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Question: The partial pressure of ethane over a solution containing \[6.56 \times {10^{ - 3}}g\]of ethane is\[...

The partial pressure of ethane over a solution containing 6.56×10−3g6.56 \times {10^{ - 3}}gof ethane is1 bar1{\text{ }}bar. If the solution contains 5×10−2g5 \times {10^{ - 2}}g of ethane, then what shall be the partial pressure of the gas?

Explanation

Solution

The partial pressure of a gas is the individual property of a gas. It is the pressure exerted by a gas if the gas is contained or occupied alone over the same volume as the mixture.

Complete step by step answer: Henry's law is one of the gas laws which states that at a constant temperature, the amount of a given gas that dissolves in a volume of liquid will be directly proportional to the partial pressure of that gas in equilibrium with that liquid.
So following Henry's law, the solubility of gas in a liquid is directly proportional to the pressure of the gas.
P∝CP \propto C
Or P=H×CP = H \times C
Where PPis the partial pressure of the gas, HH is a constant called Henry solubility constant and CC is the solubility of the gas at definite temperature in a solvent.
In terms of mole fraction of a gas is directly proportional to the mass or concentration of the gas. The above equation in terms of mole fraction can also be written as:
χ=Kh×P\chi = {K_h} \times P
Where χ\chi the mole fraction of that gas, Kh{K_h} is Henry's volatility constant.
The mole fraction of ethane will be directly proportional to its mass of ethane (mm).
Thus
m=Kh×Pm = {K_h} \times P
For two different solutions the relation will be
m1P1=m2P2\dfrac{{{m_1}}}{{{P_1}}} = \dfrac{{{m_2}}}{{{P_2}}}
m1{m_1} and P1{P_1} are the mass and pressure of the first solution. m2{m_2} and P2{P_2} are the mass and pressure of the second solution.
Given,m1=6.56×10−3g{m_1} = 6.56 \times {10^{ - 3}}g ,P1=1 bar{P_1} = 1{\text{ }}bar
m2=5×10−2g{m_2} = 5 \times {10^{ - 2}}g, P2 = ?{P_2}{\text{ = }}?
Inserting in the equation,
6.56×10−31=5×10−2P2\dfrac{{6.56 \times {{10}^{ - 3}}}}{1} = \dfrac{{5 \times {{10}^{ - 2}}}}{{{P_2}}}
P2=5×10−26.56×10−3{P_2} = \dfrac{{5 \times {{10}^{ - 2}}}}{{6.56 \times {{10}^{ - 3}}}}
P2=7.62bar.{P_2} = 7.62bar.
Hence, the partial pressure of gas containing 5×10−2g5 \times {10^{ - 2}}g of ethane will be7.62bar7.62bar.

Note: Concentration, mass and mole fraction are the parameters of gas. They give the information about the amount of gas present in the solution. The sum of partial pressure of each component of gases present in a mixture is equal to the total pressure of the mixture.