Ideal Gases — Question 4

Question 4

Gas density identifies a binary mixture composition

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. An ideal mixture has density $1.00\,\mathrm{kg/m^3}$ at 100 kPa and 300 K. It contains only species A and B with molar masses .0200 and .0400 kg/mol.

1. Infer its mean molar mass and mole fraction of B.

2. Find the partial pressure and mass fraction of B. Explain why its mass fraction exceeds its mole fraction.

Question 4 - Solution

Gas density identifies a binary mixture composition

(a) Method and calculation

$$\rho=\frac{pM_{\mathrm{mix}}}{RT}\ \Rightarrow\ M_{\mathrm{mix}}=\frac{1.00(8.314)(300)}{100000}=.024942\,\mathrm{kg/mol}.$$$$M_{\mathrm{mix}}=(1-x_B)(.0200)+x_B(.0400)\ \Rightarrow\ x_B=.247100.$$

(b) Interpretation and further calculation

$$p_B=x_Bp=24.7100\,\mathrm{kPa}.$$$$w_B=\frac{x_BM_B}{M_{\mathrm{mix}}}=\frac{.2471(.0400)}{.024942}\approx.396279.$$The heavier species carries more of the mass than of the molecules.

Physical check

An inferred molar mass outside the interval between the two species would reject the assumed mixture model.