 
  • #3
You know the initial temperature, pressure, and volume. So how many moles of vapor do you have? You know the final temperature, pressure, and volume. So how many moles of vapor do you have?
 
  • #4
John Park said:
You're told the volume of the container, that its value is fixed, and that the volume of liquid is negligible. . . .
Yes i know, but after the process both liquid and gas mole fractions changed. I need to know, that variation, specifically in the gas one.
 
  • #5
FranciscoSili said:
Yes i know, but after the process both liquid and gas mole fractions changed. I need to know, that variation, specifically in the gas one.
The vapor occupies virtually the entire volume. From the ideal gas law, how many moles of vapor do you have in the final state? You know the temperature, the pressure, and the volume.
 
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Likes FranciscoSili
  • #6
Chestermiller said:
The vapor occupies virtually the entire volume. From the ideal gas law, how many moles of vapor do you have in the final state? You know the temperature, the pressure, and the volume.
Oh, yes. Oh my god, it was very simple :) thank you very much! :D
 

1. What is the Clapeyron Equation?

The Clapeyron Equation is a thermodynamic equation that relates the change in temperature and pressure to the change in volume of a substance during a phase change.

2. How is the Clapeyron Equation used in thermodynamics?

The equation is used to calculate the change in enthalpy during a phase change, which is important in determining the conditions under which a substance will change from one phase to another.

3. What are the assumptions made in the Clapeyron Equation?

The Clapeyron Equation assumes that the substance is in a closed system and that the phase change occurs at a constant temperature and pressure.

4. Can the Clapeyron Equation be used for all types of phase changes?

No, the Clapeyron Equation is only applicable for phase changes between two equilibrium states, such as solid to liquid or liquid to gas.

5. How is the Clapeyron Equation derived?

The Clapeyron Equation is derived from the combination of the ideal gas law and the Clausius-Clapeyron equation, which relates the change in vapor pressure to the change in temperature during a phase change.

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