CHAPTER 4. THERMODYNAMICS: THE FIRST LAW

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4-10 held constant, all of the heat goes into producing a corresponding change in temperature (i.e., it all goes into changing the internal energy), whereas when the pressure is held constant, some of the heat goes into work of expansion or contraction. Therefore, it is to be expected that since a given q results in a smaller T. P C P > CV When a monatomic ideal gas is heated at constant volume, all of the heat goes into increasing the translational kinetic energy of the atoms. However, in the case of molecules, there are other energy modes available in addition to kinetic energy. Specifically, molecules may have rotational and vibrational energy. Therefore, when a molecular gas is heated, the resulting energy increase is distributed among all three energy modes. Statistically, it may be shown that the amount of energy per mole that goes into each energy mode is U m = N m(½RT), where N is related to the number of degrees of freedom and the number of different types of energy m (e.g., kinetic and potential) associated with the energy mode, and R is the gas constant. This result, which assumes that all energy modes have continuous energy distributions, is known as the equipartition theorem. The number of degrees of freedom for a molecule, D , is related to the m number of coordinates required to specify the position of each atom. The degrees of freedom associated with the various energy modes of a molecule and the corresponding values of N are m summarized below; N = D , N = D , N = 2D tran tran rot rot vib vib D = 3n tot D = 3 tran D = 2 (linear molecule) or 3 (nonlinear) rot D = 3n !5 (linear molecule) or 3n ! 6 (nonlinear) vib The equipartition theorem may be used to estimate the heat capacities of gas-phase molecules at high temperatures. Example: Estimate the value of C for CO . V 2
4-11 In this case, n = 3 and so D = 9. This molecule is linear; therefore D = 2 and D tot rot vib = 9 ! 5 = <strong>4.</strong> Hence U = U tran + U rot + U vib = 3(½RT) + 2(½RT) + 8(½RT) = 13/2 RT. The estimated heat capacity is therefore C V ' dU dT ' 13 R ' 6.5 R. The 2 measured value at 298 K is 3.46 R. Exercise: Estimate the value of C for CH . V 4 In this case, n = 5 and so D tot = 15. This molecule is nonlinear; therefore D rot = 3 and D vib = 15 ! 6 = 9. Hence U = U tran + U rot + U vib = 3(½RT) + 3(½RT) + 18(½RT) = 12RT. The estimated heat capacity is therefore C V ' dU ' 12 R. The measured dT value at 298 K is 3.25 R. Because the energies of rotation and vibration are not really continuous, the values of C obtained V for molecules from the equipartition theorem are reached only at very high temperatures. The relationship between C V and C P can be derived for an ideal gas as follows: H = U + PV MH ' MU % P MV MT P MT P MT P , but for an ideal gas, the internal energy only depends on temperature U = f(T) (recall that U = trans (3/2)RT), and so MU ' MU MT P MT ' C V . V Also V m ' RT P , so MV m MT ' R . Therefore P P . Exercise: Determine the values of C V and C P for Ar assuming it to be an ideal gas. U = (3/2)RT so C V = dU/dT = (3/2)R and C P = (5/2)R.
Page 1 and 2: 4-1 CHAPTER 4. THERMODYNAMICS: THE
Page 3 and 4: is = 4-3 Mechanical work is defined
Page 5 and 6: expansion processes which start at
Page 7 and 8: 4-7 when heat is added to a system,
Page 9: 4-9 !3 = 30.8 kJ/mol - (8.314x10 kJ
Page 13 and 14: F. Thermochemistry 4-13 A very impo
Page 15 and 16: 4-15 Applying this formula directly
Page 17 and 18: orbitals orbitals 4-17 o H (C ) = 7
Page 19 and 20: ). w = !nRT ln (V /V ) 2 1 4-19 whe
Page 21 and 22: 4-21 An adiabatic expansion is show
Page 23 and 24: 4-23 Review Questions 1. Define ope
Page 25 and 26: Answers: 4-25 4. (a) w = 0, (b) w =
Page 27: = 4-27 where where c = C /R. V P 1

CHAPTER 4. THERMODYNAMICS: THE FIRST LAW

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