Question

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A mixture of gases contains 0.750 mol N2, 0.300 mol O2, and 0.150 mol CO2. If the total pressure of the mixture is 1.56 atm then

a. calculate the mole fraction for each gas

b. Calculate the partial pressure for each gas.

Answer 1

a)
n(N2),n1 = 0.75 mol
n(O2),n2 = 0.3 mol
n(CO2),n3 = 0.15 mol

Total number of mol = n1+n2+n3
= 0.75 + 0.3 + 0.15
= 1.2 mol

Mole fraction of each components are
X(N2) = n1/total mol
= 0.75/1.2
= 0.625

X(O2) = n2/total mol
= 0.3/1.2
= 0.25

X(CO2) = n3/total mol
= 0.15/1.2
= 0.125

b)
n(N2),n1 = 0.75 mol
n(O2),n2 = 0.3 mol
n(CO2),n3 = 0.15 mol

Total number of mol = n1+n2+n3
= 0.75 + 0.3 + 0.15
= 1.2 mol

Partial pressure of each components are
p(N2),p1 = (n1*Ptotal)/total mol
= (0.75 * 1.56)/1.2
= 0.975 atm

p(O2),p2 = (n2*Ptotal)/total mol
= (0.3 * 1.56)/1.2
= 0.39 atm

p(CO2),p3 = (n3*Ptotal)/total mol
= (0.15 * 1.56)/1.2
= 0.195 atm