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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 $\mathrm{kg} / \mathrm{mol} .$

0.742

Calculus 1 / AB

Chapter 3

DIFFERENTIATION

Section 2

The Derivative as a Function

Derivatives

Differentiation

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Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

04:28

$\bullet$ Oxygen $\left(\m…

09:02

For diatomic carbon dioxid…

03:20

06:10

The average speed of molec…

03:22

The average molecular velo…

02:16

For diatomic cation dioxid…

03:42

Show that the rms speed of…

01:10

The most probable speed fo…

02:39

a) At what temperature do …

00:54

At what temperature is the…

all right. So taking the derivative with respects to tee off eight or divided by pi and to the power over 1/2 multiplied by t to the power of one house and just think driven with respect ity of all this, we end up with a part of it. But I am to the power of 1/2 multiplied by 1/2 times T to the power of negative 1/2. So substitute given value or T, is equal to 300 cakes to end up when 25.71 multiplied by 0.0 point 0289 and this is approximately equal to 0.742

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