[wallaby]

I'm not sure about this problem at all...

A sample of 0.50 g of water is placed in a 5.0 L flask, and the entire flask heated to 60 degrees C.  What is the vapor pressure?

[wallaby]

I forgot to mention that my teacher said that you can just use the ideal gas law for this problem (i.e. plug in the given V, n (mass/molar mass of water) and T to find P), but I am skeptical because how can we know that the water is all gas?  It doesn't make sense to me...

[ARGOS++]

Dear wallaby;

Do the calculation with the ideal gas law, and if "your" final pressure would be higher than 19.86 kP (= partial pressure of water at 60°C) then not all water will be evaporated.

For an even more exact control you may use: 
            http://en.wikipedia.org/wiki/Vapor_pressure#Water_vapor_pressure

Hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[wallaby]

Thank you Argos.  So looking back at the original question, it asks "What is the vapor pressure of the water?"  The vapor pressure is actually 149 mm Hg (or 19.86 kP ask you said) because that is always the vapor pressure of water at 60 degrees C.

I believe my teacher has used the term "vapor pressure" incorrectly here because she says the answer to the question is 120 mm Hg, which is the pressure value found from using the ideal gas law: P(5.0 L)=(.50 g)/(18.02 g/mol)*(62.364 L mm Hg/molK)*(333 K).  I tried explaining this to my teacher but she is very stubborn and does not like to be questioned about these sorts of things.  So... am I correct here, did my teacher make a mistake?

[ARGOS++]

Dear wallaby;

Your re-arranged formula is nearly correct, except that you forgot the volume of the flask!
If you do it correct, then you will get a pressure lower than 149 Torr.

But now we have a bad situation, because we know only your question; if the flask was evaporated before the addition of the water and still closed at 60°C, then your calculation will come correct; if not, then we need more information about.

Good Luck!
                    ARGOS⁺⁺

[wallaby]

Hmmm, I'm afraid I don't understand what you're saying.

The equation I listed: P(5.0 L)=(.50 g)/(18.02 g/mol)*(62.364 L mm Hg/molK)*(333 K) results in P=120 mm Hg.  I have included the volume of the flask (5.0 L) in the calculation (on the left side), I was just putting the equation in the form PV=nRT because that's how the ideal gas law was presented to us in class.

My teacher states that the answer to the problem (which I wrote out in my original post) is 120 mm Hg.
I am questioning whether my teacher is correct here: the problem asks for the "vapor pressure of the water", so shouldn't the correct answer be 149 mm Hg, NOT 120 mm Hg because the vapor pressure of water at 60 degrees C is 149 mm Hg?

Can you confirm what I am saying?

[ARGOS++]

Dear wallaby;

I'm sorry!,  -  that I haven’t full recognized the left side of your equation!

In my calculation I get also 116 Torr!
That’s the correct pressure, if only vapour of 0.5g water and no air is present in the volume of 5 Liter.
In this case there is no bad situation!

Good Luck!
                    ARGOS⁺⁺

[wallaby]

Okay, so the ideal gas law calculation is correct, that's good.  The problem doesn't explicitly state whether the flask is sealed and vacuous, but I would assume that that is the intended meaning.

However, as I have stated, I think there is a problem with the question itself.  My chemistry book (and wikipedia for that matter) says that "vapor pressure" is the same as "equilibrium vapor pressure".  Can you corroborate this statement?  If so, doesn't that mean that the answer to the problem is in fact 149 mm Hg because that is the equilibrium vapor pressure of water at 60 degrees C?

[ARGOS++]

Dear wallaby;

Equilibrium exists only if both phases of water (gas/liquid) co-exist under given conditions.
This definition is used for partial pressure.

Good Luck!
                    ARGOS⁺⁺

[wallaby]

Okay, so lets say there was 2.0 g of water in the flask instead of .5 g.  What would be the vapor pressure then?

[ARGOS++]

Dear wallaby;

For 2.0g water in 5 L at 60°C:  p ~ 4.0 * 116 Torr =  >149 Torr, and that means, that two phases will coexist and the pressure will be the partial pressure of water at 60°C and that’s 149 Torr.

Good Luck!
                    ARGOS⁺⁺

[wallaby]

I think I get it now.  So if it is all vapor (like the original problem) then the "vapor pressure" isn't really "equilibrium vapor pressure" because it isn't in equilibrium with anything.  So in the that case we are just talking about the pressure of a gas (which can easily be calculated with the ideal gas law).  But if it is not all vapor, then the "vapor pressure" does refer to "equilibrium vapor pressure" because the vapor and liquid are both in the flask.  In that case we can just use a chart to find water's vapor pressure at the given temperature.

Am I correct now?

[ARGOS++]

Dear wallaby;

Correct.

Good Luck!
                    ARGOS⁺⁺

[wallaby]

Thank You!!

You have been very helpful to me.  Especially since I have a chemistry test tomorrow 

[ARGOS++]

Dear wallaby;

You 're welcome!   ─   Soon again.

Good Luck!
                    ARGOS⁺⁺