## Related questions with answers

Calculate the rms speeds of CO and

$Cl_2$

molecules at 300 K.

Solutions

VerifiedIn this exercise, we need to determine the rms speed of CO and Cl_{2} molecules at 300 K.

We need to calculate for the rms speed of CO and Cl$_2$ in a given temperature.

#### (b)

Let us calculate the root mean square speed of $\mathrm{CO}$ and $\mathrm{Cl_2}$ molecules at $T = 300 \mathrm{K}$.

The gas constant $R = 8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} \times \frac{1 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}}{1 \mathrm{J}} = 8.314 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \cdot \mathrm{mol} \cdot \mathrm{K}$

From part a), we know that the molar mass of

$\begin{align*} M_{CO} &= 28.01 \mathrm{g/mol} =28.01 \cdot 10^{-3} \mathrm{kg/mol} \\ M_{Cl_2} &= 70.906 \mathrm{g/mol} = 70.906 \cdot 10^{-3} \mathrm{kg/mol}\\ \end{align*}$

Now we can calculate the root mean square speed

$\begin{align*} \mu_{rms-CO} &= \sqrt { \frac { 3RT } { M_{CO} } }\\ &= \sqrt { \frac { 3 \cdot 8.314 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \cdot \mathrm{mol} \cdot \mathrm{K} \cdot 300 \mathrm{K} } { 28.01 \cdot 10^{-3} \mathrm{kg/mol} } }\\ &= {\color{#4257b2} 516.86 \mathrm{m/s}}\\ \ \\ \mu_{rms-Cl_2} &= \sqrt { \frac { 3RT } { M_{Cl_2} } }\\ &= \sqrt { \frac { 3 \cdot 8.314 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} \cdot \mathrm{mol} \cdot \mathrm{K} \cdot 300 \mathrm{K} } { 70.906 \cdot 10^{-3} \mathrm{kg/mol} } }\\ &= {\color{#4257b2} 324.85 \mathrm{m/s}} \end{align*}$

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