Physical Chemistry water vapor/phases present question?

[qnzbabi91]

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! =]

 

[Borek]

23.76 * 133.322/10^5

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

Please elaborate.

 

[qnzbabi91]

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.

 

[Borek]

$\frac{RT}{nP}$ 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.

 

[DeeNos]

A) Based on the given information, the phases present in the container are gas and liquid. This is because the container contains water at a temperature where both the gas and liquid phases can exist simultaneously.

B) To calculate the volume at which only the gas phase would be present, we can use the ideal gas law, PV = nRT, where P is the vapor pressure of water, V is the volume, n is the number of moles of water, R is the gas constant, and T is the temperature. Rearranging the equation to solve for V, we get V = nRT/P. Plugging in the given values, we get (0.08314*298)/(23.7/18 * 23.76 * 133.322/10^5) = 594 L. Therefore, at a volume of 594 L, only the gas phase of water would be present.

C) To calculate the volume at which only the liquid phase would be present, we can use the equation for the density of water, which is mass/volume. Rearranging the equation to solve for volume, we get volume = mass/density. Plugging in the given values, we get 0.018* 23.7/18.02 = 0.024 L. Therefore, at a volume of 0.024 L, only the liquid phase of water would be present.