Temperature: Measure of random kinetic energy of molecules
Here's what I'm trying to say we have two book, and we place on book on moving truck and other on ground, so according to definition our instructor gave, shouldn't kinetic energy of particles increase? as "Molecules" are the part of "System"(book)
There is work done on system by truck, so kinetic energy of the system should increase and molecules are part of system so should their kinetic energy, if there kinetic energy increases then then temperature should also increase.
Might sound like obvious question, but I couldn't explain this
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Qmechanic
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2Does this answer your question? Why does the temperature of a gas inside a moving container not increase with velocity? – BioPhysicist Sep 08 '21 at 02:23
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1Or this one – BioPhysicist Sep 08 '21 at 02:24
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2Or also this one – BioPhysicist Sep 08 '21 at 02:25
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If temperature depended on non-random KE, which book should be hotter, the one on the truck, or the one on the ground? Remember, velocity is relative, and each book is stationary in its own frame. – PM 2Ring Sep 08 '21 at 10:11
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You actually answered the question in the question: "random kinetic energy", so at rest the book has a distribution of velocities $\vec v_i$ for $i=(1, O(N_A))$ such that $T\propto \langle ||v_i||^2\rangle$, or the average random kinetic energy.
Since the velocities are random, they are uncorrelated:
$$\langle \vec v_i\cdot \vec v_j\rangle =0\ \ \ \ \ i\ne j$$
Now accelerate the book on the truck to $v_{\rm truck}$. Then
$$ \langle \vec v_i\cdot \vec v_j\rangle = v^2_{\rm truck}\ \ \ \ \ i\ne j$$
The velocities are correlated, which is not random, and that needs to be subtracted out in order to compute the temperature from the microscopic coordinate velocities. Of course, subtracting it out means measuring them in their shared rest frame where:
$$\sum_i \vec v_i = 0$$
JEB
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