Question

Consider a mixture of hydrocarbons 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 separation is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture.

Answer 1

`\Delta P_m=6\text{kPa}\\\Delta P_e=3\text{kPa}\\\Delta P_p=-9\text{kPa}`

Step-by-step explanation:

mole fraction of propane after passing through the separator is
`\beta_p`

`(\beta)/(0.6+0.3+\beta)=0.01`

`\beta =9.09* 10^-^3`

mole fractions of ethane
`\beta _e` and methane
`\beta_m` after passing through separator are:

`\beta_e =(0.3)/(0.3+0.6+0.00909)=0.66\\\beta_m=(0.6)/(0.3+0.6+0.00909)=0.33`

Change in partial pressures then can be written as:

`\Delta P=(y_2-y_1)\cdot P` where
`y_2` and
`y_1` are mole fractions after and before passing through the separator

Hence,

`\Delta P_m=(0.66-0.6)\cdot 100\text{k}=6\text{kPa}\\\Delta P_e=(0.33-0.3)\cdot 100\text{k}=3\text{kPa}\\\Delta P_p=(0.01-0.1)\cdot 100\text{k}=-9\text{kPa}`