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

Aug 1, 2017

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

$P M = D R T$,

where:

• $P$ is pressure in $\text{bar}$.
• $M$ is molar mass in $\text{g/mol}$.
• $R = 0.083145$ is the universal gas constant in $\text{L"cdot"bar/mol"cdot"K}$.
• $T$ is temperature in $\text{K}$.
• Hence the density $D$ shall be in $\text{g/L}$.

And we obtain:

$\textcolor{b l u e}{D} = \frac{P M}{R T}$

$= \left(\left(\text{1 bar")("18.015 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("298.15 K}\right)\right)$

$=$ $\textcolor{b l u e}{\underline{\text{0.7267 g/L}}}$

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

$P = \frac{R T}{\overline{V} - b} - \frac{a}{\overline{V}} ^ 2$,

where:

• $\overline{V} = \frac{V}{n}$ is the molar volume.
• $a$ accounts for intermolecular forces, and shall be in units of ${\text{bar"cdot"L"^2//"mol}}^{2}$.
• $b$ is the excluded volume, and shall be in units of $\text{L/mol}$.

The van der Waals constants of water are $a = {\text{5.536 bar"cdot"L"^2//"mol}}^{2}$ and $b = \text{0.03049 L/mol}$. To solve for the mass density, we would have to rearrange this into the cubic molar volume form.

Get common denominators:

$P = \frac{R T {\overline{V}}^{2} - a \left(\overline{V} - b\right)}{\left(\overline{V} - b\right) \left({\overline{V}}^{2}\right)}$

Multiply through...

$P \left(\overline{V} - b\right) \left({\overline{V}}^{2}\right) = R T {\overline{V}}^{2} - a \left(\overline{V} - b\right)$

Distribute...

$P {\overline{V}}^{3} - b P {\overline{V}}^{2} = R T {\overline{V}}^{2} - a \overline{V} + a b$

Rearrange to get:

$\underline{P {\overline{V}}^{3} - \left(b P + R T\right) {\overline{V}}^{2} + a \overline{V} - a b = 0}$

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

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

In this case, ${x}_{n} = {\overline{V}}_{n}$, and $f \left({\overline{V}}_{n}\right)$ is the cubic equation shown above. The form of the ratio is:

$\frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)} = \frac{P {\overline{V}}^{3} - \left(b P + R T\right) {\overline{V}}^{2} + a \overline{V} - a b}{3 P {\overline{V}}^{2} - 2 \left(b P + R T\right) \overline{V} + a}$

So, to solve this, let:

• $\overline{V} = X = \text{take a guess number}$
• $P = 1$
• $b P + R T = 24.82017175$
• $a = 5.536$
• $a b = 0.16879264$

with the appropriate units. Thus, put the following into your TI calculator to solve for $\overline{V}$ at ${25}^{\circ} \text{C}$, or $\text{298.15 K}$:

$\left(X - \frac{{X}^{3} - 24.82017175 {X}^{2} + 5.536 X - 0.16879264}{3 {X}^{2} - 49.6403435 X + 5.536}\right) \to X$

• 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 $\text{24.5954 L/mol}$, or a mass density of $\textcolor{b l u e}{\underline{\text{0.7325 g/L}}}$.
• Starting at $x = 5$, I get $\text{0.188372 L/mol}$.
• Starting at $x = 20$, I get $\text{0.0364321 L/mol}$.

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

$D = M / \overline{V}$.)