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Topic: Van der Waals Equation Problem  (Read 4481 times)

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Offline Cooper

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Van der Waals Equation Problem
« on: September 20, 2014, 05:34:17 PM »
Hi, I was doing a textbook problem regarding the van der Waals equation and my textbook is giving me an answer I don't think is right. Can anyone verify?

Q. Suppose 10.0 mol ethane gas is confined to 4.860 L at 300 K. Predict the pressure exerted by ethane from (a) the perfect gas...given: a = 5.507 L^2 atm mol^-2, b = 0.0651 L mol^-1.

So when they say "confined to 4.860 L," I am supposed to take that as V_measured is 4.860 L and therefore

[tex]V_{ideal}=V_{measured}-nb=4.860 L-(10.0 mol)(0.0651\frac{L}{mol})=4.209L[/tex]?

Therefore...

[tex]p_{ideal}=\frac{nRT}{V_{ideal}}=\frac{(10.0 mol)(0.08206 \frac{L*atm}{mol*K})(300 K)}{4.209 L} = 58.4 atm[/tex]

And the answer is 58.4 atm, correct?

In the answer key to the book they used 4.860 L to calculate the pressure of the ideal gas and they get 50.7 atm. Is this not incorrect to do since 4.860 L is not the volume of the ideal gas?

Thanks
~Cooper :)

Offline Cooper

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Re: Van der Waals Equation Problem
« Reply #1 on: September 20, 2014, 06:28:02 PM »
After looking about at Youtube videos doing problems like this I believe the book is right but I am not sure why. How can the V_ideal be the size of the container (4.860 L) for part (a) and then also be defined as V_measured - nb in the van der Waals Equation? Isn't V_measured the size of the container?

It actually makes sense to me why you would use the size of the container, because for a perfect gas you are assuming they don't take up any space so the volume they can be in is the entire volume of 4.860 L. But then I don't understand what you solve for with V_ideal = V_measured - nb.
~Cooper :)

Offline Borek

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Re: Van der Waals Equation Problem
« Reply #2 on: September 21, 2014, 03:42:01 AM »
What do other points ask for? Ideal gas means "a=b=0". Are you sure you are not overcomplicating?
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Offline Cooper

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Re: Van der Waals Equation Problem
« Reply #3 on: September 21, 2014, 11:26:14 AM »
What do other points ask for? Ideal gas means "a=b=0". Are you sure you are not overcomplicating?

Thanks for responding. Perhaps this video (https://www.youtube.com/watch?v=1LArvAZx0aQ) will show what I am confused about.

You can see on the left side of his board he has define ideal volume as...

[tex]V_{ideal}=V_{measured}-nb[/tex]

Yet at 0:28 and 1:20 he says the volume for the ideal gas law equation is the volume of the container (shouldn't V_ideal be used? AKA box volume - nb?). But then later in the video at 4:25 he says the volume of the container goes in for the variable V_measured and then calculates V_ideal using the equation displayed above.

At 1:20 why does he use the volume of the container for the ideal gas law equation? But then defines V_ideal as something different at 4:25?
~Cooper :)

Offline Borek

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Re: Van der Waals Equation Problem
« Reply #4 on: September 21, 2014, 03:23:44 PM »
First, he does the calculation assuming a=b=0, that's the ideal gas approximation. As a=b=0, Videal=Vmeasured.

He then calculates the volume and pressure from VdW equation.

I agree his naming can be confusing.
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Offline Cooper

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Re: Van der Waals Equation Problem
« Reply #5 on: September 21, 2014, 08:15:13 PM »
Thanks :) I need some time to think about it to see if I fully understand.
~Cooper :)

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