The correct answer is (1) 0.33.
To find the mole fraction of substance A in the vapor phase when substances A and B have partial pressures in the ratio 1:2 and mole fractions in the ratio 1:2, you can use Dalton's law of partial pressures. According to Dalton's law, the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of individual gases.
Given:
\(P^o_A : P^o_B = 1 : 2\) (partial pressure ratio)
Mole fraction of A : Mole fraction of B = 1 : 2
Let's assume that the total pressure in the vapor phase is \(P_{\text{total}}\).
Now, we can write the partial pressures of A and B as follows:
Partial pressure of A (\(P_A\)) = Mole fraction of A (\(X_A\)) × Total pressure (\(P_{\text{total}}\))
Partial pressure of B (\(P_B\)) = Mole fraction of B (\(X_B\)) × Total pressure (\(P_{\text{total}}\))
We are given that \(P^o_A : P^o_B = 1 : 2\), which means the ratio of partial pressures of A to B is 1:2 when both are at their standard states. Therefore,
\(P_A^o : P_B^o = 1 : 2\)
Now, we can use the mole fraction ratios to express \(X_A\) and \(X_B\) in terms of each other:
\(X_A : X_B = 1 : 2\)
This means \(X_A = \frac{1}{3}\) and \(X_B = \frac{2}{3}\).
Now, we can use Dalton's law to relate the mole fractions to the partial pressures:
\(P_A = X_A \cdot P_{\text{total}} = \frac{1}{3} \cdot P_{\text{total}}\)
\(P_B = X_B \cdot P_{\text{total}} = \frac{2}{3} \cdot P_{\text{total}}\)
Since \(P_A^o : P_B^o = 1 : 2\), we can write:
\(\frac{P_A}{P_B} = \frac{P_A^o}{P_B^o} = \frac{1}{2}\)
Now, substitute the expressions for \(P_A\) and \(P_B\) in terms of \(P_{\text{total}}\):
\(\frac{\frac{1}{3} \cdot P_{\text{total}}}{\frac{2}{3} \cdot P_{\text{total}}} = \frac{1}{2}\)
Now, solve for \(P_{\text{total}}\):
\(\frac{1}{2} = \frac{\frac{1}{3} \cdot P_{\text{total}}}{\frac{2}{3} \cdot P_{\text{total}}}\)
\(\frac{1}{2} = \frac{1}{2}\)
This equation is satisfied, which means that the total pressure in the vapor phase (\(P_{\text{total}}\)) is not dependent on the ratio of partial pressures. It cancels out in the ratio calculation.
Now, we can calculate the mole fraction of A in the vapor phase (\(X_{A, \text{vapor}}\)):
\(X_{A, \text{vapor}} = \frac{P_A}{P_{\text{total}}} = \frac{\frac{1}{3} \cdot P_{\text{total}}}{P_{\text{total}}} = \frac{1}{3}\)
So, the mole fraction of substance A in the vapor phase is (1) \(0.33\) . |
