Page 1 of 1

### Reversible and Irreversible

Posted: Sun Feb 02, 2020 8:01 pm
Can anyone please clarify the differences between the reversible and irreversible process? thank you

### Re: Reversible and Irreversible

Posted: Sun Feb 02, 2020 8:05 pm
In a reversible process, at each point along the process path, the system is only slightly removed from being in thermodynamic equilibrium with its surroundings. So the path can be considered as a continuous sequence of thermodynamic equilibrium states. For an irreversible process, the system is not close to thermodynamic equilibrium with its surroundings at each point along the path.

At any point along the path of a reversible process, both the system and the surroundings can be returned to their original states without significantly affecting anything else. This cannot be accomplished if the process is irreversible.

### Re: Reversible and Irreversible

Posted: Sun Feb 02, 2020 8:07 pm
In an irreversible reaction, reactants will create a product that cannot be then turned back into reactants.

### Re: Reversible and Irreversible

Posted: Sun Feb 02, 2020 9:52 pm
Also, irreversible processes almost always do more work than reversible processes.

### Re: Reversible and Irreversible

Posted: Sun Feb 02, 2020 11:02 pm
Michael Du 1E wrote:Can anyone please clarify the differences between the reversible and irreversible process? thank you

Yes, I am confused on this as well.

### Re: Reversible and Irreversible

Posted: Mon Feb 03, 2020 10:14 am
Juana Abana 1G wrote:
Michael Du 1E wrote:Can anyone please clarify the differences between the reversible and irreversible process? thank you

Yes, I am confused on this as well.

I think I understand it now. When you're given a constant external pressure, it would be considered irreversible because once the gas expands the piston outward, the external pressure wouldn't be able to push it back inward as the pressure is constant. We would use the equation of W=-PdeltaV for this.
As for reversible, the pressure CAN change, meaning if the gas pushes the piston outward and the pressure decreases, there will also be a change in the external pressure as it will decrease in order to reach equilibrium. The equation we would use for this is w=-nRTln(V2/V1)