Question

The total pressure observed by mixing two liquids A and B is 350 , mm Hg when their mole fractions are 0.7 and 0.3 respectively The total pressure becomes 410 ,mm ,Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B The vapour pressure of pure A is _________mm ,Hg (Nearest integer) Consider the liquids and solutions behave ideally.

Hint:

Use Raoult's law: \(P_{\text{total}} = P_A^0 X_A + P_B^0 X_B\) for ideal solutions. Solve simultaneous equations for unknown vapour pressures.

Answer 1

Correct Answer: 314

To solve this problem, we apply Raoult's law, which is valid for ideal solutions. According to Raoult's law, the total vapor pressure \(P\) of a solution can be computed as:

\(P = x_A \cdot P_A^0 + x_B \cdot P_B^0\) where:

• \(x_A\) and \(x_B\) are the mole fractions of components A and B.
• \(P_A^0\) and \(P_B^0\) are the vapor pressures of pure liquids A and B.

We have the following two equations from the problem:

1. For \(x_A = 0.7\) and \(x_B = 0.3\), total pressure \(P = 350\) mm Hg:

\(0.7 \cdot P_A^0 + 0.3 \cdot P_B^0 = 350\)

2. For \(x_A = 0.2\) and \(x_B = 0.8\), total pressure \(P = 410\) mm Hg:

\(0.2 \cdot P_A^0 + 0.8 \cdot P_B^0 = 410\)

We solve these simultaneous equations to find \(P_A^0\) and \(P_B^0\).

Multiply equation (1) by 4 and equation (2) by 7 to eliminate \(P_B^0\):

Equation (1):

\((4 \cdot 0.7) \cdot P_A^0 + (4 \cdot 0.3) \cdot P_B^0 = 4 \cdot 350\)

\(2.8 \cdot P_A^0 + 1.2 \cdot P_B^0 = 1400\)

Equation (2):

\((7 \cdot 0.2) \cdot P_A^0 + (7 \cdot 0.8) \cdot P_B^0 = 7 \cdot 410\)

\(1.4 \cdot P_A^0 + 5.6 \cdot P_B^0 = 2870\)

Subtract equation (1) from equation (2):

\((1.4 - 2.8) \cdot P_A^0 + (5.6 - 1.2) \cdot P_B^0 = 2870 - 1400\)

\(-1.4 \cdot P_A^0 + 4.4 \cdot P_B^0 = 1470\)

Re-arranging:

\(4.4 \cdot P_B^0 = 1.4 \cdot P_A^0 + 1470\)

\(P_B^0 = \frac{1.4 \cdot P_A^0 + 1470}{4.4}\)

Substitute \(P_B^0\) back into equation (1):

\(0.7 \cdot P_A^0 + 0.3 \cdot \left(\frac{1.4 \cdot P_A^0 + 1470}{4.4}\right) = 350\)

Simplifying:

\(0.7 \cdot P_A^0 + \frac{0.42 \cdot P_A^0 + 441}{4.4} = 350\)

\(0.7 \cdot P_A^0 + \frac{0.42 \cdot P_A^0 + 441}{4.4} = 350\)

\(0.7 \cdot P_A^0 \cdot 4.4 + 0.42 \cdot P_A^0 + 441 = 1540\)

\(3.08 \cdot P_A^0 + 0.42 \cdot P_A^0 = 1540 - 441\)

\(3.5 \cdot P_A^0 = 1099\)

\(P_A^0 = \frac{1099}{3.5}\)

\(P_A^0 ≈ 314\) mm Hg.

The calculated vapor pressure of pure A, \(314\) mm Hg, satisfies the expected range of 314,314. Therefore, the vapor pressure of pure A is \(314\) mm Hg.