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?

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)  

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

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

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

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

it should be p1 v1 = p2 v2

6.4 cm (max) - 25 cm (min)