Chapter 12 What Is Temperature? What Is Heat Capacity?

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Mercy Blair
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1 Chapter 2 What Is Temperature? What Is Heat Capacity?. Heat capacity peaks from phase transitions. The peak heat capacity in the figure below shows that helium gas adsorbed on graphite undergoes a transition from an ordered state at low temperature to a disordered state at high temperature. Use the figure to estimate the energy associated with the transition. C/Nk T (K) Source: HJ Kreuzer and ZW Gortel, Physisorption Kinetics, Springer Verlag, Heidelberg, 986. Data are from RL Elgin and DL Goodstein, Phys Rev A 9, (974). 52

2 The peak in the heat capacity suggests that some kind of bond or interaction breaks at a temperature of about T 3 K. A characteristic energy can be computed from ε kt (.987 cal K mol )(3 K) 6 cal mol. Since this heat capacity peak is much narrower than in the two-state model, it shows that the two-state model would not be an adequate way to model this. We also need to account for cooperativity. 2. Two-state model of a hydrogen bond. Suppose a hydrogen bond in water has an energy of about 2 kcal mol. Suppose a made bond is the ground state in a two-state model and a broken bond is the excited state. At T 300 K, what fraction of hydrogen bonds are broken? Use Equation (2.4), ( ) fground f excited ε kt 2000 cal mol (2 cal K mol )(300 K) 3.33 f ground f excited 29, so f excited f excited + f ground +f ground /f excited % of the bonds are broken. 53

3 3. A random energy model of glasses. Glasses are materials that are disordered and not crystalline at low temperatures. Here s a simple model. Consider a system that has a Gaussian distribution of energies E, according to Equation (2.6): [ ] (E E )2 p(e) p 0 exp, 2ΔE 2 where E is the average energy and ΔE characterizes the magnitude of the fluctuations, that is, the width of the distribution. (a) Derive an expression showing that the entropy S(E) is an inverted parabola: S(E) S 0 k(e E )2 2ΔE 2. (b) An entropy catastrophe happens (left side of the curve in the figure below) where S 0. For any physical system, the minimum number of accessible states is W, so S k W implies that the minimum entropy is S 0. At the point of the entropy catastrophe, the system has no states that are accessible below an energy E E 0. Compute E 0. (c) The glass transition temperature T g is the temperature of the entropy catastrophe. Compute T g from this model. S S 0 0 E 0 E (a) Substitute p(e) into S(E) k p(e) to get S k p (E E )2 0. 2ΔE 2 54

4 Defining S 0 k p 0, this becomes S(E) S 0 k (E E )2 2ΔE 2. (b) At S 0, the entropy equation gives 0 S 0 k (E 0 E ) 2 2ΔE 2 (E 0 E ) 2 2ΔE2 S 0 k () E 0 E ΔE 2S0 k. (c) Substitute the entropy equation S(E) into the definition of temperature, evaluated at the point E E 0 : S T g E E0 k (E E ) ΔE 2. E0 Now substituting the square root of Equation () into this expression gives T g k 2ΔE2 S 0. (2) ΔE 2 k (The minus sign has disappeared because Equation () has two roots: one positive and one negative. Since our interest is in the left side of the S(E) curve, T g > 0.) Rearranging Equation (2) gives T g ΔE 2 2kS 0. For more details, see JN Onuchic, Z Luthey-Schulten, and PG Wolynes, Ann Rev Phys Chem 48, (997). 55

5 4. Fluctuations in enthalpy. The mean-square fluctuations in enthalpy for processes at constant pressure are given by an expression similar to Equation (2.7) for processes at constant pressure: ΔH 2 (H H ) 2 kt 2 C p. What is the root-mean-square enthalpy fluctuation ΔH 2 /2 for water at T 300 K? H 2 /2 (kt 2 C p ) /2 [(2 cal K mol )(300 K) 2 (8 cal K mol )] /2.8 kcal mol. 5. Oxygen vibrations. Oxygen molecules have a vibrational frequency of 580 cm (see Example.3). If the relative populations are f ground /f excited 00 in the Schottky model, what is the temperature of the system? T T ( ) ( ) k fground ε 0 f excited ( ) ( ) k fground hν f excited ( ) ( ) fground θ vib f excited 2274 (000) K. (000) 56

6 7. Schottky model for hydrogen. Hydrogen molecules have a vibrational wavenumber of 4300 cm. If 0.5% of the molecules are in the excited state, what is the temperature of the system? T ( ) 99.5, where θ vib hν θ vib 0.5 k, θ vib ( )(4300)( ) So T 68 K 8. Finding the C V of a gas. A gas is placed in an airtight container. A piston decreases the volume by 0% in an adiabatic process, and the temperature is observed to increase by 0%. What is the constant-volume heat capacity of the gas? T 2 T ( ) Nk/Cv V, V 2 ( ) 0 00 Nk/Cv, Take the logarithm of both sides: log(.0) log(.) R C v, C V (2 cal K mol )2.2 cal K mol. 58

2013, 2011, 2009, 2008 AP

2013, 2011, 2009, 2008 AP Lecture 15 Thermodynamics I Heat vs. Temperature Enthalpy and Work Endothermic and Exothermic Reactions Average Bond Enthalpy Thermodynamics The relationship between chemical reactions and heat. What causes