How does the condensation of steam into water create a vacuum?

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I know the ideal gas law applies, but how?
 

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Drakkith
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Off the top of my head I would say that if an enclosed space filled only with water vapor was forced to condense, then instead of having a gas filling the space you have a much smaller amount of that space filled with water and the rest is a vacuum.
 
  • #3
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It does not create a perfect vacuum, of course, there is no perfect vacuum.
It simply reduces the pressure according to the saturation pressure law for water (which is an equilibrium between water and vapor).
 
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Actually you cannot get vacuum in that way. The enclosed space will always be filled with water vapor if there is also water.
 
  • #5
russ_watters
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Well compared with atmospheric pressure you get a partial vacuum: at room temp its about 98%.
 
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When water turns into steam, it occupies about 1600 times the volume (at standard temperature and pressure).

So, if you take a 1600 liter jug filled with steam and try to condense it back into liquid, the resulting water will only want to take up 1 liter of space. The rest is, well, nothing. Vacuum.

Of course that's not going to happen because water in a vacuum will inevitably boil. But the point is that some equilibrium will be reached where the pressure is much less than atmospheric.

A condenser in a power plant is made pretty robustly in order to withstand the "vacuum".
 
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  • #7
DrClapeyron
So, if you take a 1600 liter jug filled with steam and try to condense it back into liquid...

...some equilibrium will be reached where the pressure in much less than atmospheric.
Multiply that that jug into a tank car.

LL_railcar.jpg
 
  • #8
Drakkith
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What exactly happened there DrClapeyron?
 
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When water turns into steam, it occupies about 1600 times the volume (at standard temperature and pressure).

So, if you take a 1600 liter jug filled with steam and try to condense it back into liquid, the resulting water will only want to take up 1 liter of space. The rest is, well, nothing. Vacuum.
At standard temperature and pressure, you are not going to have steam. You will have water vapor. Steam is defined as water vapor at temperatures above 100°C.

If the volume remains constant at 1600 liters, the only way you can produce condensation is by lowering the temperature. Once condensation starts, you will have a water/vapor interface until that water turns to ice. You will eventually have an ice/vapor interface and that interface will continue to exist no matter how low you drop the temperature.

This is because the ice surface molecules will have a Boltzmann distribution of kinetic energies of translation at every temperature. No matter how cold the ice gets, some of its surface molecules will have sufficient kinetic energy of translation to escape the surface and become vapor molecules.

The net result is that you cannot condense all of the vapor without reducing the volume of the container to the volume that liquid water or solid ice will have at that temperature.
 
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At standard temperature and pressure, you are not going to have steam. You will have water vapor. Steam is defined as water vapor at temperatures above 100°C.

If the volume remains constant at 1600 liters, the only way you can produce condensation is by lowering the temperature. Once condensation starts, you will have a water/vapor interface until that water turns to ice. You will eventually have an ice/vapor interface and that interface will continue to exist no matter how low you drop the temperature.

This is because the ice surface molecules will have a Boltzmann distribution of kinetic energies of translation at every temperature. No matter how cold the ice gets, some of its surface molecules will have sufficient kinetic energy of translation to escape the surface and become vapor molecules.

The net result is that you cannot condense all of the vapor without reducing the volume of the container to the volume that liquid water or solid ice will have at that temperature.
good explanation

but how can we explain that process use "ideal gas equation"
condensation is at isothermal process. ( T = constants)
but it will give wrong explanation because P1.v1 = P2.v2

is there any better explanation use thermodynamic equation ?
 
  • #11
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good explanation

but how can we explain that process use "ideal gas equation"
condensation is at isothermal process. ( T = constants)
but it will give wrong explanation because P1.v1 = P2.v2

is there any better explanation use thermodynamic equation ?
Most gas laws require conditions of equilibrium--at the very least at the beginning and end of the process. Condensation is--by definition--a non-equilibrium process. The enthalpy of condensation is being continuously added to the condensate during the process. At the same time, the temperature of the remaining vapor is increasing due to the "selecting out" of cooler molecules to be condensed. To keep the process isothermal requires that you somehow remove the enthalpy of condensation from the condensate and some measure of enthalpy from the remaining vapor.

I am most comfortable with kinetic gas theory and statistical mechanics. I will leave the classical thermodynamic equations to others better qualified.
 

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