a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
[tex]\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}[/tex]
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
[tex]\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s[/tex]
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
[tex]\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }[/tex]
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
[tex]\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5[/tex]
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
