Heat Capacities: Q ∝ ∆T Q = C ∆T T ↑ ∆U T ↑ ∆H Q = C V ∆T Q = C V

Heat Capacities: Q ∝ ∆T Q = C ∆T T ↑ ∆U T ↑ ∆H Q = C V ∆T Q = C V

Heat Capacities: q ∝∝∆∆Tq =C ∆T The value of C depends on the process, Closed system Closed system Constant pressure Constant volume ↑ T ↑ T 1 atm Experimentally: q = C ∆∆T q = C ∆∆T qvv =Cvv T qpp =Cpp T ∆U ∆H Heat capacity of an “ideal” gas: ClassicallyClassically heat heat energy energy can can be be stored stored in in a a gas gas as as internal internal energy energy with: with: ROTATION ½ kT per degree of freedom kT per degree of freedom VIBRATION TRANSLATION ½ kT per degree of freedom WhenWhen quantization quantization of of energy energy levels levels is is taken taken into into account, account, contributions contributions to to the the heat capacity can be considered classically only if E << kT. Energy levels heat capacity can be considered classically only if Enn << kT. Energy levels with E ≥≥kT contribute little, if at all, to the heat capacity. with Enn kT contribute little, if at all, to the heat capacity. Since vibrational energies are generally comparable to or greater than kT, only rotations and translations store much energy under normal conditions. Classical internal energy of an “ideal” gas: ATOMS: 3 Translations per atom ATOMS: # of moles × U = n 3 RT U = N(3 ½ kT) 2 # of atoms (or molecules) LINEARLINEAR MOLECULES: MOLECULES: 3 Translations per molecule 2 Rotations per molecule U = N 3 kT + N(2 × ½kT) U = n 5 RT 2 2 NON-LINEARNON-LINEAR MOLECULES: MOLECULES: 3 Translations per molecule 3 Rotations per molecule U = N 3 kT + N(3 × ½kT) 2 U = n 3RT Molar heat capacity of an “ideal” gas: q = C ∆T ∆ p p qv = Cv T PV = nRT ∆H = C ∆T ∆U = C ∆T p v P∆V = nR∆T ∆U + P∆V = C ∆T ∆ ∆ p Cv = U/ T ∆U + R∆T = C ∆T 1 mole p Constant Volume ∆ ∆ Cp = U/ T + R Constant Pressure 1.1. ATOMS: 20.8 J K-1/mol U = n 3 RT ∆ ∆ 3 ∆ ∆ 5 2 Cv = U/ T = 2 R Cp = U/ T + R = 2 R 1 mole 2.2. LINEAR MOLECULES: 29.1 J K-1/mol 5 ∆ ∆ 5 C = ∆U/∆T + R = 7 R U = n 2 RT Cv = U/ T = 2 R p 2 3.3. NON-LINEAR MOLECULES: 33.3 J K-1/mol ∆ ∆ ∆ ∆ U = n 3RT Cv = U/ T = 3R Cp = U/ T + R = 4R SampleSample Problem:Problem: When 229 J of energy is supplied as heat to 3.00 mol Ar (g) at constant pressure the temperature of the sample increases by 2.55 K. Calculate the molar heat capacity Cp for Ar (g). How well does this compare to the ideal gas limit? Estimate Cv for Ar (g). SampleSample Problem:Problem: 10.4 kcal of heat are evolved at constant pressure∅ when 1.00g of water combines 3+ 3+ ∆ with Al in forming Al(H2O)6 . Calculate H for the reaction: 3+ → 3+ Al (aq) + 6 H2O (l) Al(H2O)6 (aq) Molar heat capacity of a monatomic solid: For vibrational motion U = kT per degree of freedom. For a solid: d.o.f = 3N - 6 ≈ 3N U = 3 NkT U = 3 nRT ∆ ∆ Cv = U/ T = 3R 24.9 Quantized Energy Levels → −−∆∆E/kT AtAt high high T T C Cv →3R3R P ∝ e mol v n / 1 - Classical Behavior: J K ∆E << kT At low T C → 0 v At low T Cv → 0 v Energy C Non-Classical Behavior: ∆ ≥ T E kT.

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