Can I use the ideal gas law for this problem?
If water is in a container and it is 75% vapor and 25% liquid and im only given the pressure. And in its second phase its 100% vapor. How can I calculate the final pressure. Can I use 100 degrees as Temperature 2 in the ideal gas law equation. P1/T1=P2/T2 And then use a value from the steam tables to find T1? Im Lost Thanks
- Anonymous9 months agoFavourite answer
No, you can not use the ideal gas equation where you have a phase change, you need a set of steam tables and only steam tables.
< If water is in a container and it is 75% vapor and 25% liquid and im only given the pressure. >
So you know the temperature from the steam tables. You know the density of the steam from the steam tables. In the second part, the mixture is 100% so all the water is now steam. Assuming the volume of water is insignificant (a very reasonable assumption), find the temperature of a saturated mixture of steam where the density is 1.33x the value in the first part (because if you convert the 25% water to steam, you will have 1.33x more mass of steam than in the previous step, density = mass / volume).
- 8 months ago
use PV=nRT to solve this question...
- Anonymous8 months ago
No but the steam tables will give you the answer.
Let's say the pressure is 50 psia. The density of steam = 0.1174 lb/ft3 (and the temperature is 281F though that doesn't matter). You didn't give a pressure so I just picked this one as an example.
75% of the water per the problem is in the vapor phase and 25% of the water is liquid. Almost all of the volume of the container is steam and we will assume the volume occupied by water is insignificant.
Heat the container. The liquid water vaporizes to make steam. Since the volume is assumed to be fixed, the new density of the steam is 1 1/3 greater than the starting value when the last bit of water vaporizes. Density is just mass divided by volume. If you started out with 1 lb of water, that was distributed as 0.75 lb of steam and 0.25 lb of water. Make it saturated steam and you have 1 lb of steam vapor, 1 1/3x greater than 0.75 lb therefore the new density is 1.33 greater than the starting density.
Now use the steam table to find what saturated steam has a density of 0.1565 lb/ft3. That is about 68 psia and 301 deg F (you can play with the numbers more to get a more accurate value).
- pisgahchemistLv 78 months ago
Water vapor .....
If you have a sealed container where liquid water is in equilibrium with water vapor, you can look up the temperature using a vapor pressure table for water (*). It will be about 92.4C
No. There simply is insufficient information, and maybe even some misunderstanding on the part of the question writer. The combined gas law.... P1V1/T1 = P2V2/T2
assumes that the moles of gas are fixed. But if some of the water is vapor, and then later all of it is in the vapor phase, the moles of gas is not constant, and you can't use the equation.
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- Roger the MoleLv 79 months ago
There is no reason to think the vapor is at 100 degrees, especially since you are not given an initial temperature.
It seems to me that all you need is 75% vapor goes to 100% vapor. That's a 1/3 increase, so the pressure would also increase by 1/3. So multiply the initial given pressure by 4/3.