Let me see, is there anything that typically happens at or just below zero degree Celsius that just might complicate aqueous concentration levels?

H_f, the specific enthalpy of water, is defined and referenced by convention to be zero at the triple point of water. At lower temperatures, the saturated water is actually a solid; the information tabulated refers to fluids only.

The specific heat of ice is close to 0.50.

Because ice or frozen water does not fit the definition of saturated water.

Saturated water being in steam or liquid state

What properties do you have in mind? An old copy of Perry gives vapour pressure of liquid water down to -16°C and water ice down to -14°C. It doesn't say how the water is kept liquid at -16°C. Says water at -14°C has VP = 1.436mmHg, and water ice at -14°C = 1.361mmHg.

Thank you sir, for your reply. I want to calculate the amount of water vapor present in a room of 12kPa total pressure (dry air+water vapor) with a temperature of -40dec C & RH of 100% inside a space of Volume ''V'?

For example at 25deg C with 100% RH and 100kPa Total pressure, i can find the amount of water vapor by finding partial pressure of water vapor(Pv) at 25 deg C by

Pv = RH%*partial pressure of water vapor at saturated condition(Pg) at 25 deg C

So @ 100% RH ----> Pv=Pg where Pg will be taken from Saturated water table.

And then mass of water vapour (mv) = Pv*V/(Rv*T)

Kindly help.... Thank you.

<...calculate the amount of water vapor present in a room of 12kPa total pressure (dry air+water vapor) with a temperature of -40dec C & RH of 100%...>

It's about 2/5f(a)³.

What size is the room?

Thank you for the reply. The room volume is 10mX15mX15m

Could you explain what you've mentioned sir, "It's about 2/5f(a)^{3 "}

For water vapour pressure I use formula Pv = A*exp(B*T/(T + C)) where

A = 6.109, B = 17.288, C = 237.6, T in °C, VP in mbar.

At -14°C it gives 2.07 mbar = 1.57 mmHg. Not quite the tabulated value, formula is optimised for temperature range ~ 0 - 100°C. I have values of A, B and C for other temperature ranges but not to hand. I wouldn't swear it works at very low temperature, but I don't see why it wouldn't give a reasonable figure, and if so at -40°C it gives 0.185 mbar = 0.14 mmHg.

BTW what is Rv in your formula? I make mass of water vapour, kg

Pv(bar)*V*18/22.7*273/(273 + T) = 0.39 kg for your room.

(I imagine PW Slack was having a laugh!)

Thanks for your reply sir.

The formula i've written is the state equation of Ideal gas. i.e. m=PV/(RT). All values are for vapor and so subscript 'v'.

Rv is the Gas constant for water vapor.

I also have got one graph, which is matching with your values sir.

Thank you sir.

OK, so what do you make the vapour density and hence the mass in the room?

I don't get the units on your graph. If you multiply vertical figures by 10 it's about right (in mbar). At 20°C it's ~ 23 mbar saturated. But it says at top 10 x 10³.

I can barely read the equation, but it looks like e (or it might be ρ - density?) = 461.50*Pv*(T + 273.15). Rv is about 461.5.

Density = m/V = Pv/(Rv*T) in right units of course, and using T in K or °C properly.

Incidentally, the air pressure is irrelevant, it contributes to the total pressure but doesn't affect the mass of water vapour.