# What is the density of water vapor at 25 C?

##### 1 Answer

Well, by assuming ideality, we can use the ideal gas law.

#PM = DRT# ,where:

#P# is pressure in#"bar"# .#M# is molar mass in#"g/mol"# .#R = 0.083145# is the universal gas constant in#"L"cdot"bar/mol"cdot"K"# .#T# is temperature in#"K"# .- Hence the density
#D# shall be in#"g/L"# .

And we obtain:

#color(blue)(D) = (PM)/(RT)#

#= (("1 bar")("18.015 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("298.15 K"))#

#=# #color(blue)ul"0.7267 g/L"#

If not assuming ideality, it can be estimated by the **van der Waals equation of state**:

#P = (RT)/(barV - b) - a/barV^2# ,where:

#barV = V/n# is the molar volume.#a# accounts for intermolecular forces, and shall be in units of#"bar"cdot"L"^2//"mol"^2# .#b# is the excluded volume, and shall be in units of#"L/mol"# .

The van der Waals constants of water are

Get common denominators:

#P = (RTbarV^2 - a(barV - b))/((barV - b)(barV^2))#

Multiply through...

#P(barV - b)(barV^2) = RTbarV^2 - a(barV - b)#

Distribute...

#PbarV^3 - bP barV^2 = RTbarV^2 - abarV + ab#

Rearrange to get:

#ul(PbarV^3 - (bP + RT) barV^2 + abarV - ab = 0)#

This is in general difficult to solve, but an iterative method exists called the **Newton-Raphson method**:

#x_(n+1) = x_n - (f(x_n))/(f'(x_n))#

In this case,

#(f(x_n))/(f'(x_n)) = (PbarV^3 - (bP + RT) barV^2 + abarV - ab)/(3PbarV^2 - 2(bP + RT) barV + a)#

So, to solve this, let:

#barV = X = "take a guess number"# #P = 1# #bP + RT = 24.82017175# #a = 5.536# #ab = 0.16879264#

with the appropriate units. Thus, put the following into your TI calculator to solve for

#(X - (X^3 - 24.82017175X^2 + 5.536X - 0.16879264)/(3X^2 - 49.6403435X + 5.536)) -> X#

Then, press enter until your answer converges. There are three "answers" you would get:

- One molar volume for the liquid phase
- One molar volume for the gas phase
- One molar volume in between them that makes no physical sense.

I got it to converge by taking some guesses...

- Starting at
#x = 0.05# , I get a molar volume of#"24.5954 L/mol"# , or a mass density of#color(blue)ul"0.7325 g/L"# . - Starting at
#x = 5# , I get#"0.188372 L/mol"# . - Starting at
#x = 20# , I get#"0.0364321 L/mol"# .

The first answer is for the gas, and that's all we're looking for here. (I got the density as

#D = M//barV# .)