Hello.
I bought the book "Quimica" (Chemistry in english). I started to read it and a notice this. In page 3 it says that in STP water in gas state has Xxxx density. How do you can have water in gas state at STP? That is what I dont understand.
Thanks!
Dear Emilio,
"STP" means "Standard Temperature and Pressure" which is defined as 0'C, 1atm. So, as long as water can exist in a vapor state at those conditions (it can - see the previous response from Alex) we can calculate (or measure) the amount water in a given volume in order to calculate a density.
Best regards,
Steven
Thank tou for your response.
In that case it would not be STP (because in our experiment water would be at a pressure of 4.6 torr, much inferior to 1 atm) That is the part I dont understand. I think is a mistake of the book, is telling us the density of the water at 0°C and 4.6 torr, not at STP.
¿What do you think?
Dear Emilio,
You may be confusing the terms here. "STP" IS a defined physical condition. Water vapor pressure is the partial pressure - or part of the total system pressure - that water contributes to the system 1atm pressure. Find a good psychrometric chart to see all of the possibilities. Thus, AT "STP" the temperature of the water is 0°C - it may be at the triple-point where solid (ice), water (liquid) and steam (vapor) coexist - and the vapor present has a pressure of 4.6 Torr. That is a small fraction of the system pressure of 760 Torr (1 atm) at STP, but that is because partial pressures are independent of the system pressure - only contributing to it, but not constrained by it. It then also has the density indicated in your text, most likely. Other tables may be located that also contain the density information.
Best regards,
Steven
P.S. Here is a psychrometric chart in SI units.
At STP water is mainly in the liquid or solid state. Water is not an ideal gas at 0 oC. However, a small amount of it is in equilibrium in the gas phase - 4.6 torr. If you run Alex's calculation and replace 1 mole by 18 g/mole, you get 2367L/ 1g or 0.004 g/L. Similarly, if you take water at 100oC, 1 mole, 18 g should have 22.4 L or 0.81 g/L at 760 torr. But at 0 oC, you have only 4.6 torr in the gas phase. 0.81 x 4.6/760 = 0.005 g/L.
The vapor pressure of water at 0°C is 4.6 torr:
Vapor Pressure of Water from 0 °C to 100 °C
You could put 4.6 torr into the Ideal Gas Law
PV=nRT
with R = 62.36 L torr/mol K, T = 273.15K and n = 1 and solve for the molar volume. Then, calculate density (18.02 grams/volume).