
Vapour pressure of ${C_6}{H_6}$ and ${C_7}{H_8}$ are 119 mm and 37 mm Hg. Calculate the molar composition of ${C_6}{H_6}$ and ${C_7}{H_8}$ in a mixture having V.P of 80 mm of Hg. Also, calculate the vapor phase composition over the mixture:
Answer
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Hint: According to Raoult’s Law, the partial pressure of the gas present in the mixture is equal to vapour pressure of pure gas multiplied by the mole fraction. The composition of the gas is calculated as the vapor pressure of gas divided by the total pressure of gas present in the mixture.
Complete step by step answer:
The expression for Raoult’s law is given as shown below.
\[{P_A} = {X_A} \times {P_A}^0\]
\[{P_B} = {X_B} \times {P_B}^0\]
Where,
${P_A}$ and ${P_B}$ are the partial pressure of gas A and gas B.
${X_A}$ and ${X_B}$ are the mole fraction of gas A and gas B.
${P_A}^0$ and ${P_B}^0$ are the vapour pressure of pure gas A and B.
The total vapor pressure is the sum total of the individual partial pressure.
$P = {P_A} + {P_B}$
$P = {X_A}.{P_A}^0 + {X_B}.{P_B}^0......(i)$
Given,
${P_A}^0$ is 119mmHg
${P_B}^0$ is 37mmHg
The total pressure is 80 mmHg
Let us assume that the mole fraction of A gas is ${X_A}$ and ${X_B}$ = $1 - {X_A}$
Substitute the values in equation (i).
$80 = 119 \times {X_A} + 37 \times (1 - {X_A})$
$\Rightarrow {X_A} = 0.524$
Therefore, the solution contains 52.4% of A and 47.5% B.
In vapour pressure, pressure because of liquid A and B
${P_A} = 119 \times 0.524$
$\Rightarrow {P_A} = 61.88$
${P_B} = 37 \times 0.47$
$\Rightarrow 17.39$
The composition of the gas in vapour phase is calculated as shown below.
$A = \dfrac{{61.88}}{{61.88 + 17.39}}$
$\Rightarrow A = 0.779$
$B = \dfrac{{17.39}}{{61.88 + 17.39}}$
$\Rightarrow B = 0.221$
Therefore, 77.9% of vapor A and 22.2 % of vapor B is present in the mixture.
Note: Raoult’s law is similar to ideal gas but the only difference is that Rault’s law is implied to the solution. When the liquid present in the mixture does not have uniform attractive forces, then the solution deviates from Raoult’s law. It can show either positive deviation or negative deviation.
Complete step by step answer:
The expression for Raoult’s law is given as shown below.
\[{P_A} = {X_A} \times {P_A}^0\]
\[{P_B} = {X_B} \times {P_B}^0\]
Where,
${P_A}$ and ${P_B}$ are the partial pressure of gas A and gas B.
${X_A}$ and ${X_B}$ are the mole fraction of gas A and gas B.
${P_A}^0$ and ${P_B}^0$ are the vapour pressure of pure gas A and B.
The total vapor pressure is the sum total of the individual partial pressure.
$P = {P_A} + {P_B}$
$P = {X_A}.{P_A}^0 + {X_B}.{P_B}^0......(i)$
Given,
${P_A}^0$ is 119mmHg
${P_B}^0$ is 37mmHg
The total pressure is 80 mmHg
Let us assume that the mole fraction of A gas is ${X_A}$ and ${X_B}$ = $1 - {X_A}$
Substitute the values in equation (i).
$80 = 119 \times {X_A} + 37 \times (1 - {X_A})$
$\Rightarrow {X_A} = 0.524$
Therefore, the solution contains 52.4% of A and 47.5% B.
In vapour pressure, pressure because of liquid A and B
${P_A} = 119 \times 0.524$
$\Rightarrow {P_A} = 61.88$
${P_B} = 37 \times 0.47$
$\Rightarrow 17.39$
The composition of the gas in vapour phase is calculated as shown below.
$A = \dfrac{{61.88}}{{61.88 + 17.39}}$
$\Rightarrow A = 0.779$
$B = \dfrac{{17.39}}{{61.88 + 17.39}}$
$\Rightarrow B = 0.221$
Therefore, 77.9% of vapor A and 22.2 % of vapor B is present in the mixture.
Note: Raoult’s law is similar to ideal gas but the only difference is that Rault’s law is implied to the solution. When the liquid present in the mixture does not have uniform attractive forces, then the solution deviates from Raoult’s law. It can show either positive deviation or negative deviation.
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