To determine the relationship between the mole fractions of components A and B in both the liquid and vapor phases, we can use Raoult's Law for ideal solutions:
Raoult's Law: For a two-component system, the partial vapor pressure of each component is given by:
\( P_{A} = P_{A}^0 \times x_A \)
\( P_{B} = P_{B}^0 \times x_B \)
where \( P_{A}^0 \) and \( P_{B}^0 \) are the vapor pressures of the pure components A and B, respectively.
Total Vapor Pressure: The total vapor pressure of the solution is:
\( P_{total} = P_{A} + P_{B} = P_{A}^0 \times x_A + P_{B}^0 \times x_B \)
Mole Fraction in Vapor Phase: The mole fractions in the vapor phase can be calculated using Dalton's Law:
\( y_A = \frac{P_{A}}{P_{total}} = \frac{P_{A}^0 \times x_A}{P_{total}} \)
\( y_B = \frac{P_{B}}{P_{total}} = \frac{P_{B}^0 \times x_B}{P_{total}} \)
Comparison: We need to compare \( \frac{x_A}{x_B} \) with \( \frac{y_A}{y_B} \).
Given that \( P_{A}^0 = 350 \) mm Hg and \( P_{B}^0 = 750 \) mm Hg, and knowing that \( P_{B}^0 > P_{A}^0 \), it follows that component B is more volatile than component A. Therefore, in the vapor phase, the mole fraction of B will be greater as compared to A, leading to:
\( \frac{x_A}{x_B} > \frac{y_A}{y_B} \)
Thus, the correct answer is:
\( \frac{x_A}{x_B}>\frac{y_A}{y_B} \)
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).