Separation units often use membranes, absorbers, and other devices to reduce the mole fraction of selected con- stituents in gaseous mixtures. Consider a mixture of hydro- carbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. The mixture pressure before and after the separa- tion is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture

ΔP(ch4) = 6kPa

ΔP(c2h6) = 3kPa

ΔP(c3h8) = -9kPa

Explanation:

first, we find the factor that is used in the determination of mole fraction of the other components using the mole fraction of propane, both at the beginning and at the end. This means

β/(0.6+0.3+β) = 0.01

β = 0.009 + 0.01β

β - 0.01β = 0.009

0.99β = 0.009

β = 0.00909

the mole fractions of the methane and ethane at the end are

yCH4 = 0.6/(0.6 + 0.3 +0.00909)

yCH4 = 0.6/0.90909

yCH4 = 0.66

yC2H6 = 0.3/(0.6 + 0.3 + 0.00909

yC2H6 = 0.3/0.90909

yC2H6 = 0.33

Now, the change in partial pressure will be determined from the mole fraction at the beginning and at the end

ΔP(ch4) = (y2 - y1)ch4 * P

ΔP(ch4) = (0.66 - 0.6) * 100

ΔP(ch4) = 6kPa

ΔP(c2h6) = (y2 - y1)c2h6 * P

ΔP(c2h6) = (0.33 - 0.3) * 100

ΔP(c2h6) = 3kPa

ΔP(c3h8) = (y2 - y1)c3h8 * P

ΔP(c3h8) = (0.01 - 0.1) * 100

ΔP(c3h8) = -9kPa

Therefore

ΔP(ch4) = 6kPa

ΔP(c2h6) = 3kPa

ΔP(c3h8) = -9kPa

h let me think

its first and third is your answer i did the qustion the fourth isnt right