Question

The partial pressure of ethane over a solution of ${\text{6}}{{.56 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{g}}$ ethane is 1 bar. If the solution contains ${\text{5}}{{.00 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{g}}$of ethane then what shall be the partial pressure of the gas?

Answer 1

In the above question, we have to find the partial pressure of ethane when mass of the ethane is given. Since we don’t have any direct relationship between partial pressure and mass of the substance, we will be using Henry’s law which gives us a relationship between partial pressure and mole fraction of the gas.
Formula Used-
${\text{p = }}{{\text{K}}_{\text{H}}}{{\chi }}$
where p = partial pressure of the gas
${{\text{K}}_{\text{H}}}$= Henry’s Law constant
${{\chi }}$= mole fraction of the gas

Complete step by step answer:
From Henry's Law, we know that the pressure of the gas in the vapour state (p) is directly proportional to the mole fraction of the gas (${{\chi }}$) in the solution. So,
${\text{p}} \propto {{\chi }}$
${\text{p = }}{{\text{K}}_{\text{H}}}{{\chi }}$ (equation 1)
We know that mole fraction can be expressed as:
${{\chi = }}\dfrac{{\text{m}}}{{\text{M}}}$
where ${{\chi }}$=mole fraction of the gas, m= given mas and M= Molar mass of the gas
So replacing the value of ${{\chi }}$in the equation 1 we get:
${\text{p = }}\dfrac{{{{\text{K}}_{\text{H}}}{\text{m}}}}{{\text{M}}}$
Taking m to the left hand side of the equation we get:
$\dfrac{{\text{p}}}{{\text{m}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{H}}}}}{{\text{M}}}$
Since, ${{\text{K}}_{\text{H}}}$is the Henry’s Law constant and M is the molar mass of the ethane which is constant, so we will always get $\dfrac{{\text{p}}}{{\text{m}}}$ ratio constant
Now, if $p_1$= partial pressure of ethane in first solution
$m_1$= mass of ethane in the first solution
$p_2$= partial pressure of ethane in second solution
$m_2$= mass of ethane in the second solution
So, we get:
$\dfrac{{{{p_1}}}}{{{{m_1}}}}{\text{ = }}\dfrac{{{{p_2}}}}{{{{m_2}}}}$
Placing the values given in the question, we get:
$\dfrac{{\text{1}}}{{{\text{6}}{{.56 \times 1}}{{\text{0}}^{{\text{ - 3}}}}}}{\text{ = }}\dfrac{{{{p_2}}}}{{{\text{5}}{{.00 \times 1}}{{\text{0}}^{{\text{ - 2}}}}}}$
By cross-multiplication, we get
$\dfrac{{{\text{5}}{{.00 \times 1}}{{\text{0}}^{{\text{ - 2}}}}}}{{{\text{6}}{{.56 \times 1}}{{\text{0}}^{{\text{ - 3}}}}}}{{ = p_2}}$
${{p_2 = 7}}{\text{.62 bar}}$
$\therefore$The partial pressure of ethane in the second solution is $7.62$ bar.

Note:
Henry’s Law is not a universal law, it has some exceptions. So, it is not applicable when:
The pressure of gas is too high or the temperature is too low.
The gas should undergo any changes in the solution.