1 12. Thermodynamics The essential concepts here are heat and temperature. Heat is a form of energy transferred between a system and an environment due to the temperature difference existing between them. A Temperature Temperature is one of SI base quantities and is measured in kelvins (K). Zeroth law of thermodynamics If body A is in thermal equilibrium with body T (a thermoscope) and body B is in thermal equilibrium with body T, then bodies A and B are in thermal equilirium with each other. We say that both bodies have the same temperature. Measurements of temperature A temperature scale must be defined. The most fundamental is the Kelvin scale but in practice some others (Celsius, Fahrenheit) are used. Selecting a certain thermometric body, the thermometric feature and assuming a relationship between this feature and temperature, a thermometer is defined. The constant volume gas thermometer is a standard against which all other thermometers are calibrated.

2 Constant volume gas thermometer The temperature is determined by measuring the pressure p of a gas in a fixed volume. Introducing a linear temperature scale (12.1) and taking the temperature of the triple point of water as a reference temperature T tr = K, one can eliminate constant A (12.2) Dividing eq.(12.1) by (12.2) one obtains (12.3) As the indications of such a thermometer depend on the kind of a gas used, one has to use smaller and smaller gas quantities and calculate the extrapolated value (12.4) The extrapolated value gives the ideal gas temperature. A A gas thermometer measuring the temperature of a liquid Figure from HRW,2 A gas thermometer measuring T of a boiling water. The different gases give only one extrapolated value K.

3 Heat is a form of energy transferred between bodies of different temperatures (it is not a feature of a system). Similarly to work it is measured in joules (J). Previously the unit calorie (cal) was used. The specific heat is defined as a heat needed to warm a unit of mass of a given body by 1 K (12.5) Above definition is more strict in a differential form, moreover determining c a body has to be warmed in well defined conditions, under certain parameter constant, e.g. pressure or volume x – a constant parameter (p, V,…) Some specific heats are listed below A Heat C[J/g·K] Al0.90 Cu0.39 Pb0.13 Water4.19 Mercury0.14

4 A Work done by a gas The work done on a gas by acting a force F on a piston with surface area A is (12.6) where dV is the differential change in the volume. The total work between volumes V 1 and V 2 is (12.7) and depends on the path of changing the volume. Lets calculate the work between the initial state A and final state E done on different paths. The work done along path ABE is along ADE is and along ACE is In each case one obtains a different result. Work is a path-dependent quantity.

5 The variation in internal energy ΔU of the system is equal to the heat Q added to it, lowered by work W done by the system (12.8) If a work is done on a system, the sign „-” in formula (12.6) is changed for „+”. The internal energy is a function of state and its variation is independent of the process path. It depends only on the positions of initial and final states ( the variations of heat and work are path dependend). For the ideal gas the variation in internal energy is only a function of temperature c V – specific heat at a constant volume Example What is a change in internal energy in adiabatic and constant-volume processes? a)adiabatic process means Q = 0 In this case ΔU = -W = W’, where W’ – work done on the system b)constant-volume (isochoric) process means V = const In this case W = 0 and then ΔU = Q. A 12.4.The first law of thermodynamics

6 The processes which obey the first law of thermodynamics are not always possible. For some processes the energy is always conserved but they are irreversible, i.e. cannot be reversed by only small changes in the environment. The direction of processes is indicated by the changes of entropy; for isolated system the entropy for irreversible processes always increases. The change in entropy is given by the integral (12.9) where i and f mean the initial anf final states, Q is the energy transferred to or from the system at temperature T. Integral (12.9) can be calculated only for reversible processes. Problem 1 Calculate the change in entropy for 1 mol of nitrogen when it expands freely in the isolated system to the double volume. The process of free expansion is irreversible. For a free expansion we have:. In this case we can replace this process by the reversible isothermal expansion between the same initial and final states and calculate ΔS with eq.(12.9). A Entropy and the second law of thermodynamics

