Boiling liquid in a closed vessel

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In a closed vessel containing liquid and vapor, heating above the boiling point increases both the vapor pressure of the liquid and the pressure of the air inside, leading to boiling when these pressures equalize. The pressure inside the vessel can be calculated using Dalton's law, considering the vapor pressure and the pressure of the air. To determine the proportions of liquid and vapor, the available volume for vapor can be calculated by subtracting the volume of the liquid from the total vessel volume. The ideal gas law can be used to estimate the number of vapor molecules, but caution is advised near the boiling point due to phase transitions. If there isn't enough liquid to reach the vapor pressure, all liquid will evaporate, resulting in a pressure lower than the vapor pressure.
Telmerk
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Hi dear Forumers,

I 've run into a little thermodynamic problem:

There is a closed vessel, containing a given amount of liquid. I heat up the vessel, higher than the boiling point of the liquid. Here it is what happens:

As a start, I know that the pressure inside the vessel consists of the vapor pressure of the liquid + the pressure of other gases (air).
Increasing the temperature, the partial pressure of the air increases, simply according to ideal gases p ~ T. And the vapor pressure also increases. At a certain temperature, vapor pressure of the liquid will be equal to the pressure of the air inside the vessel, am I right? Then the liquid begins to boil. Question: How can we determine the portion of liquid and vapor?
I tried to find it out by simply using ideal gas law. Volume of the vessel is given, so I can determine, how many liquid need to become vapor in order to reach the given vapor pressure at a given temperature. Then I can substract this amount from the volume of the liquid. And then I reach to the result, hopefully:shy:.
Is this OK? I 'm afraid, because close to the boiling point it is not a wise thing to use ideal gas law.
Please check my thoughts. . .
 
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I am afraid I was not clear. Just think about a vessel that contains some kind of a refrigerant, or LN2. Above it there is some vapor and air. The vessel is closed. The temperature is above the boiling point of the system.
-What is the pressure inside?
-How many gas molecules are in the vapor phase, and how many stays as a liquid?

Cheers,
TtM
 
If you find the equlibrium temperature by experiment, you can look up the properties (ie, saturation pressure) in a steam (or refrigerant) table.
 
attempt

I can look out the saturation(vapor) pressure from a table. What do you mean about equ. temperature?

Here's my attempt to answer the questions:

Known quantities:
V_vessel - volume of the vessel
M_refr - molar mass of the gas
m_refr - mass of the refrigerant filled in the vessel
N_refr - number of refrigerant molecule in the vapor state
m_air - mass of the air that stays in the vessel
T - temperature
P_vap - vapor pressure of the refrigerant
rho_refr - density of the refrigerant in a liquid state

- What is the pressure inside? The pressure inside the vessel comes from the vapor pressure and the pressure of the air, according to Dalton's law. In order to determine the amount of the refrigerant (in a vapor state) inside the vessel, it is enough to consider what causes the vapor pressure inside the vessel? How many refrigerant need to boil in order to sustain the vapor pressure?
First, I have to determine the available volume for the refrigerant vapor. It is: V_vap=V_vessel - V_refr, where V_refr is the volume of the refrigerant when it is in the liquid state. And here there is an assumption, that the refrigerant needed to be boiled is much less than the whole refrigerant quantity. V_refr = m_refr / rho_refr

V_vap=V_vessel - m_refr / rho_refr This volume is filled with refrigerant vapor and air molecules. The pressure here is p_air+p_refr, where p_refr should be the vapor pressure at the given temperature. From this, using the ideal gas law, we can determine the amount of refrigerant molecules, that are inside this volume. We don't have to deal with the air molecules, because of Dalton's law.

p_vap * V_vap = N_refr * k * T

N_refr = (p_vap *V_vap) / (k * T) where N_refr is the number of refrigerant molecules in the vapor state, k is the Boltzmann constant.
From N_refr, we can calculate the mass of the refr. molecules in a vapor state:

m_vap = N_refr * M_refr, and from this, we can obtain the volume of this amount in a liquid state: V_refr_lq = m_vap / rho_refr. We substract this amount from the total volume of the refrigerants, and then we arrived. We will know the quantity of refrigerant in a liquid and in a vapor state.
(Our assumption was that m_refr_vap << m_refr_lq)

I hope this is OK.

Another question: what if there is not enough liquid to reach the vapor pressure? Then all of the liquid should evaporate/boil, giving a pressure below the vapor pressure, and this pressure can be calculated again from ideal gas law.
Two more questions:
What can we do, if our assumption is not right? (m_refr_vap cca.= m_refr_lq)
And:
Is it right to use the ideal gas law? I mean that there are phase transitions!

Cheers, and thanks for any comments,
TtM
 
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