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:
\( \frac{x_A}{x_B}>\frac{y_A}{y_B} \)
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: