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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?

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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

User Roykasa
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