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Topic: Intermolecular Forces (Vapor Pressure)  (Read 7001 times)

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Offline kbought1

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Intermolecular Forces (Vapor Pressure)
« on: January 11, 2011, 08:18:25 PM »
I have a series of questions that build off of each other, and I have answered all of them but the final one incorrectly.
 Question #1: If a volatile liquid is added to a sealed flask, the total pressure inside the flask is a sum of the air pressure and the vapor pressure of the liquid. Consider a flask at 298 K with an air pressure of 732. mmHg to which some n-butanol is added. If the measured pressure after the addition is 737. mmHg, what is the vapor pressure of n-butanol at 298 K?
Answer #1: 5mmHg
Question #2: Now, if the temperature inside the same flask is increased to 328 K the measured pressure will increase to read 849. mmHg. This is because both the air pressure and the vapor pressure of n-butanol will increase. The air pressure at 328 K can be calculated from its value at298 K by using Pi/Ti=Pf/Tf.
(a) What is the value of the air pressure at 328 K (in mmHg)?
Answer (a) 806 mmHg
(b) Using that last result and measure pressure at 328 K, what is the vapor pressure of n-butanol at 328 K?
I can't seem to figure out the formula for this. I don't want the answer, I just would like to be explained on HOW to find the answer.

Offline rabolisk

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #1 on: January 11, 2011, 08:31:33 PM »
Wouldn't you just subtract the air pressure from total pressure? What was the given answer?

Offline kbought1

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #2 on: January 11, 2011, 09:21:28 PM »
The total pressure was not given.

Offline opti384

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #3 on: January 11, 2011, 09:29:20 PM »
What rabolisk is saying is that the measured pressure inside the flask at 328K, 849.mmHg, is the sum of the pressure of the air and n-butanol.

Offline kbought1

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #4 on: January 11, 2011, 09:31:27 PM »
What rabolisk is saying is that the measured pressure inside the flask at 328K, 849.mmHg, is the sum of the pressure of the air and n-butanol.
How did you get the 849. mmHg?

Offline opti384

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #5 on: January 11, 2011, 09:34:54 PM »
Read the question carefully.
Question #2: Now, if the temperature inside the same flask is increased to 328 K the measured pressure will increase to read 849. mmHg.

Offline kbought1

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #6 on: January 11, 2011, 09:37:38 PM »
Read the question carefully.
Question #2: Now, if the temperature inside the same flask is increased to 328 K the measured pressure will increase to read 849. mmHg.
I mean I understand that, but what equation did you use (I'm horrible at figuring out what equation to use for what)

Offline opti384

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Re: Intermolecular Forces (Vapor Pressure)
« Reply #7 on: January 11, 2011, 09:46:29 PM »
I think the pressure 849. mmHg is given from the question as you can see. The focus of the second question will not be how you get the total pressure 849.mmHg when the temperature is increased but to understand that the total pressure inside the flask equals to the sum of the partial pressure of air and the vapor pressure.

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