Understanding Adiabatic Compression

[Kurokari]

Hi, I have a little problem understanding adiabatic compression.

Let me start with the definition of adiabatic process from wikipedia, "In thermodynamics, an adiabatic process or an isocaloric process is a thermodynamic process in which the net heat transfer to or from the working fluid is zero."

My problem is, when we compress a certain gas in a closed container, we inject our kinetic energy to decrease the volume of the container, so shouldn't this means there is a net heat change, or a change in the total energy of the system?

or this kind of injection of energy does not categorize under heat transfer, I am quite confused.

I give my greatest thanks in advance! :)

 

[Doc Al]

My problem is, when we compress a certain gas in a closed container, we inject our kinetic energy to decrease the volume of the container, so shouldn't this means there is a net heat change, or a change in the total energy of the system?

When you do work on the gas, you definitely add energy. But that's not 'heat'. Heat is the flow of energy due to temperature difference.

Read this: https://www.physicsforums.com/showpost.php?p=1595186&postcount=1

 

[Kurokari]

Thank you for the link, it helped me a lot!

One more small thing, why is it that the rotation of a diatomic molecule along the atom-atom bond not counted as one of the degree of freedom?

 

[Ken G]

One more small thing, why is it that the rotation of a diatomic molecule along the atom-atom bond not counted as one of the degree of freedom?

That's actually a quantum mechanical effect. The kinetic energy for something undergoing cyclical motion (vibration or rotation) is characterized by mx²w², where x is the spatial size scale and w is the frequency. But here's where a big difference between vibrations and rotations appears-- for vibrations, x is a variable, and can be as large as it needs to be to get ~kT of energy into the mode in question. But for rotations, x is fixed by the size scale of the rotating object, the "lever arm" of the appropriate rotation. So when x is extremely small, as in the case you mention, it would require huge w to get kT of energy into that mode, of order w=(kT/m)^1/2x^-1. However, huge w, coupled with the quantum mechanical minimum action h, means you won't excite that mode, since here homework >> kT, because (kT/m)^1/2x^-1>> kT/h whenever x << h/(mkT)^1/2. So we only exite modes like that when T is very high, and it generally isn't that high in the applications you have in mind.