Physical Chemistry water vapor/phases present question?

AI Thread Summary
The discussion addresses the phases of water present in a container with 23.7 g of water at 298.15 K and a vapor pressure of 23.76 Torr. It confirms that both gas and liquid phases are present initially. To determine the volume for only the gas phase, one participant calculated 594 L using the ideal gas law, while another pointed out that the formula used was incorrect. For the liquid phase, a calculation yielded 0.024 L, but confusion arose regarding the methodology. The consensus is that a volume entirely occupied by liquid would mean no gas is present.
qnzbabi91
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23.7 g of water is in a container of 23.7 L at 298.15 . The vapor pressure of water at this temperature is 23.76 Torr.
A) What phases are present?
B) At what volume would only the gas phase be present?
C) At what volume would only the liquid phase be present?



A) I know that only the gas and liquid are present.

B) I used RT/(nP) to find the volume and got (0.08314*298)/(23.7/18 * 23.76 * 133.322/10^5) = 594 L

C) I used 0.018* 23.7/18.02 = 0.024 L

Thanks for any help! =]
 
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qnzbabi91 said:
23.76 * 133.322/10^5

C) I used 0.018* 23.7/18.02 = 0.024 L

Please elaborate.
 
Okay, so 23.76 is the P in torr so I converted to Pa by multiplying by 133.322 and then I divided by 10^5 to convert Pa to bar. In my textbook the authors used bar so I thought I should as well.

For the second part, I found the 0.018 Vm of gas in my physical chemistry textbook, so I just multiplied it by the number of moles in question.

I am semi-confused.
 
\frac{RT}{nP} is an incorrect formula.

And I have no idea what you did in C. However, the only way to have no gas is to have whole volume occupied by liquid.
 
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