Question

[14] Use the Maxwell distribution of speeds to estimate the fraction of CO2 molecules at 300 K that have speeds in the range 200 to 250 m s Approximate the integral by calculating the area of the region as a rectangle whose height equals that of the full distribution at the center of the range.

Answer 1

Step-by-step explanation:

Speed (c) will be calculated as follows.

c =
`(200 m/s + 250 m/s)/(2)` = 225 m/s

Speed range (dc) will be calculated as follows.

dc = (250 - 200)
`ms^(-1)`

= 50
`ms^(-1)`

Boltzmann constant =
`1.38 * 10^(-23) kg m^(2)/K s^(2)`

T = 300 K, mass of
`CO_(2)` molecule =
`7.31 * 10^(-26) kg`

Calculate the fraction of molecules in range dc as follows.

`(dN)/(N) = 4 \pi c^(2) ((m)/(2 \pi K_(B)T))^{(3)/(2)} e^{(-mc^(2))/(2K_(B)T)} dc`

=
`4 * 3.14 * (225)^(2) ((7.31 * 10^(-26) kg)/(2 * 3.14 * 1.38 * 10^(-23) kg m^(2)/K s^(2) * 300 K))^{(3)/(2)} e^{(-7.31 * 10^(-26) kg * (225)^(2))/(2 * 1.38 * 10^(-23) kg m^(2)/K s^(2) * 300 K)} dc`

= 0.001917

Thus, we can conclude that fraction of
`CO_(2)` molecules present are 0.001917.