I have a gas mixture (steam and air) contained in a tank. I know the tank pressure ($P$), temperature ($T$) and the massic air fraction ($x_a$).

I would like to determine the density of both gases to perform some calculations afterwards.

Intuitively, I would have done the following :

$$ \rho_{steam} = IAPWS97('rho\_PT', P,T) $$

i.e. asking IAPWS tables to provide me the density of steam at the tank conditions.

However, it seems based on available ressources that this is wrong. The correct way do do it would rather be

$$ \rho_{steam} = IAPWS97('rho\_PT', P_{steam},T) $$

Therefore taking into account the partial steam pressure to compute its density within the tank rather than the total tank pressure

Why is that ? Intuitively, I would use total tank pressure because it seems to me that both steam and air experiences the total tank pressure and not only their partial pressure and therefore I would use total tank pressure to compute their properties.


1 Answer 1


Considering that air and steam behave as ideal gases, you can write the following equations: $$ P V = nRT $$ $$ P_A V = n_A RT$$ $$ P_W V = n_W RT $$ where I use subscripts $_A$ and $_W$ for air and steam respectively. Taking $M_A$ and $M_W$ as the molar masses of air and water, the density of each of the gases would be $$ P_A V = (m_A/M_A) RT \rightarrow \underbrace{(m_A/V)}_{\rho_A} =\frac{P_A M_A}{RT} $$
$$ P_W V = (m_W/M_W) RT \rightarrow \underbrace{(m_W/V)}_{\rho_W} =\frac{P_W M_W}{RT} $$

Hence if you want to find the density of each of the gases, you only care of the partial pressure because it relates with the amount of matter of each of the gases.

If you wanted to find the density of the mixture of gases, you would use the total pressure and an "equivalent molar mass" for the mixture

  • $\begingroup$ Thanks. Steam is not an ideal gas but I guess your argument still proves that density of one gas relates to its partial pressure and not the total pressure $\endgroup$ Aug 8, 2017 at 12:47
  • 1
    $\begingroup$ Yes, steam is not an ideal gas as it can condense but for a relatively large range of pressure and temperature, the volumetric properties of steam are pretty well described by the ideal gas law. $\endgroup$
    – Toulousain
    Aug 8, 2017 at 13:22
  • $\begingroup$ I seem to remember the "dryness fraction" coming into play here (1 - x) but don't have access to those notes at the moment... $\endgroup$
    – Solar Mike
    Aug 8, 2017 at 15:36
  • $\begingroup$ @SolarMike IIRC dryness factor comes into play when near the condensation temperature for the humidity level in question. $\endgroup$ Aug 8, 2017 at 18:07
  • $\begingroup$ @CarlWitthoft I was measuring steam boiler output at 170 Deg C. $\endgroup$
    – Solar Mike
    Aug 8, 2017 at 18:10

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