# Question #83193

Sep 8, 2017

If the percentage composition is by volume then the ratio of number of moles of component gases will be ratio of their respective volume.
Hence the ratio of number of moles of ${H}_{2} , {O}_{2} \mathmr{and} H e$ will be as follows

${n}_{{H}_{2}} : {n}_{{O}_{2}} : {n}_{H e} = 20 : 15 : 65 = 4 : 3 : 13$

So mole fractions of component gases

${X}_{{H}_{2}} = \frac{4}{4 + 3 + 13} = \frac{1}{5} = 0.2$

${X}_{{O}_{2}} = \frac{3}{4 + 3 + 13} = \frac{1}{20} = 0.15$

${X}_{H e} = \frac{65}{4 + 3 + 13} = \frac{13}{20} = 0.65$

And the partial pressures of component gases

${p}_{{H}_{2}} = {X}_{{H}_{2}} \times P = 0.2 \times 4 a t m = 0.8 a t m$

${p}_{{O}_{2}} = {X}_{{O}_{2}} \times P = 0.15 \times 4 a t m = 0.6 a t m$

${p}_{H e} = {X}_{H e} \times P = 0.65 \times 4 a t m = 2.6 a t m$

When the percentage composition is by weight then the ratio of number of moles of component gases ${H}_{2} , {O}_{2} \mathmr{and} H e$ will be as follows

${n}_{{H}_{2}} : {n}_{{O}_{2}} : {n}_{H e}$

$= \frac{20}{\text{molar mass of " H_2):15/("m. mass of " O_2):65/("m.mass of } H e}$

$= \frac{20}{2} : \frac{15}{32} : \frac{65}{4}$

$= \frac{4}{2} : \frac{3}{32} : \frac{13}{4}$

$= 64 : 3 : 104$

Now we are to proceed similar way as above

And the partial pressures of component gases

${p}_{{H}_{2}} = {X}_{{H}_{2}} \times P = \frac{64}{171} \times 4 a t m \approx 1.5 a t m$

${p}_{{O}_{2}} = {X}_{{O}_{2}} \times P = \frac{3}{171} \times 4 a t m \approx 0.07 a t m$

${p}_{H e} = {X}_{H e} \times P = \frac{104}{171} \times 4 a t m \approx 2.43 a t m$
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