Question 4A5 from the textbook

[Ernie Lee 3F]

Can someone explain to me the step by step process for Section b. I do not understand why they did not just use the equation w=-P(delta)V as I thought that the pressure was constant.

[Gregory_Kislik_2C]

For part b, it is given that it is reversible and isothermic, meaning that the reaction happens in infinitely small steps, and the volume changes in infinitely small steps, which we will call dV. This means that the work equation cannot be w = -PdeltaV, since "delta" is not appropriate anymore (since the changes in volume are infinitely small), and instead must be dw = -PdV. This means that each incremental "step" of work done is infinitely small, just as the change in volume is done in infinitely small, reversible steps.

From the ideal gas law, we know PV = nRT, so P = nRT/V.

Substituting in, we get dw = -nRTdV/V

Integrate the left side from 0 to w, and the right from initial to final volume and you get:

w = -nRTln(V2/V1).

Plugging in the givens from the problem gives: w = -0.2(8.3145(298)ln(2.4/1.2) = 343.4 J

The reason for the work being greater in part b than in part a is because when a system is at equilibrium and the volume changes in infinitely small steps, the work is more efficient. The formula derived above can be used for isothermic, reversible reactions of gasses, but not if the gas is pushing against an external pressure irreversibly.

I hope this helps and please let me know if there are any errors.