7 The change in entropy in an isothermal process is Q – heat transferred to the gas The work done by a gas during expansion in this process R – gas constant n – number of moles In this process because Finally one obtains As ΔS is positive, the entropy increases what could be expected for an irreversible process of a free expansion in a closed system. Analysing the reversible process of isothermal expansion one has to take into account a heat that is exchanged between the gas and the heat reservoir (in contrast to a free expansion for which Q=0). For the whole closed system gas – heat reservoir the total change in entropy is equal to zero, as the change in entropy of a reservoir is of reverese sign vs. that of a gas. A Changes in entropy Isothermal expansion process for ideal gas

8 There are several equivalent formulations of the second law. One of them is: In a process that occurs in an isolated system, the change in entropy is greater than zero for irreversible processes and equal to zero for reversible processes. (12.10) In a real world almost all processes are irreversible and reversibility is an idealization. In living organisms many processes are irreversible but the entropy of the organism does not increase because the processes involving the surroundings cause an increase of the entropy of the environment. A Second law of thermodynamics The Carnot cycle It is a thrmodynamic cycle in which the conversion of heat into work is done with a maximum efficiency. Heat Q 1 is absorbed during isothermal expansion AB and Q 2 is discharged during isothermal compression CD. Processes AB and CD are reversible. The full cycle consists of two isothermal processes AB, CD and two adiabatic process BC, DA. Positive work is done by the gas in processes AB and BC. In processes CD and DA the environment does work on the gas. The net work W is the area enclosed by the cycle.

9 The changes in entropy in a Carnot cycle are process AB:(12.11) process BC:(12.12) process CD:(12.13) process DA:(12.14) The net change in entropy in a complete cycle is zero (entropy is a state function). From (12.11) – (12.14) one obtains (12.15) The efficiency of any thermal engine is defined as the ratio of work done per cycle to the thermal energy absorbed per cycle (12.16) For the Carnot engine with the use of (12.15) we thus obtain A The Carnot cycle, cont. No series of processes is possible whose sole result is the tranfer of heat from a thermal reservoir and the complete conversion of this heat to work – second law of thermod.

10 Using work an engine can transfer heat from a low temperature reservoir to a high temperature reservoir. Examples: houshold refrigerator, air conditioner, heat pump. This can be done by reversing the Carnot cycle A The Carnot refrigerator Carnot refrigerator From (12.15) one obtains and what can be written as The coefficient of performance of a Carnot refrigerator is then (12.17) Example: What is the coefficient of performance of a Carnot refrigerator working between temperatures t 2 = C and t 1 = 27 0 C. For real fridges

11 To calculate the changes of entropy for the free expansion of an ideal gas, the statistical mechanics will be used. We analyse the distribution of gas particles in the two identical halves of a box. As an example we take 3 gas molecules and calculate the number of microstates corresponding to a given configuration Configurations Multiplicity W (number of microstates) First configuration 1 Second configuration 3 Third configuration 3 Fourth configuration 1 A The statistical interpretation of entropy Multiplicity of configuration W we calculate from where (n! – „n factorial”) For the 2 nd conf. we have

12 Different configurations have different number of microstates, then the configurations are not equally probable. In the example discussed configurations 2 and 3 are more probable than 1 and 4. The probability for configuration 2 is 3/8 and for configuration 1 is 1/8. Example 2 We take 100 molecules. What is the number of microstates for configurations: a)n 1 = 100, n 2 = 0 b)n 1 = 50, n 2 = 50 a) b) The most probable are therefore these configurations where the molecules distibute equally between the halves of the box. L. Boltzmann formulated a relationship between the entropy of a given configuration and the multiplicity W of that configuration where the Boltzmann constant As all states are equally probable, then configurations with a large number of microstates are the most frequent and related entropy takes the highest values. A The statistical interpretation of entropy, cont. for large N one can use the Stirling’s approximation

13 Problem We solve again the problem of change in entropy for 1 mol of nitrogen when it expands freely in the isolated system to the double volume but using the statistical theory. In n moles we have N molecules. Initially (the molecules occupy the left side of a container) the muliplicity of configuration (N,0) is 1. When nitrogen occupies the full volume, i.e. the configuration is (N/2, N/2) one obtains The entropy for initial state The final entropy Because Nk B = nR one gets We obtained the same result as when the change in entropy was calculated in terms of temperature and heat transfer. A The statistical interpretation of entropy, cont.