Respuesta :
The rate of effusion of a gas is inversely proportion to the to the square root of the mass of its particles.
the ratio of effusion of hydrogen gas to krypton gas is calculated as follows
Re of hydrogen/ Re of krypton = square root of molar mass of krypton/molar mass of hydrogen
that is sqrt of 83.793/2 g/mol =6.47
rate of hydrogen =6.47x rate of krypton
the ratio of effusion of hydrogen gas to krypton gas is calculated as follows
Re of hydrogen/ Re of krypton = square root of molar mass of krypton/molar mass of hydrogen
that is sqrt of 83.793/2 g/mol =6.47
rate of hydrogen =6.47x rate of krypton
The ratio of effusion rate of hydrogen gas to the effusion rate of krypton gas is [tex]\boxed{6.447}[/tex].
Further explanation:
Graham’s law of effusion:
This law describes the relationship between rate of effusion and molar mass of gases. Effusion is the process of motion of gas particles through a small hole from region of high pressure to region of low pressure. According to this law, the effusion rate of a gas is inversely related to the square root of the molar mass of gas.
The expression for Graham’s law is written below.
[tex]{\text{R}} \propto \dfrac{1}{{\sqrt {{\mu }} }}[/tex]
Where,
R is the rate of effusion of gas.
[tex]{\mu }}[/tex] is the molar mass of gas.
The expression for rate of effusion of is as follows:
[tex]{{\text{R}}_{{{\text{H}}_2}}} \propto \dfrac{1}{{\sqrt {{{\mu }}_{{{\text{H}}_2}}}} }}[/tex] …… (1)
Where,
[tex]{{\text{R}}_{{{\text{H}}_{\text{2}}}}}[/tex] is the rate of effusion of [tex]\text{H}_{2}[/tex].
[tex]{{\mu }}_{{{\text{H}}_{\text{2}}}}}[/tex] is the molar mass of [tex]\text{H}_{2}[/tex].
The expression for rate of effusion of Kr is as follows:
[tex]{{\text{R}}_{{\text{Kr}}}} \propto \dfrac{1}{{\sqrt {{{\mu }}_{{\text{Kr}}}}} }}[/tex] …… (2)
Where,
[tex]{{\text{R}}_{{\text{Kr}}}}[/tex] is the rate of effusion of Kr.
[tex]{{\mu }}_{{\text{Kr}}}}[/tex] is the molar mass of Kr.
Dividing equation (1) by equation (2), we get:
[tex]\dfrac{{{{\text{R}}_{{{\text{H}}_2}}}}}{{{{\text{R}}_{{\text{Kr}}}}}} = \sqrt {\dfrac{{{\mu }}_{{\text{Kr}}}}{{{{\mu }}_{{{\text{H}}_2}}}}}}[/tex] ...... (3)
Molar mass of [tex]\text{H}_{2}[/tex] is 2.01588 g/mol.
Molar mass of Kr is 83.798 g/mol.
Substitute these values in equation (3).
[tex]\begin{aligned}\dfrac{{{{\text{R}}_{{{\text{H}}_2}}}}}{{{{\text{R}}_{{\text{Kr}}}}}} &= \sqrt {\frac{{{\text{83}}{\text{.798 g/mol}}}}{{2.01588{\text{ g/mol}}}}}\\&= 6.447 \\\end{aligned}[/tex]
Learn more:
- Which statement is true for Boyle’s law: https://brainly.com/question/1158880
- 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, Graham’s law of effusion, molar mass, H2, Kr, 6.447, 2.01588 g/mol, 83.798 g/mol.
