### Entropy of Isothermal Processes

Posted:

**Fri Jan 15, 2016 2:31 pm**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(V

Since V

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(V

_{f}/V_{i}), the equation for the change in entropy of an isothermal process.Since V

_{f}and V_{i}both exist, how come work is equal to 0? w= -PΔV, and ΔV exists.