Thermodynamics: The First Law

1. Thermodynamics: The First Law 자연과학대학 화학과 박영동 교수 Classical thermodynamics

2. Thermodynamics: the first law 2.1 The conservation of energy 2.1.1 Systems and surroundings 2.1.2 Work and heat 2.1.3 The measurement of work 2.1.4 The measurement of heat 2.1.5 Heat influx during expansion 2.2 Internal energy and enthalpy 2.2.6 The internal energy 2.2.7 Internal energy as a state function 2.2.8 The enthalpy 2.2.9 The temperature variation of the enthalpy

3. Thermodynamic Systems, States and Processes Objectives are to: • define thermodynamics systems and states of systems • explain how processes affect such systems • apply the above thermodynamic terms and ideas to the laws of thermodynamics

4. Thermodynamic universe The Universe The universe consists of the system and surroundings. surroundings the system isthe region of interest; its region is defined by the boundary. the rest of the world is its surroundings. The surroundings are where observations are made on the system. system

5. Various Systems open isolated Some other thermodynamics terms state, state functions, path, process, extensive properties, intensive properties. closed

6. A State Function and Paths The altitude is a state property, because it depends only on the current state of the system. The change in the value of a state property is independent of the path between the two states. The distance between the initial and final states depends on which path (as depicted by the blue and red lines) is used to travel between them. So it is not a state function. pressure, temperature, volume, ... heat, work

7. The First Law of Thermodynamics -- Heat and work are forms of energy transfer and energy is conserved. U = Q + W change in total internal energy work done on the system heat added to system State Function Process Functions

8. Calculating the change in internal energy • Suppose someone does 622 kJ of work on an exercise bicycle and loses 82 kJ of energy as heat. What is the change in internal energy of the person? Disregard any matter loss by perspiration. • Solution • w = −622 kJ (622 kJ is lost by doing work on the bicycle), • q = −82 kJ (82 kJ is lost by heating the surroundings). • Then the first law of thermodynamics givesus • U = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ • We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating.

9. Increase in volume, dV Work, and the expansion(p-V) work Work = force • distance pex=F/A force acting on the piston = pex × Area distance when expand = dy F F +y dW (p-V work)

10. Total Work Done To evaluate the integral, we must know how the pressure depends (functionally) on the volume. We will consider the following cases 0. Constant volume work Free expansion Expansion against constant pressure Reversible isothermal expansion

11. 0. Work for the constant volume process For the constant volume process, there is no p-V work

12. Heat and Internal Energy For the constant volume process, there is no p-V work, so U = q+ w = qV If we add heat q to the system, the temperature of the system increases by T , and the internal energy increases by U. Constant volume heat capacity

13. Internal Energy of a Gas A pressurized gas bottle (V = 0.05 m3), contains helium gas (an ideal monatomic gas) at a pressure p = 1×107 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

14. molar constant volume heat capacity of monatomic gases (at 1 atm, 25 °C)

15. Temperature and Energy distribution The temperature is a parameter that indicates the extent to which the exponentially decaying Boltzmann distribution reaches up into the higher energy levels of a system. (a) When the temperature is low, only the lower energy states are occupied (as indicated by the green rectangles). (b) At higher temperatures, more higher states are occupied. In each case, the populations decay exponentially with increasing temperature, with the total population of all levels a constant.

16. A constant-volume bomb calorimeter. The constant-volume heat capacity is the slope of a curve showing how the internal energy varies with temperature. The slope, and therefore the heat capacity, may be different at different temperatures.

17. 1. Work for free expansion case For the free expansion process, there is no p-V work U = q+ w = q

18. 2. Work for expansion against constant pressure U = q+ w = q - p V qp = U + p V = (U + p V ) If we define Enthalpy as, H≡ U + pV qp = H Enthalpy change is the heat given to the system at constant pressure.

19. Cp, Cv for an ideal gas For an ideal gas, U and H do not depend on volume or pressure. For an example, for an ideal monatomic gases, U = (3/2) nRT

20. Heat capacity of CO2 and N2 The heat capacity with temperature as expressed by the empirical formula Cp,m/(J K-1 mol-1) = a+bT+c/T2. The circles show the measured values at 298 K.

21. molar heat capacities,Cp,m/(J K-1 mol-1) = a+bT+c/T2 *calculated results are given in unit of R(8.314 J K-1 mol-1)

22. An exothermic process Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) When hydrochloric acid reacts with zinc, the hydrogen gas produced must push back the surrounding atmosphere (represented by the weight resting on the piston), and hence must do work on its surroundings. This is an example of energy leaving a system as work. exothermic and endothermic processes

23. Work and Heat Work is transfer of energy that causes or utilizes uniform motion of atoms in the surroundings. For example, when a weight is raised, all the atoms of the weight (shown magnified) move in unison in the same direction. Heat is the transfer of energy that causes or utilizes chaotic motion in the surroundings. When energy leaves the system (the green region), it generates chaotic motion in the surroundings (shown magnified).

24. The expansion(p-V ) Work When a piston of area A moves out through a distance h, it sweeps out a volume ΔV = Ah. The external pressure pex opposes the expansion with a force pexA.

25. 3. Work for reversible isothermal expansion for an ideal gas

26. Reversible Isothermal Expansion Work for a perfect gas The work of reversible, isothermal expansion of a perfect gas. Note that for a given change of volume and fixed amount of gas, the work is greater the higher the temperature.

27. Molecular Basis for Heat Capacity The heat capacity depends on the availability of levels. (a) When the levels are close together, a given amount of energy arriving as heat can be accommodated with little adjustment of the populations and hence the temperature that occurs in the Boltzmann distribution. This system has a high heat capacity. (b) When the levels are widely separated, the same incoming energy has to be accommodated by making use of higher energy levels, with a consequent greater change in the ‘reach’ of the Boltzmann distribution, and therefore a greater change in temperature. This system therefore has a low heat capacity. In each case the green line is the distribution at low temperature and the red line that at higher temperature.

28. Temperature dependence of enthalpy(H) and internal energy(U) The enthalpy of a system increases as its temperature is raised. Note that the enthalpy is always greater than the internal energy of the system, and that the difference increases with temperature.

29. Temperature dependence of enthalpy(H) and internal energy(U) Note that the heat capacityies depend on temperature, and that Cp is greater than CV.

30. a DSC(differential scanning calorimeter) sample reference A thermogram for the protein ubiquitin. The protein retains its native structure up to about 45°C and then undergoes an endothermic conformational change. (Adapted from B. Chowdhry and S. LeHarne, J. Chem. Educ. 74, 236 (1997).)

31. Cpand CV

32. The Maxwell relations

33. 4. Work for adiabatic expansion qad = 0 U= q+ w = w If the heat capacity is independent of temperature, U= CV T = w CVdT = -pdV

34. 4. Work for adiabatic expansion For an adiabatic process for an ideal gas,

35. For an ideal monatomic gasof n moles,calculate q, w, ΔU for each process. w12 = -pb(Vb-Va) ΔU12= CV(T2-T1) = CV(pbVb/nR - pbVa/nR) = CVpb(Vb - Va) /nR q12 = ΔU12- w12 w2 3= 0 ΔU23 = CV(T3-T2) = CV(pbVa/nR- pbVb/nR) = CVpb(Va - Vb) /nR = - ΔU12 q12 = ΔU23 w31= -nRT3ln(Va/Vb) = -pbValn(Va/Vb) ΔU31 = 0 q12 = - w31 T2 = pbVb/nR T1 =T3= pbVa/nR= paVb/nR pbVa= paVb