By Evans L.C.

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**Example text**

6) 2. Claim # 1. For each α, β ∈ Σ: (7) M (α, β) = A(α, β) or M (α, β) = M + (α, β) or M (α, β) = M − (α, β). / To prove this, ﬁx α, β ∈ Σ, Γ1 , Γ2 ∈ M (α, β). Assume ﬁrst Γ1 ∈ M + (α, β), but Γ2 ∈ + − + − − ˆ M (α, β). Then Γ2 ∈ A(α, β) ∪ M (α, β), and so Q (Γ2 ) = Q (Γ1 ) = 0. Hence Q (Γ2 ∗ ˆ 2 ) + Q− (Γ1 ) = Q+ (Γ2 ) + Q− (Γ1 ) = 0, but Γ ˆ 2 ∗ Γ1 is not adiabatic. Thus Γ1 ) = Q− (Γ ˆ 2 ∗ Γ1 ) = ξ(Γ = = ˆ 2 ∗Γ1 ) Q + (Γ ˆ 2 ∗Γ1 ) T + (Γ ˆ 2 )+Q + (Γ1 ) Q + (Γ ˆ 2 ∗Γ1 ) T + (Γ Q − (Γ2 )+Q + (Γ1 ) ˆ 2 ∗Γ1 ) T + (Γ > 0.

Thus we can deﬁne π(α, β) := ξ(∆) (∆ ∈ I(α, β)). Then according to Claim # 4 ξ(Γ) ≤ π(α, β) (Γ ∈ P (α, β)), and so to derive (11) we must show that we can write (12) π(α, β) = φ(β) − φ(α) for all α, β ∈ Σ. For this ﬁx a state γ ∈ Σ and a temperature level θ. Owing to the Main Hypothesis, there exist Γ1 ∈ M (α, β), Γ2 ∈ M (β, γ) such that T ± (Γ1 ) = θ on t+ (Γ1 ) ∪ t− (Γ1 ) T ± (Γ2 ) = θ on t+ (Γ2 ) ∪ t− (Γ2 ). Then Γ2 ∗ Γ1 ∈ I(α, γ) and ξ(Γ2 ∗ Γ1 ) = + Q + (Γ2 ∗Γ1 ) 2 ∗Γ1 ) − TQ− (Γ T + (Γ2 ∗Γ1 ) (Γ2 ∗Γ1 ) − − (Γ ) Q + (Γ2 )+Q + (Γ1 ) 1 − Q (Γ2 )+Q θ θ = = ξ(Γ2 ) + ξ(Γ1 ).

We parameterize Γ by writing Γ = {(T (t), V (t)) for a ≤ t ≤ b}, where a < b and V, T : [a, b] → R are C 1 . ✷ Deﬁnitions. (i) We deﬁne the working 1-form d− W = P dV (2) and deﬁne the work done by the ﬂuid along Γ to be d− W = W(Γ) = (3) Γ P dV. Γ (ii) We likewise deﬁne the heating 1-form d− Q = CV dT + ΛV dV (4) and deﬁne the net heat gained by the ﬂuid along Γ to be (5) d− Q = Q(Γ) = Γ CV dT + ΛV dV. Γ Remarks. (a) Thus b W(Γ) = P (T (t), V (t))V˙ (t)dt a 30 ˙= d dt and b Q(Γ) = CV (T (t), V (t))T˙ (t) + ΛV (T (t), V (t))V˙ (t)dt.