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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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University of Michigan - Ann Arbor
Idaho State University
Boston College
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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