The order of decreasing rate of effusion for the given gases is [tex]\boxed{{\text{He}}>{\text{{\rm A}r}}>{{\text{C}}_3}{{\text{H}}_8}}[/tex].
Further explanation:
Graham’s law of effusion:
Effusion is the process by which molecules of gas travel through a small hole from high pressure to the low pressure. According to Graham’s law, effusion rate of a gas is inversely proportional to the square root of the molar mass of gas.
The expression for Graham’s law is as follows:
[tex]{\text{R}} \propto\frac{1}{\sqrt{\mu}}[/tex]
Here,
R is the rate of effusion of gas.
is the molar mass of gas.
Higher the molar mass of the gas, smaller will be the rate of effusion and vice-versa.
The rate of effusion of Ar is expressed as
[tex]{{\text{R}}_{{\text{Ar}}}} \propto\frac{1}{{\sqrt\mu}}_{{\text{Ar}}}}}}}[/tex] …… (1)
The rate of effusion of is expressed as
[tex]{{\text{R}}_{{{\text{C}}_3}{{\text{H}}_8}}} \propto \frac{1}{{\sqrt {{{{\mu }}_{{{\text{C}}_3}{{\text{H}}_8}}}}}}[/tex] …… (2)
The rate of effusion of He is expressed as
[tex]{{\text{R}}_{{\text{He}}}} \propto \frac{1}{{\sqrt {{{{\mu }}_{{\text{He}}}}}}}[/tex] …… (3)
The formula to calculate the molar mass of [tex]{{\text{C}}_3}{{\text{H}}_8}[/tex] is as follows:
[tex]{\text{Molar mass of }}{{\text{C}}_3}{{\text{H}}_8} = \left[ {3\left( {{\text{Atomic number of C}}} \right) + 8\left( {{\text{Atomic number of H}}} \right)} \right][/tex] …… (4)
The atomic mass of C is 12.01 g.
The atomic mass of H is 1 g.
Substitute these values in equation (4).
[tex]\begin{aligned}{\text{Molar mass of }}{{\text{C}}_3}{{\text{H}}_8}&=\left[ {3\left( {{\text{12 g}}} \right) + 8\left( {{\text{1 g}}} \right)} \right]\\&=36{\text{ g}}+{\text{8 g}}\\&={\text{44 g/mol}}\\\end{aligned}[/tex]
The molar mass of Ar is 40 g/mol.
The molar mass of He is 4 g/mol.
The molar mass of [tex]{{\text{C}}_3}{{\text{H}}_8}[/tex] is 44 g/mol.
The molar mass of He is the lowest so its rate of effusion will be the highest among the given gases.
The molar mass of [tex]{{\text{C}}_3}{{\text{H}}_8}[/tex] is the highest so its rate of effusion will be the lowest.
The molar mass of Ar is more than that of He but less than [tex]{{\text{C}}_3}{{\text{H}}_8}[/tex] so its rate of effusion is more than that of [tex]{{\text{C}}_3}{{\text{H}}_8}[/tex] and less than He.
So the order of decreasing rate of effusion for the given gases is as follows:
[tex]{\text{He}}>{\text{{\rm A}r}}>{{\text{C}}_3}{{\text{H}}_8}[/tex]
Learn more:
1. Which statement is true for Boyle’s law: https://brainly.com/question/1158880
2. Calculation of volume of gas: https://brainly.com/question/3636135
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ideal gas equation
Keywords: Effusion, rate of effusion, argon, C3H8, He, rate of effusion of He, rate of effusion of C3H8, rate of effusion of Ar, molar mass, Graham’s law, inversely proportional, square root, decreasing rate of effusion, order.
