Change in internal energy of gas (thermodynamics)

[axe34]

Hi,If the change in U between 2 points (A,B) in a thermodynamic process is always the same despite the path, then please help with the following:

Say I have adiabatic conditions so that delta U = Work done only (Q=0)
Surely I can go through multiple paths between A+B with increased or decreased amounts of work? Then how can the change in internal energy be the same each time?Thanks

 

[Chestermiller]

Hi,If the change in U between 2 points (A,B) in a thermodynamic process is always the same despite the path, then please help with the following:

Say I have adiabatic conditions so that delta U = Work done only (Q=0)
Surely I can go through multiple paths between A+B with increased or decreased amounts of work? Then how can the change in internal energy be the same each time?Thanks

Actually, you can't go adiabatically through multiple paths between A and B with increased or decreased amounts of work. If you think you can, please provide one specific example.

Chet

 

[axe34]

Say I want to decrease pressure and volume from A to B.
Could I not drop the pressure immediately, then reduce the volume at this constant pressure. The area (work) of this graph would be less if I did a smooth transition between A + B, would it not?

 

[Chestermiller]

Say I want to decrease pressure and volume from A to B.
Could I not drop the pressure immediately, then reduce the volume at this constant pressure. The area (work) of this graph would be less if I did a smooth transition between A + B, would it not?

You can't decrease both pressure and volume adiabatically. But, suppose you wanted to decrease the imposed force per unit area immediately, and then let the volume increase adiabatically at this force per unit area. Option 2 is to do a smooth gradual transition between the initial pressure and the force per unit area you used in Option 1. The final temperature and the final volume under Option 2 would be different from the final temperature and the final volume under Option 1, the work would be different, and the change in internal energy (determined by the change in temperature) would be different. You could not reach the same final equilibrium state under Option 2 as you did under Option 1.

Chet