The Laws of Thermodynamic

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1 The Laws of Thermodynamic Zeroth Law If A is in thermal equilibrium with C and B in thermal equilibrium with C then A and B have to be in thermal equilibrium. No heat flows! First Law of Thermodynamics (or energy conservation) Principle of Conservation of Energy: Energy can neither be created nor destroyed, only transformed Energy may be transformed from one form to another, but the total energy of any body or system of bodies is a quantity that can be neither increased nor diminished How can we change the internal energy of a system? 1. Add or remove heat 2. Work done on the system or from system on the external world 1

2 First Law of Thermodynamics The change in a systems internal energy is related to the heat and the work. U= U2 - U1 = Q - W Where: U1 = internal energy of start U2 = internal energy of end Q = net thermal energy flowing into system during process Positive when system gains heat Negative when system loses heat W = net work done by the system Positive when work done by the system Negative when work done on the system The state of a system is described by its pressure, temperature and volume.thermal processes can change the state of a system. We assume that thermal processes have no friction or other dissipative forces. In other words: All processes are reversible (Reversible means that it is possible to return system and surroundings to the initial states) REALITY: irreversible 2

3 Processes with constant pressure (Isobar) F = P 0 A W = Fs = P 0 A(x f -x i ) = P 0 (V f -V i )= P 0 V (=area under PV-plot) U= Q - P V Processes with constant volume (Isochore) W=0 (no change of volume) U= Q Processes with constant temperature (Isotherm) T const. no change in the internal energy PV = NkT = constant Area under PV-curve (calculus) W = NkT ln(v f /V i ) = nrt ln(v f /V i ) Q=W U= 0 3

4 Adiabatic ( not passable ) processes (no heat is gained or lost by the system Q=0, i.e. system perfectly isolated ) Q=0 U= -W PV = constant (isothermal) PV γ = constant (adiabatic) Specific Heats under constant pressure and constant volume (amount of heat depends on type of process) Specific heat Q = m c T Molar specific heat Q = n C T Constant Volume: C V Constant Pressure : C P Which one is larger? 4

5 With U= = Q W and Q = n C T e.g for a monoatomic gas U = 3/2 n R T Constant volume: U= Q 3/2 n R T = n C V T C V = 3/2 R Constant pressure: Q = U + W n C V T = 3/2 n R T + n R T C P = 5/2 R C V C P = R (always valid for any ideal gas) Calculus shows that γ = C P /C V For a monoatomic gas therefore γ = 5/3 PV γ = constant Or P = const. 1/ V γ Summary Thermal processes Constant pressure W = P V Q = U + P V Constant volume W = 0 Q = U Isothermal (constant temperature) W = Q U = 0 Adiabatic (no heat flow) W = U Q = 0 5

6 The second law of thermodynamics (the entropy law or law of entropy) Heat can never pass spontaneously from a colder to a hotter body. As a result of this fact, natural processes that involve energy transfer must have one direction, and all natural processes are irreversible. This law also predicts that the entropy of an isolated system always increases with time. Entropy is the measure of the disorder or randomness of energy and matter in a system. Because of the second law of thermodynamics both energy and matter in the universe are becoming less useful as time goes on. Perfect order in the Universe occurred the instance after the Big Bang when energy and matter and all of the forces of the Universe were unified. As we know from other processes Physicist Lord Kelvin stated it technically as follows: "There is no natural process the only result of which is to cool a heat reservoir and do external work." Convert heat into mechanical work: Heat engines W = Q h Q C Efficiency: e = W/Q h = (Q h Q C )/ Q h = 1 - Q C / Q h (e.g. a efficiency of e=0.8 means 80% of the heat is converted to mechanical work) 6

7 Carnot s Theorem/Principle To get maximal efficiency is must be a heat engine where all processes are reversible. An irreversible heat engine operating between two heat reservoirs at constant temperatures cannot have efficiency greater than that of a reversible heat engine operating between the same two temperatures. Reversible engines operating between the same temperatures have the same efficiency. Maximum efficiency of a heat engine: e = = 1 - T C / T h Maximum work done by a heat engine: W max =e max Q h = (1 - T C / T h ) Q h 7