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Equations and constants for Exam 2, CHM 3400, Fall 2015 You are responsible for knowing the conditions under which the equations apply. 1 atm = 760 Torr = 1.01325 bar R = 8.31451 J K -1 mol -1 = 0.0831451 L bar K -1 mol -1 = 0.08206 L atm K -1 mol -1 dU = TdS – pdV ΔH = C p ΔT Δ r H(T’) = Δ r H(T) + ΔC p ΔT = ! = !"# !"## = − ∆ = ! ! ∆ = ! ∆ = ! ! ! ∆ = ∆ !"#$% ΔS = -nR(x A lnx A + x B lnx B ) dG = -SdT + Vdp ! = − ! = ∆ = ∆ − ∆ ΔG = VΔp ΔG = nRT(x A lnx A + x B lnx B ) = ∆ !"#$% ∆ !"#$% ! ! = ∆ !"# ! 1 ! − 1 ! f = c – p + 2 p A = x A p A * p B = x B K , p B = m B K’ = ! ! + ! ! ! = ! ∗ + ln ! ! = ! ∗ + ln ! a j = γ j x j Π = [B]RT ΔT f = K f m B ΔT b = K b m B ± = ! ! ! ! ! ! ! ! (ℎ = ! + ! ) !" ± = − ! ! !/! = 1 2 ! ! !
