# Calculating Steam Pressure in Closed Container

by jackal67347
Tags: container, pressure, steam
 P: 6 I am trying to calculate the volume of liquid water i need to place in a sealed container in order to obtain 10 psi of steam pressure in that closed container. Here are the numbers: Temp: 816 C Volume of steel pipe: 154.497 ml Final pressure: 10 psi If I left out a required number please let me know. If you could show how you solved it or what equation you're using that' be great. Thanks in advance to whoever helps me out!
 P: 718 Welcome to PF! Is this homework or coursework? There is another section of the forum for that. Perhaps the mentors will move it. In any case, it is required on PF that you make an attempt to solve problems on your own and show us you work. List any equations you think may be relevant and show your best attempt at a solution.
P: 6
 Quote by LastOneStanding Welcome to PF! Is this homework or coursework? There is another section of the forum for that. Perhaps the mentors will move it. In any case, it is required on PF that you make an attempt to solve problems on your own and show us you work. List any equations you think may be relevant and show your best attempt at a solution.
I'm working on a DIY project and i'm trying to steam some catalyst. It's not related to schoolwork.

Here is my attempt:

pv=nrt

(.6805 atm)(v)=( .05555 moles h2o)( .08206) (1089.15 K)

7.296 Liters of steam required

So from the steam table I looked at .001003217 specific volume of h2o at room temp
49.526 specific volume of steam at 800C

49526 ratio

so i need .147 ml according to this calculation

P: 718

## Calculating Steam Pressure in Closed Container

You're on the right track with the ideal gas law. I'm not going to check your unit conversions, I will assume you've done them right. Few things:
- For that numerical value, the gas constant (R) has units ##m^3 atm K^{-1} kg \cdot mol^{-1}##. These do not match the units you are using for volume and mass is your calculation.
- Your volume is known: 154.497 mL. You want gas at this volume at a certain pressure and temperature. The quantity you are solving for is ##n##, how many moles of water vapour are required. I'm not sure where you got the number you input for ##n##.
- The number of moles of water vapour is the same as the number of moles of liquid water needed to produce it (conservation of particle number). So, you will want to convert the number moles of water to the corresponding mass of the water. See here for a worked example. Then converting the mass of the water to its volume is a straightforward density calculation.
 Sci Advisor HW Helper Thanks PF Gold P: 4,504 Is that going to be 10 psi gage or 10 psi absolute? And, is there going to be air in the closed container also? What is the total pressure in the container before it is sealed?
P: 6
 Quote by Chestermiller Is that going to be 10 psi gage or 10 psi absolute? And, is there going to be air in the closed container also? What is the total pressure in the container before it is sealed?
10 psi gauge

no air in the container
P: 6
 Quote by LastOneStanding You're on the right track with the ideal gas law. I'm not going to check your unit conversions, I will assume you've done them right. Few things: - For that numerical value, the gas constant (R) has units ##m^3 atm K^{-1} kg \cdot mol^{-1}##. These do not match the units you are using for volume and mass is your calculation. - Your volume is known: 154.497 mL. You want gas at this volume at a certain pressure and temperature. The quantity you are solving for is ##n##, how many moles of water vapour are required. I'm not sure where you got the number you input for ##n##. - The number of moles of water vapour is the same as the number of moles of liquid water needed to produce it (conservation of particle number). So, you will want to convert the number moles of water to the corresponding mass of the water. See here for a worked example. Then converting the mass of the water to its volume is a straightforward density calculation.
Thanks for your help. I redid the calculation with your suggestions but as I continued researching I discovered steam is not an ideal gas. Is this correct?

Could I use the specific volume of room temperature water vs specific volume of steam at 816 C to determine the ratio? I could then use the ratio and compare that to ml of water. Is my thinking correct here?
 Sci Advisor HW Helper Thanks PF Gold P: 4,504 You need to get the specific volume of water vapor at 816 C and 1.6805 atm absolute (the ideal gas law and other more accurate equations of state are couched in terms of absolute pressure). This is above the critical temperature of water. But you can find the specific volume either from steam tables, thermodynamic diagrams for water, or equations of state for water (e.g., z factor). You then divide the volume of your vessel by the specific volume under these conditions. This gives you the mass of water required.
P: 6
 Quote by Chestermiller You need to get the specific volume of water vapor at 816 C and 1.6805 atm absolute (the ideal gas law and other more accurate equations of state are couched in terms of absolute pressure). This is above the critical temperature of water. But you can find the specific volume either from steam tables, thermodynamic diagrams for water, or equations of state for water (e.g., z factor). You then divide the volume of your vessel by the specific volume under these conditions. This gives you the mass of water required.
.001003217 specific volume of water at room temp
2.9662 specific volume of steam at 816C
both in m3/kg

2957.328 ratio for specific volumes

So 154.497/2957.328 = .052 ml of water

Does that sound right?