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Heat, work and entropy change of an ideal gas in PV diagram

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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 ...

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