# A mixture of propane and methane is contained in a vessel of unknown volume V at a temperature T and exerts a pressure of 320 mm Hg. The gas is burnt in excess O2 and all the carbon is recovered as CO2. The CO2 is found to have a pressure of 448 mm Hg in a volume V1 at the same temperature T. Calculate mole fraction of propane in mixture.

Q: A mixture of propane and methane is contained in a vessel of unknown volume V at a temperature T and exerts a pressure of 320 mm Hg. The gas is burnt in excess O2 and all the carbon is recovered as CO2. The CO2 is found to have a pressure of 448 mm Hg in a volume V1 at the same temperature T. Calculate mole fraction of propane in mixture.

Solution: Let the moles of propane (C3H8) = n1

Moles of methane CH4 = n2

We know that PV = nRT

320 × V = (n1 + n2) RT …(i)

after combustion

$\large C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2 O$

n1                               3n1

$\large CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2 O$

n2                             n2

total moles of CO2 formed = (3n1 + n2)

Once again we have PV = nRT

$\large 448 \times V = (3n_1 + n_2)\frac{R T}{V}$ ….(ii)

Dividing eqn (ii) by (i) we have

$\large \frac{448}{320} = \frac{3n_1 + n_2}{n_1 + n_2}$

$\large \frac{n_1}{n_2} = 0.25$

mol fraction of C3H8 is

$\large = \frac{n_1}{n_1 + n_2}$

$\large = \frac{0.25 n_2}{0.25 n_2 + n_2} = 0.2$