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

 

[sardar]

for reaching out with your question. I can explain the concept of change in internal energy of a gas in thermodynamics.

First, let's define internal energy. It is the sum of the kinetic and potential energies of the particles that make up a system. In a gas, the particles are constantly moving and colliding with each other, resulting in changes in internal energy.

In thermodynamics, the change in internal energy (ΔU) of a gas is equal to the heat added to the system (Q) minus the work done by the gas (W). This can be expressed as ΔU = Q - W. Therefore, if the heat added to the system is zero (Q=0), then the change in internal energy is equal to the work done by the gas (ΔU = W).

Now, let's consider the adiabatic conditions you mentioned, where there is no heat transfer (Q=0). In this case, the change in internal energy of the gas is solely determined by the work done by the gas. This work can be done in various ways, such as by compressing the gas or by stirring it. The important thing to note is that the change in internal energy depends on the initial and final states of the gas, not the path taken to get there.

In other words, the change in internal energy is independent of the path taken between two points (A and B) in a thermodynamic process, as long as the initial and final states are the same. This is because the internal energy of a gas is a state function, meaning it only depends on the current state of the system and not the path taken to reach that state. Therefore, no matter how much or how little work is done on the gas, the change in internal energy will be the same.

I hope this explanation helps to clarify your question. If you have any further doubts, please feel free to reach out.