In a mixture of liquids A and B, if the mole fractions of component A in vapour phase and liquid mixture are \(x_1\) and \(x_2\) respectively, then the total vapour pressure of liquid mixture is (where P\(_A^0\) and P\(_B^0\) are the vapour pressures of pure A and B)
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For ideal binary solutions, remember the key connection between liquid phase mole fraction (X) and vapor phase mole fraction (Y) for a component A:
\(P_A = Y_A P_{Total}\) (Dalton's Law) and \(P_A = X_A P_A^0\) (Raoult's Law).
Equating these gives \(Y_A P_{Total} = X_A P_A^0\). This allows you to solve for any of the four variables if the others are known.
This problem combines Raoult's Law and Dalton's Law of partial pressures.
Let \( P_T \) be the total vapour pressure of the liquid mixture.
Let \( X_A \) and \( X_B \) be the mole fractions in the liquid phase. From the question, \( X_A = x_2 \).
Let \( Y_A \) and \( Y_B \) be the mole fractions in the vapour phase. From the question, \( Y_A = x_1 \).
According to Raoult's Law, the partial pressure of a component in the vapour phase (\(P_A\)) is equal to the product of its mole fraction in the liquid phase (\(X_A\)) and the vapour pressure of the pure component (\(P_A^0\)).
\( P_A = X_A P_A^0 \).
Using the notation from the question, this is:
\( P_A = x_2 P_A^0 \). (Equation 1)
According to Dalton's Law, the partial pressure of a component in the vapour phase (\(P_A\)) is also equal to the product of its mole fraction in the vapour phase (\(Y_A\)) and the total vapour pressure (\(P_T\)).
\( P_A = Y_A P_T \).
Using the notation from the question, this is:
\( P_A = x_1 P_T \). (Equation 2)
Now we have two expressions for the partial pressure \(P_A\). We can equate them.
\( x_2 P_A^0 = x_1 P_T \).
We are asked to find the total vapour pressure, \(P_T\). We can rearrange the equation to solve for \(P_T\).
\( P_T = \frac{P_A^0 x_2}{x_1} \).
This matches option (C).
