When water is boiled, and the dissolved oxygen that was in the water

[jools111]

When water is boiled, and the dissolved oxygen that was in the water escapes, why does this raise the entropy of the system? The O2 gas has a higher entropy than the H20, so when it departs the liquid, shouldn't the entropy decrease as the gas is no longer present? Thanks.

 

[Mapes]

How are you defining your system?

 

[jools111]

How are you defining your system?

H20(l) + Heat ---> H20(g), in an open system. Is that what you're asking? Thanks.

 

[Mapes]

If the system includes both the water and the adjacent gas, I don't see a problem with entropy decreasing in one region as long as the increase in the adjacent region is at least as large.

 

[jools111]

If the system includes both the water and the adjacent gas, I don't see a problem with entropy decreasing in one region as long as the increase in the adjacent region is at least as large.

But with a portion of the dissolved oxygen escaped into the atmosphere, shouldn't the H20(l) that is was dissolved in now have a lower entropy, as gas has escaped?

 

[espen180]

As long as the total entropy of the universe increases, there is no problem.

 

[jools111]

As long as the total entropy of the universe increases, there is no problem.

Thanks for the replies. I think I might be asking my question wrong. Let's look at it from the other end. O2(g) has a higher entropy than H2O(l), so how come when O2(g) is dissolved into water, the entropy decreases. Should it not go up, as O2(g) contributes a higher entropy to the reaction?

 

[Mapes]

...how come when O2(g) is dissolved into water, the entropy decreases.

I don't believe it does. Entropy will always increase when two pure substances are mixed together.

 

[jools111]

I don't believe it does. Entropy will always increase when two pure substances are mixed together.

That's what I thought as well. My text is telling me otherwise though.

 

[Mapes]

If you'd like to relay what the text says, you'll likely get helpful comments. Please include as much as possible so the context is clear.

 

[jools111]

If you'd like to relay what the text says, you'll likely get helpful comments. Please include as much as possible so the context is clear.

"Large negative values for (delta)H and large positive values for (delta)S are associated with high solubilities. This is consistent with the observation tat acetone and chloroform are completely miscible"

Also, when decribing the effect of a carbonated soda bottle being opened:

"Entropy increases, since the dissolved gas is allowed to escape, creating a more random condition."

It's a little confusing to me...

 

[Mapes]

"Large negative values for (delta)H and large positive values for (delta)S are associated with high solubilities.

The $\Delta S$ here is for the dissolution process itself, not the entropy of either of the constituents.

"Entropy increases, since the dissolved gas is allowed to escape, creating a more random condition."

Total entropy increases, even though the entropy of the liquid decreases because of the departure of (some, not all of) the gas. The gas escapes until its partial pressure in the liquid equals its partial pressure above the liquid. At that point, the entropy benefit of escaping into the atmosphere exactly equals the entropy penalty of no longer being dissolved in the liquid. Does this make sense?

 

[jools111]

The $\Delta S$ here is for the dissolution process itself, not the entropy of either of the constituents.



Total entropy increases, even though the entropy of the liquid decreases because of the departure of (some, not all of) the gas. The gas escapes until its partial pressure in the liquid equals its partial pressure above the liquid. At that point, the entropy benefit of escaping into the atmosphere exactly equals the entropy penalty of no longer being dissolved in the liquid. Does this make sense?

It does... But with it being an open system, it not the entropy of the liquid the only factor we would be concerned with? (In reference the soda bottle)

 

[Mapes]

It does... But with it being an open system, it not the entropy of the liquid the only factor we would be concerned with? (In reference the soda bottle)

We have to consider all possible changes in entropy, everywhere, when predicting whether a process will be spontaneous or not.

 

[jools111]

We have to consider all possible changes in entropy, everywhere, when predicting whether a process will be spontaneous or not.

Ok. Thanks for your help!

 

[igurmendez]

Remember that the entropy tells you whether your system is in disorder or not.

If initially you had molecules of O2 and H2O (having no interaction at all), you could say that it tends to a bigger disorder than having molecules of O2 and H2O together. As they are working now like one only molecule, thus decreasing the disorder of the system.

Good Luck