Molecular work in an Adiabatic process ?

[karen_lorr]

Molecular "work" in an Adiabatic process ?

I have studied Adiabatic cooling in connection with a rising parcel of air. I’m confused. I understand that temperature is a macroscopic principle (not a Microscopic); molecules do not have a temperature, only velocity and Kinetic energy. I have read that there is no heat loss (heat being the amount of work available from a system)

I have read the molecules in the packet are doing “work” as the air parcel expands – what work are they doing ?

If the total energy remains the same (1st law T’m’d) surely the molecules are expending energy?

I have read so many pages concerning the adiabatic process that you’d think I’d have it by now, but I just confused.

Can anyone give any insights into this topic?

Thank you

 

[LostConjugate]

I have studied Adiabatic cooling in connection with a rising parcel of air. I’m confused. I understand that temperature is a macroscopic principle (not a Microscopic); molecules do not have a temperature, only velocity and Kinetic energy. I have read that there is no heat loss (heat being the amount of work available from a system)

I have read the molecules in the packet are doing “work” as the air parcel expands – what work are they doing ?

If the total energy remains the same (1st law T’m’d) surely the molecules are expending energy?

I have read so many pages concerning the adiabatic process that you’d think I’d have it by now, but I just confused.

Can anyone give any insights into this topic?

Thank you

You are losing heat, just not to your surroundings. Adiabatic cooling just means to reduce the temp by reducing the pressure rather than through heat transfer (transfer of momentum between molecules).

With cooling the molecules do work to expand the size of the parcel which decreases the temp. With heating external work is done to increase the pressure which increases the temp.

 

[Andrew Mason]

...
I have read the molecules in the packet are doing “work” as the air parcel expands – what work are they doing ?

Just to add to what LostConjugate has said, the expanding gas does work on the surrounding air mass because it has to push the surrounding air over some distance: This work is W = \int PdV (= \int (F/A)Ads = \int Fds). Since there is no heat flow into or out of the air, this work comes at the expense of its internal energy: dQ = dU + dW. dQ = 0 -> dU = - dW. So W = -nCv\Delta T

If the total energy remains the same (1st law T’m’d) surely the molecules are expending energy

The total energy of the expanded gas does not remain the same. It decreases. The molecules in the original sample expend some of their kinetic energy to do work on the surrounding air. So the temperature of that original mass of air decreases. That energy has gone into the surroundings in some form. The form will depend on how the nature and configuration of the surroundings.

AM

 

[karen_lorr]

Ahh, I have it now

Thank you both