Question

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

b) What is the rate of effusion of sulfur dioxide compared to nitrogen triodide at the same temperature? __________________

Answer 1

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

Further explanation

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)

`\large {\boxed {\bold {v_ {rms} = \sqrt {\frac {3RT} {Mm}}}}`

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

`\tt v=\sqrt{(3* 8.314* 373)/(0.064) }\\\\v=381.27~m/s`

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

`\rm (r_1)/(r_2)=\sqrt{(M_2)/(M_1) }`

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

`\tt (r_1)/(r_2)=\sqrt{(395)/(64) }=(20)/(8)=2.5`

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide