Reversible and Isothermal


Moderators: Chem_Mod, Chem_Admin

Yen Vuong 2I
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am

Reversible and Isothermal

Postby Yen Vuong 2I » Sun Jan 22, 2017 4:04 pm

How does a reversible piston and an isothermal system affect work?

Courteney Hedicke 3J
Posts: 6
Joined: Wed Sep 21, 2016 2:56 pm

Re: Reversible and Isothermal

Postby Courteney Hedicke 3J » Sun Jan 22, 2017 5:45 pm

I think Isothermal means that the temperature doesnt change.

Posts: 19
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

Re: Reversible and Isothermal

Postby TiengTum2D » Sun Jan 22, 2017 6:33 pm

I think isothermal system is a system that has a consistent heat source around it and it can affect work by either expansion or compression with the piston. If the gas expands (piston moves out), the system is doing work but heat around the area will replenish the lost energy as heat, therefore keeping it at the same temperature. I hope that makes sense!

Coco Hailey 2E
Posts: 32
Joined: Fri Sep 25, 2015 3:00 am

Re: Reversible and Isothermal

Postby Coco Hailey 2E » Sun Jan 22, 2017 6:51 pm

When calculating the work of a reversible or isothermal system, do we use different equations. Basically if its a reversible system do we use w= -P∆V or w = -nRTln(V2/V1)??

Pauline Tze 3B
Posts: 57
Joined: Sat Jul 09, 2016 3:00 am
Been upvoted: 1 time

Re: Reversible and Isothermal

Postby Pauline Tze 3B » Mon Jan 23, 2017 1:34 am

Hi Coco!

A reversible system is typically isothermal because the heat absorbed by the system is canceled out by the heat released by the system (as is the nature of systems at equilibrium). We use those two specific equations you listed as follows:
w= -P∆V is to find work in a system where there's constant pressure
w = -nRTln(V2/V1) is to find work when there's variable pressure (or if the system is, for example, expanding/increasing in volume). Of the two equations, w = -nRTln(V2/V1) is appropriate for reversible reactions because the external pressure changes forcing the system to "reverse" that action by moving toward equilibrium.

I hope that helps!

Return to “Calculating Work of Expansion”

Who is online

Users browsing this forum: No registered users and 1 guest