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### Topic: thermodynamic - gaz - henry's law & open system  (Read 5630 times)

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#### shrx

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##### thermodynamic - gaz - henry's law & open system
« on: February 16, 2005, 04:43:26 AM »
Hello,

I am a organic chemist (PhD) and my thermodynamics is not what it used to be 10 years ago...

So here is my problem (modelized from an actual issue):

initial state : I have a small (1ml) container with water and dissolved air at room temperature and atmospheric pressure.
intermediate state : the container is closed, with no bubbles and put under pressure (say 10 atm).
final state  : assuming there is little capillary leaks and no mass loss, how can I explain theoretically the large augmentation of dissolved air ? I known the henry's law but how can I explain that more atmospheric air is dissolved in a recipient which is under a much higher pressure???

shrx

#### eugenedakin

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##### Re:thermodynamic - gaz - henry's law & open system
« Reply #1 on: April 18, 2005, 01:34:51 AM »
Hello Dr. shrx,

Thats a great question.  I had to dig out some of my old books to remind me of the answer.  Lets use the example of the equilibrium of dissolved oxygen and oxygen gas is O2(aq)<-->O2(g).  The equilibrium constant is K = p(O2)/c(O2). The concentration of dissolved oxygen is directly proportional to the partial pressure of oxygen.  Since K is a constant, increasing the pressure proportionally increases the concentration of dissolved oxygen.

A general method of explaining this to a student might be:  Added pressure increases the energy available to the gaseous oxygen.  This added energy is used to dissolve more oxygen into the liquid.

I hope this helps,

Eugene Dakin Ph.D., P.Chem.
There are 10 kinds of people in this world: Those who understand binary, and those that do not.

#### ksr985

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##### Re:thermodynamic - gaz - henry's law & open system
« Reply #2 on: May 16, 2005, 01:32:50 AM »
The equilibrium between the dissolved gas and the undissolved gas will result in a chemical potential equality. Solve this, it is a very small and exhaustive derivation of henrys law. You will understand the relationship of the amount dissolved with the partial pressure quite easily.

refer to any undergraduate level physical chemistry book.
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#### GCT

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##### Re:thermodynamic - gaz - henry's law & open system
« Reply #3 on: May 16, 2005, 09:44:55 AM »
Hello,

I am a organic chemist (PhD) and my thermodynamics is not what it used to be 10 years ago...

So here is my problem (modelized from an actual issue):

initial state : I have a small (1ml) container with water and dissolved air at room temperature and atmospheric pressure.
intermediate state : the container is closed, with no bubbles and put under pressure (say 10 atm).
final state  : assuming there is little capillary leaks and no mass loss, how can I explain theoretically the large augmentation of dissolved air ? I known the henry's law but how can I explain that more atmospheric air is dissolved in a recipient which is under a much higher pressure???

shrx

I'm not quite sure what you're getting to here, it seems to me that henry's law should be sufficient with explaning the pressure to solubility correlation.  Did you want a more detailed explanation?

In an elementary sense you can explain the higher solubility by considering the evaporation/condensation ratio.  The rate of evaporation is constant with a given temperature, the rate of condensation can depend upon the pressure, or by varying the volume of the container.