## Delta S equation

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

JukaKim_1D
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

### Delta S equation

How is Delta S equal to nRln(V1/V2)?

Ronald Yang 2F
Posts: 86
Joined: Fri Sep 25, 2015 3:00 am

### Re: Delta S equation

Okay, so ΔS=$\frac{q(rev)}{T}$. The type of expansion that occurs is isothermal, reversible expansion. Thus, since it's isothermal, ΔU=0, since ΔU=(3/2)*n*R*ΔT and ΔT=0, making ΔU=q+w=0 and q=-w. The work for an isothermal, reversible expansion is -nRTln(V2/V1), and since q(rev)=-w, q(rev)=-(-nRTln(V2/V1))=nRTln(V2/V1). Plugging this into the first equation, this makes ΔS=$\frac{q(rev)}{T}$=$\frac{nRTln(V2/V1)}{T}$, and the temperature values cancel out to get nRln(V2/V1).
Last edited by Ronald Yang 2F on Sat Jan 23, 2016 3:20 pm, edited 1 time in total.

Brenda Melano 3H
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

### Re: Delta S equation

This question goes back to change in w=-p*change in V. Then you integrate both sides: w= -integral from V2 to V1 of (P*dV). You then substitute nRT/V for P and take the nRT out of the integral since they are constant. Then you integrate 1/V from V2 to V1 which simplifies into ln(V2/V1). In the end you should end up with, w=-nRT*ln(V2/V1). And when the change of internal energy equals 0, q=-w. and since Delta S=q/T, you can plug in the equation we just derived in for q. q=nRT*ln(V2/V1).
So, Delta S=(nRT*ln(V2/V1))/T. T then cancels and you are left with Delta S= nR*ln(V2/V1).
I hope that helped!