Question

Find the average molecular weight M and specific constant R for air saturated with water vapor at 0oC and 1 atm of total pressure. The vapor pressure of water at 0 degreeC is 6.11mb.

Answer 1

$P_{sat} = 6.11\; mb = 6.11 \times 10^{-3} \;b = 6.0301 \times 10^{-3} \;atm$

Total pressure $P_T = 1\; atm$

The molar fraction of each gas $N_i =\frac{n_i}{n} = \frac{p_i}{p}$

$N_W = \frac{6.0301 \times 10^{-3} }{1} = 6.0301 \times 10^{-3}$

The mean molecular weight of the mixture $M_m$ is defined by

$M_m = \frac{Σn_iM_i}{n} = ΣN_iM_i$

$M_m = 18.015\times(6.0301 \times 10^{-3}) + 28.97\times (1-6.0301 \times 10^{-3})$

$M_m=28.903\; g/mol$

Specific constant $R_i=\frac{R}{M_M}$

R is the universal gas constant 8314 J/kg K

$R_i=\frac{8314}{28.903} = 287.65\; J/kg \;K$

Answer: 28.903 g/mol and 287.65 J/kg K