## Isothermal and reversible reaction with changing P and V

$\Delta U=q+w$

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### Isothermal and reversible reaction with changing P and V

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?

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### Re: Isothermal and reversible reaction with changing P and V

The heat will not be 0. An isothermal process has no temperature change. Therefore, $\Delta U = 0 = q+w$. 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.