Question

6. Find the partial pressures of the gases in a mixture with a total pressure of 101.3 kPa, if there are 7.8 mole of
Ng, 2.1 mole of 0, 0.090 mole of Ar, and 0.010 mol of Co,
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Answer 1

Answer:

Explanation:

Total mole of gases = 7.8 + 2.1 + .09 + .01 = 10

the partial pressures of the gases in a mixture

= mole fraction x Total pressure

mole fraction = mole of a gas in the mixture  / total mole

partial pressure of Ng = $\frac{7.8}{10} \times 101.3 kPa$

= 79.014 kPa

partial pressure of O = $\frac{2.1}{10} \times 101.3 kPa$

= 21.273 kPa .

partial pressure of  Ar = $\frac{.09}{10} \times 101.3 kPa$

= .9117 kPa .

partial pressure of  Co = $\frac{.01}{10} \times 101.3 kPa$

= .1013 k Pa .