Isothermal and reversible reaction with changing P and V


Moderators: Chem_Mod, Chem_Admin

HimaniMadnawat3L
Posts: 15
Joined: Fri Sep 26, 2014 2:02 pm

Isothermal and reversible reaction with changing P and V

Postby HimaniMadnawat3L » Tue Jan 20, 2015 9:40 pm

The question states: If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands isothermally and reversibly from 6.00 L to 18.00 L and has a final pressure of 1.20 atm, what is w, q, and delta U.

I understand that q is 0 since the reaction is isothermal, and that w=-n(5/2R)Tln(V2/V1) for the changing volume. Do I add another term to the work which is -n(3/2R)Tln(P2/P1) to give me the total work and delta U?

Chem_Mod
Posts: 17543
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 393 times

Re: Isothermal and reversible reaction with changing P and V

Postby Chem_Mod » Tue Jan 20, 2015 9:52 pm

The heat will not be 0. An isothermal process has no temperature change. Therefore, . The energy and enthalpy are both directly proportional to the temperature, so they don't change in an isothermal process.

Then, q=-w, and w=-nRTln(Vf/Vi), so q=nRTln(Vf/Vi).

Conceptually, if a gas expands, it would lose energy since it did work. But since the temperature didn't change, this lost energy must have been made up by adding heat. So, work is negative, heat is positive, and they cancel each other exactly.


Return to “Concepts & Calculations Using First Law of Thermodynamics”

Who is online

Users browsing this forum: No registered users and 1 guest