Question

A mixture of 40.0g of O2 gas and 55.0g of CO2 gas in a rigid, closed container at 300K. The total pressure in the container is 3.15 atm. Calculate the partial pressure of each gas in the mixture in ppm and SI Units.

Answer 1

Solution.

$p = \frac{R \times T}{V}(n(O2) + n(CO2))$

$p(O2) = \frac{\frac{40}{32} \times 8.31 \times 300}{V}$

$p(CO2) = \frac{\frac{55}{44} \times 8.31 \times 300}{V}$

$3.15 \times 101325 = \frac{6232.5}{V}$

V = 0.0195 m3

$p(O2) = \frac{1.25 \times 8.31 \times 300}{0.0195} = 159586.88 \ Pa$

$p(CO2) = \frac{1.25 \times 8.31 \times 300}{0.0195} = 159586.88 \ Pa$

$w(O2) = w(CO2) = 50 \%$

$p(O2) = p(CO2) = 500000 \ ppm$

Answer:

p(O2) = 159586.88 Pa = 500000 ppm

p(CO2) = 159586.88 Pa = 500000 ppm