Question

A) Find the gas speed of ethane at 200.0 degrees Celsius. ____________

Answer 1

a. 627.1 m/s

b.  the rate of effusion of ethane = 1.7 faster than hexane

Further explanation

Given

T = 200 + 273 = 473 K

Required

a. the gas speed

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

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 Ethane = 30 g/mol = 0.03 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 473}{30} }=627.1~m/s$

 

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

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass ethane =30

M₂ =  molar mass hexane = 86

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{86}{30} }=1.7$

the rate of effusion of ethane = 1.7 faster than hexane