Question

Ethyl alcohol vapors and air are mixed. The partial pressure of the ethyl alcohol is 0.550 kPa. The pressure and temperature of the ideal gas mixture are 96.0 kPa and 25.0oC respectively. Calculate the mol fraction (PPM), the mass concentration (mg/m3) and the molar concentration (mol/m3) of the ethyl alcohol.

Answer 1

The partial pressure of ethanol in air is 0.550 kPa while the air pressure is 96.0 kPa, both at 25⁰C.

Therefore, mole fraction of ethanol = (partial pressure of ethanol)/(air pressure) = (0.550 kPa)/(96.0 kPa) = 5.729*10^-3 ≈ 5.73*10^-3.

Therefore, ppm_v (mole fraction in ppm) = 5.73*10^-3*10⁶ = 5730 (ans).

We need to convert the air pressure to atmosphere. Given, 1 kPa = 0.009869 atm, we have the air pressure as (96.0 kPa)*(0.009869 atm/1 kPa) = 0.9474 atm.

Find the moles of ethanol corresponding to 5730 ppm_v as

1 ppm_v = (1 m³ ethanol)/(1*10⁶ m³ air)

Therefore, 5730 ppm_v = (5730 m³ ethanol)/(1*10⁶ m³ air).

Moles of ethanol = (5730 m³ ethanol)/(1*10⁶ m³ air)*(0.9474 atm)/(0.000082 m³-atm/mol.K)*(298 K) = (0.2221 mole ethanol)/(1 m³ air)

Molar mass of ethanol, C₂H₅OH = 46 g/mol.

Concentration of ethanol in mg/m³ = (0.2221 mole/1 m³)*(46 g/1 mole)*(1000 mg/1 g) = 10216.6 mg/m³ (ans).

We have 0.221 mole ethanol; therefore, molar concentration of ethanol is (0.2221 mole)/(1 m³) = 0.2221 mol/m³ (ans).