Question

A $1mol$ of an ideal gas $A$ with $300mm$ of $Hg$ is separated by $2mol$ of another ideal gas $B$ with $30mm$ of $Hg$ in a closed container at the same temperature. If the separation is removed then total pressure?
1.$200mm$ of $Hg$
2.$300mm$ of $Hg$
3.$500mm$ of $Hg$
4.$600mm$ of $Hg$

Answer 1

This question gives the knowledge about Raoult's law. Raoult’s law is defined as the vapor pressure of any component at a defined temperature is equivalent to the mole fraction of the component present in the solution which is multiplied to the vapor pressure of the component in the uncontaminated state.

Formula used: The formula used to determine the pressure using Raoult’s law is as follows:
${P_t} = {X_A}P_A^0 + {X_B}P_B^0$
Where ${P_t}$ is the total pressure, ${X_A}$ is the mole fraction of component $A$ , ${X_B}$ is the mole fraction of component $B$, $P_A^0$ is the pressure of pure component $A$ and $P_B^0$ is the pressure of pure component $B$.

Complete step-by-step answer: According to Raoult’s law, the formula used is as follows:
$\Rightarrow {P_t} = {X_A}P_A^0 + {X_B}P_B^0$
Substitute ${P_t}$ as $500mm$ of $Hg$, ${X_A}$ as $\left( {\dfrac{1}{{1 + 2}}} \right)$ and ${X_B}$ as
$\left( {\dfrac{2}{{1 + 2}}} \right)$
$\Rightarrow 500 = \left( {\dfrac{1}{{1 + 2}}} \right)P_A^0 + \left( {\dfrac{2}{{1 + 2}}} \right)P_B^0$
On simplifying, we get
$\Rightarrow 500 = \left( {\dfrac{1}{3}} \right)P_A^0 + \left( {\dfrac{2}{3}} \right)P_B^0$
Consider this as an equation $1$ .
After adding $1mol$to the component $B$, we have ${X_A}$ as $\left( {\dfrac{1}{{1 + 3}}} \right)$ and ${X_B}$ as $\left( {\dfrac{3}{{1 + 3}}} \right)$ .
According to Raoult’s law, the formula used is as follows:
$\Rightarrow {P_t} = {X_A}P_A^0 + {X_B}P_B^0$
Substitute ${P_t}$ as $525mm$ of $Hg$, ${X_A}$ as $\left( {\dfrac{1}{{1 + 3}}} \right)$ and ${X_B}$ as
$\left( {\dfrac{3}{{1 + 3}}} \right)$
$\Rightarrow 525 = \left( {\dfrac{1}{{1 + 3}}} \right)P_A^0 + \left( {\dfrac{3}{{1 + 3}}} \right)P_B^0$
On simplifying, we get
$\Rightarrow 525 = \left( {\dfrac{1}{4}} \right)P_A^0 + \left( {\dfrac{3}{4}} \right)P_B^0$
Consider this as an equation $2$ .
On multiplying equation $1$ with $3$ , we get
$\Rightarrow 1500 = P_A^0 + 2P_B^0$
Consider this as an equation $3$ .
On multiplying equation $2$ with $4$ , we get
$\Rightarrow 2100 = P_A^0 + 3P_B^0$
Consider this as an equation $4$ .
On subtracting equation$4$ from equation $3$, we get
$\Rightarrow - 600 = - P_B^0$
Therefore, the pressure of pure component $B$ is $600mm$ of $Hg$.
Substitute $P_B^0$ as $600$in equation $3$
$\Rightarrow 1500 = P_A^0 + 2\left( {600} \right)$
On simplifying, we get
$\Rightarrow 300 = P_A^0$
Therefore, the pressure of pure component $A$ is $300mm$ of $Hg$.

Hence, option $2$ is correct.

Note: Raoult’s law is the special case of Henry’s law. Raoult’s law is basically the vapor pressure of any component at a particular temperature which is equivalent to the product of mole fraction of the component present in the solution and to the vapor pressure of the component in the unadulterated state.