Question

For diatomic carbon dioxide gas (CO₂, molar mass 44.0 g/mol) at T = 300 K, calculate (a) the most probable speed v_mp; (b) the average speed v_avg; (c) the root-mean-square speed v_rms

Short answer

The most probable speed$v_{mp}=336.729m/s$

The average speed $v_{avg}=379.958m/s$

The rms speed $v_{rms}=412.407m/s$

Step-by-step solution

Data and Equation

Given data;

Molar mass of carbon dioxide$m=44\times {10}^{3}kg/mol$

Temperature T = 300K

Gas constant $K=83.14JH/molK$

Formula;

The most probable speed

$v_{mp}=\sqrt{\frac{2Kt}{m}}$ .......... (1)

The average speed

$v_{avg}=\sqrt{\frac{8Kt}{\mathrm{\pi }m}}$ .......... (2)

The rms speed

$v_{rms}=\sqrt{\frac{3KT}{m}}$ .......... (3)

Find the most probable speed

Put value in equation (1)

$\begin{array}{r}{\mathrm{v}}_{\mathrm{mp}}=\sqrt{\frac{2\times 8.314\mathrm{J}/\mathrm{molJK}\times 300\mathrm{K}}{44\times {10}^{-3}\mathrm{kg}/\mathrm{mol}}}\\ =336.729\mathrm{m}/\mathrm{s}\end{array}$

Hence, the most probable speed$v_{mp}=336.729\mathrm{m}/\mathrm{s}$

Find the average speed

Put the value in equation (2)

$\begin{array}{r}{\mathrm{v}}_{\text{avg}}=\sqrt{\frac{8\times 8.314\mathrm{J}/\mathrm{molK}\times 300\mathrm{K}}{\pi \times 44\times {10}^{-3}\mathrm{kg}/\mathrm{mol}}}\\ =379.958\mathrm{m}/\mathrm{s}\end{array}$

Hence, the average speed$v_{avg}=379.958\mathrm{m}/\mathrm{s}$

Find the rms speed

$\begin{array}{r}{\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\times 8.314\mathrm{J}/\mathrm{molJK}\times 300\mathrm{K}}{44\times {10}^{-3}\mathrm{kg}/\mathrm{mol}}}\\ =412.407\mathrm{m}/\mathrm{s}\end{array}$

Hence, the rms speed$v_{rms}=412.407\mathrm{m}/\mathrm{s}$