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361 Lec 6 Thur, 8sep16 “spontaneous” process: (happens without input of work) Life: enzymes coupling spontaneous reactions with non-spontaneous (like a heavy weight lifting a lighter one) through a pulley system enzymes are the “pulleys” of biochemistry.” (~ 75,000 enzymes busily doing their thing in the human body) 1 361 Lec 6 Thur, 8sep16 pext = 1 atm pext = 1 atm p = 1 atm IRREVERSIBLE p = 3 atm p = changes 3 atm to 1 atm IRREVERSIBLE but pext = 1 atm and constant REVERSIBLE pext = p during entire expansion by removing grains of sand from piston. Adding one grain of sand reverses direction pext = 3 atm p = 3 atm pext = 1 atm p = 1 atm REVERSIBLE pext = p always 2 What are signs of: ∆T, Adiabatic (q=0) : Isothermal : - q, w, 0 0 + ∆U? - - 0 ∆T, q, w, ∆U? + 0 + + 0 - + 0 3 361 Lec 6 Thur, 4sep14 How does the energy of an ideal gas change with volume at constant temperature? U is independent of V! U =U(T) For ideal gas isothermal process ∆U = 0 4 Reversible Isothermal PV work (ideal gas) pext = p = nRT/V (balanced forces) An integral is just a Sum V2 V2 V2 nRT w = − ∫ pext dV = − ∫ pdV = − ∫ dV V1 V1 V1 V w = −nRT ∫ V2 V1 V2 dV = −nRT ln V1 V Sum of fractional changes = natural logarithm ONLY if p = pext = and ideal gas (but is common case) What are ∆U and q? ∆U =0 (isothermal), therefore q = -w 5 Equation of State example, ideal gas: pV = nRT n = mols, p = pressure, V = volume, T = absolute temp. R = “gas constant” = 8.3145 J mol-1 K-1 = 0.083145 L bar mol-1 K-1 =0.0821 L atm mol-1 K-1 nRT = V(n, T, P) P nRT P= = P(n, T, V) V PV n= = n(T, P, V) RT PV T= = T(n, P, V) nR Note that V = or or or 6 van der Waals Equation of state for real gases 2 n ( p + 2 a )(V - b) = nRT V b is the effective volume of the molecule n/V is mol L-1 i.e., the molar concentration a is the effective equilibrium constant for association of two molecules of gas What does this equation become at low concentration? p V = nRT 7 Some useful state functions are DEFINED from other state functions. We are immediately concerned with Enthalpy ≡ H = U + pV (by definition) Why invent yet another state function? Because ∆U = q + w. q is easy to measure, but most chemical reactions have a small work term -pext∆V (due to expansion or contraction against the atmosphere) I call this “obligatory work”. We want to tabulate state functions, NOT q and w. Under the very common conditions: p=pext = constant and only the obligatory pV work on the atmosphere, the above definition yields: ∆H = q ( nice!) Beware, if the reaction is harnessed so that non-obligatory pV work (i.e., useful work) is done, then ∆H = q + wuseful 8