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

Explanation:

Speed (c) will be calculated as follows.

                  c = $\frac{200 m/s + 250 m/s}{2}$ = 225 m/s

Speed range (dc) will be calculated as follows.

               dc = (250 - 200) $ms^{-1}$

                    = 50 $ms^{-1}$

Boltzmann constant = $1.38 \times 10^{-23} kg m^{2}/K s^{2}$

     T = 300 K,       mass of $CO_{2}$ molecule = $7.31 \times 10^{-26} kg$

Calculate the fraction of molecules in range dc as follows.

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

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

                              = 0.001917

Thus, we can conclude that fraction of $CO_{2}$ molecules present are 0.001917.