## Entropy of Isothermal Processes

$\Delta S = \frac{q_{rev}}{T}$

Jar-Yee 3L
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

### Entropy of Isothermal Processes

For an ideal gas, ΔU=0 when ΔT=0, evidently because an ideal gas only has kinetic energy, so without heat there would be no change in internal energy.

ΔU = 0 = q+w,
q = -w.
With ΔT = 0, then q = 0. As a result, w = -q = 0 as well.

Eventually, a proof with this concept would give us ΔS=nRln(Vf/Vi), the equation for the change in entropy of an isothermal process.

Since Vf and Vi both exist, how come work is equal to 0? w= -PΔV, and ΔV exists.

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

### Re: Entropy of Isothermal Processes

When ΔT = 0, that doesn't mean q is also zero. In the example given today in lecture, for a reversible, isothermal expansion, heat must constantly be put into the system to keep the temperature constant, hence to keep ΔT = 0. The heat put in by the heat reservoir is used as work in expansion, so ΔV is not equal to zero, hence work is not zero.

Chem_Mod
Posts: 19135
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 820 times

### Re: Entropy of Isothermal Processes

The above discussion is correct! It is noteworthy that when $\Delta T$=0, it DOES NOT mean q=0, only $\Delta U$=0