CHEMISTRY COMMUNITY

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_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.

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.

The above discussion is correct! It is noteworthy that when =0, it DOES NOT mean q=0, only =0