Question
At what temperature is the average speed of carbon dioxide molecules $(M=44.0 \mathrm{g} / \mathrm{mol}) 510 \mathrm{m} / \mathrm{s} ?$
Step 1
This is done by multiplying the given molar mass by $10^{-3}$ kg/g. So, $M = 44.0 \times 10^{-3} \, \text{kg/mol}$. Show more…
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Suppose that the average velocity $\left(v_{\text { rms }}\right)$ of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be $1.05 \times 10^{5} \mathrm{m} / \mathrm{s} .$ What temperature does this represent?
Suppose that the typical speed $\left(v_{\mathrm{rms}}\right)$ of carbon dioxide molecules (molar mass is $44.0 \mathrm{g} / \mathrm{mol}$ ) in a flame is found to be $1350 \mathrm{m} / \mathrm{s} .$ What temperature does this indicate?
The average speed (in meters per second) of a gas molecule is $$ v_{\mathrm{avg}}=\sqrt{\frac{8 R T}{\pi M}} $$ where $T$ is the temperature (in kelvins), $M$ is the molar mass (in kilo-grams per mole), and $R=8.31 .$ Calculate $d v_{\text { avg }} / d T$ at $T=300 \mathrm{K}$ for oxygen, which has a molar mass of 0.032 $\mathrm{kg} / \mathrm{mol} .$
DIFFERENTIATION
The Derivative as a Function
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