Question

Suppose that the average velocity (vrmsvrms) of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be 1.05×105m/s1.05×105m/s. What temperature does this represent?

Answer 1

Temperature is 19464105 K or 1.94×
`10^(7)`K

Step-by-step explanation:

Given data

mass=44
`(g)/(mol)`

Vrms=
`1.05*10^(5)(m)/(s)`

Temperatur=?

Solution

`Vrms=\sqrt{(3kT)/(m) }\\ k=1.38*10^(-23)(j)/(K)\\  Vrms^(2)=(3kT)/(m)\\  T=(mVrms^(2) )/(3k)`

`T=((44*((1kg)/(1000g) )*(1)/(6.02*10^(23)molecules ) )*(1.05*10^(5) ))/(3*1.38*10^(-23) )\\ T=19464105K\\T=1.9*10^(7) K`