Question

If a mixture of total pressure 50 psia obeys Raoult's law and a species has a vapour presse of 20psia, what is the DePriester K-value for the species in the mixture?

Answer 1

Answer : The DePriester K-value for the species in the mixture is, 0.4

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

`p_i=X_i* p` ........(1)

According to the Raoult's law, the partial pressure exerted in gas phase by a component is equal to the product of the vapor pressure of that component and its mole fraction for an ideal liquid solution.

`p_i=Y_i* p_v` ........(2)

When the gas and the liquid are in equilibrium then these partial pressures must be the same.

`X_i* p=Y_i* p_v`

or,

`(X_i)/(Y_i)=(p_v)/(p)`

This ration is called as equilibrium ratio
`K_i` of the i-th component.

`K_i=(X_i)/(Y_i)=(p_v)/(p)` .......(3)

As we are given that,

Total pressure =
`p=50psia`

The vapor pressure =
`p_v=20psia`

According to the relation (3), we get

`K_i=(X_i)/(Y_i)=(p_v)/(p)=(20)/(50)=0.4`

Therefore, the DePriester K-value for the species in the mixture is, 0.4