Thermodynamics: The First Law 자연과학대학 화학과 박영동 교수 Classical thermodynamics

Thermodynamics: the first law 2.1 The conservation of energy Systems and surroundings Work and heat The measurement of work The measurement of heat Heat influx during expansion 2.2 Internal energy and enthalpy The internal energy Internal energy as a state function The enthalpy The temperature variation of the enthalpy

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

Thermodynamic universe the system is the 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. The universe consists of the system and surroundings. system surroundings The Universe

Various Systems systemmatterenergy open( 열린 계 ) allowed closed( 닫힌 계 ) forbiddenallowed isolated( 고립 계 ) forbidden Some other thermodynamics terms state, state functions, path, process, extensive properties, intensive properties. closed open isolated

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

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

Calculating the change in internal energy We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating. 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 gives us  U = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ

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

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 1.Free expansion 2.Expansion against constant pressure 3.Reversible isothermal expansion

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

Heat and Internal Energy  U = q + w = q V For the constant volume process, there is no p-V work, so 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

Internal Energy of a Gas A pressurized gas bottle (V = 0.05 m 3 ), contains helium gas (an ideal monatomic gas) at a pressure p = 1×10 7 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

molar constant volume heat capacity of monatomic gases (at 1 atm, 25 °C) Monatomic gasC V, m (J/(mol·K))C V, m /R He Ne Ar Kr Xe

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.

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.

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

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

C p, C v 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

Heat capacity of CO 2 and N 2 The heat capacity with temperature as expressed by the empirical formula Cp,m/(J K -1 mol -1 ) = a+bT+c/T 2. The circles show the measured values at 298 K.

molar heat capacities, Cp,m/(J K -1 mol -1 ) = a+bT+c/T 2 ab/(l0 -3 K -1 )c/(10 5 K 2 Monatomic gases Other gases Br Cl CO F2F H2H I2I N2N NH O2O H 2 O(l) C(s, graphite) *calculated results are given in unit of R(8.314 J K -1 mol -1 )

An exothermic process 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. Zn(s) + 2 HCl(aq) → ZnCl 2 (aq) + H 2 (g) exothermic and endothermic processes

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).

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 p ex opposes the expansion with a force p ex A.

3. Work for reversible isothermal expansion for an ideal gas

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.

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.

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.

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

a DSC (differential scanning calorimeter) 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).) samplereference

C p and C V

The Maxwell relations

4. Work for adiabatic expansion  U = q + w = w  U = C V  T = w q ad = 0 If the heat capacity is independent of temperature, C V dT = -pdV

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

For an ideal monatomic gas of n moles, calculate q, w, ΔU for each process. w 12 = -p b (V b -V a ) ΔU 12 = C V (T 2 -T 1 ) = C V (p b V b /nR - p b V a /nR) = C V p b (V b - V a ) /nR q 12 = ΔU 12 - w 12 T 2 = p b V b /nR T 1 = T 3 = p b V a /nR = p a V b /nR p b V a = p a V b w 2 3 = 0 ΔU 23 = C V (T 3 -T 2 ) = C V (p b V a /nR - p b V b /nR) = C V p b (V a - V b ) /nR = - ΔU 12 q 12 = ΔU 23 w 31 = -nRT 3 ln(V a /V b ) = -p b V a ln(V a /V b ) ΔU 31 = 0 q 12 = - w 31