See attached file.
Let's first calculate, using the known formula for the entropy of the ideal gas, what the change in entropy should be. The entropy is given by:
S = Nk [log[V/N (U/N)^(3/2)]+5/2 + 3/2 Log((4 pi m)/(3 h^2))] (1)
Here U is the internal energy: U = 3/2N k T
If you change V and P then in (1), V and U will change. We can treat all the other variables as constants. It is then convenient to write (1) as:
S = N k[ Log(V) + 3/2 Log(T)] + const. (2)
If we go from (P1, V1) to (P2, V2), the temperature changes according to:
T2/T1 = P2 V2/(P1 V1)
So, the change in entropy is:
S2 - S1 = N k[ Log(V2) + 3/2 Log(T2)] - N k[ Log(V1) + 3/2 Log(T1)] =
N k[ Log(V2/V1) + 3/2 Log(P2 V2/(P1 V1))] =
N k[ 5/2 Log(V2/V1)+3/2 Log(P2/P1)] (3)
Let's now consider the three paths from (P1, V2) --> (P2,V2) and calculate the change in entropy along these paths. We need to integrate dS = dQ/T, so we need to know dQ for an infinitesimal step along the paths. This you can find using the first law ...
A detailed solution is given.