Respuesta :
Answer:
2.02 times
Explanation:
According to Graham's law, the rate of effusion of two gases can be expresed as:
[tex]\frac{Rate_{1}}{Rate_{2}} =\sqrt{\frac{M_{2}}{M_{1}} }[/tex]
Where M are the atomic/molecular masses of the gases.
In this problem, the subscript 1 refers to O₂ gas (32 g/mol), and the subscript 2 refers to Xe gas (131.293 g/mol).
Putting the data in the equation gives us:
[tex]\frac{RateO_{2}}{RateXe}=\sqrt{\frac{131.293}{32} } =2.02[/tex]
So O₂ effuses at a rate that is 2.02 times faster than the rate of Xe.
[tex]{{\text{O}}_{\text{2}}}[/tex] (g) effuses at a rate that is [tex]\boxed{{\text{2}}{\text{.02}}}[/tex] times that of Xe (g) under the same conditions.
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, the 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]\boxed{\text{R}\propto \dfrac{1}{\sqrt{\mu}}}[/tex]
Here,
R is the rate of effusion of gas.
[tex]\mu[/tex] 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 [tex]{{\text{O}}_{\text{2}}}[/tex] is expressed as follows:
[tex]{{\text{R}}_{{{\text{O}}_{\text{2}}}}} \propto \dfrac{1}{{\sqrt {{{{\mu}}_{{{\text{O}}_{\text{2}}}}}} }}[/tex] ......(1)
Here,
[tex]{{\text{R}}_{{{\text{O}}_{\text{2}}}}}[/tex] is the rate of effusion of [tex]{{\text{O}}_{\text{2}}}[/tex].
[tex]{{{\mu }}_{{{\text{O}}_{\text{2}}[/tex] is the molar mass of [tex]{{\text{O}}_{\text{2}}}[/tex].
The rate of effusion of Xe is expressed as follows:
[tex]{{\text{R}}_{{\text{Xe}}}} \propto \dfrac{1}{{\sqrt {{{\mu }}_{{\text{Xe}}}}} }}[/tex]
......(2)
Here,
[tex]{{\text{R}}_{{\text{Xe}}}}[/tex] is the rate of effusion of Xe.
[tex]{{\mu }}_{{\text{Xe}}}}[/tex] is the molar mass of Xe.
On dividing equation (1) by equation (2),
[tex]\dfrac{{{{\text{R}}_{{{\text{O}}_{\text{2}}}}}}}{{{{\text{R}}_{{\text{Xe}}}}}}=\sqrt{\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{\mu }}_{{{\text{O}}_{\text{2}}}}}}}}[/tex] ......(3)
Rearrange equation (3) to calculate [tex]{{\text{R}}_{{{\text{O}}_{\text{2}}}}}[/tex].
[tex]{{\text{R}}_{{{\text{O}}_{\text{2}}}}}=\left( {\sqrt {\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{{\mu }}_{{{\text{O}}_{\text{2}}}}}}}} } \right){{\text{R}}_{{\text{Xe}}}}[/tex]
......(4)
The molar mass of [tex]{{\text{O}}_{\text{2}}}[/tex] is 32 g/mol.
The molar mass of Xe is 131.29 g/mol.
Substitute these values in equation (4).
[tex]\begin{aligned}{{\text{R}}_{{\text{Ne}}}}&=\left( {\sqrt {\frac{{{\text{131}}{\text{.29}}}}{{{\text{32}}}}} } \right){{\text{R}}_{{\text{Xe}}}}\\&= \left( {\sqrt {4.1028} } \right){{\text{R}}_{{\text{Xe}}}}\\&=2.02{{\text{R}}_{{\text{Xe}}}}\\\end{aligned}[/tex]
Therefore the rate of effusion of [tex]{{\mathbf{O}}_{\mathbf{2}}}[/tex] is 2.02 times the rate of effusion of Xe.
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, molar mass, O2, Xe, 2.02 times, Graham’s law, inversely proportional, square root.
