Okay, so ΔS=
. 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=
, and the temperature values cancel out to get nRln(V2/V1).