1 Handout 12: Thermodynamics Zeroth law of thermodynamics When two objects with different temperature are brought into contact, heat flows from the hotter body to a cooler one Heat flows until the temperatures of the two bodies are equal When the temperatures of two objects are equal, these objects are in thermal equilibrium The zeroth law of thermodynamics states that if A and B are separately in thermal equilibrium with C, A and B must be in thermal equilibrium with each other This means that A, B and C have the same temperature First law of thermodynamics The first law of thermodynamics states that when heat Q enters the system, part of it increases the internal energy U of the system and part of it is converted into work W The statement of first law is simply the conservation of energy: Figure 1: Heat flows from the hotter to the cooler objects that are in contact Q = U + W The first law is true for all phases (solid, liquid, gas) of substances Here we focus on ideal gas For ideal gas, U depends on the change in temperature T Work W is related to the change in volume of the gas Figure 2: Three objects that are in thermal equilibrium with one another The sign of Q, U and W are summarized: Q + when heat enters the system, Q when heat leaves the system, U + when temperature of the gas increases, U when temperature of the gas decreases, W + when volume of the gas increases, W when volume of the gas decreases Thermodynamics processes 1 Isochoric process (constant volume) In this process, W = 0 Therefore, from equation, Q = U; all the heat is used to increase the internal energy and temperature of the gas Consider monatomic gas:
2 Q = U nc V T = 3 2 nr T, where c V is the molar specific heat capacity at constant volume: c V = 3 2 R 2 Isobaric process (constant pressure) In this process, work done by gas W = p V = nr T From equation, for monatomic gas, Figure 3: p-v diagrams for thermodynamic process nc p T = 3 2 nr T + nr T = 5 2 nr T, where c p is the molar specific heat capacity at constant pressure: c p = 5 2 R The general relation is that c p = c V + R The expressions for c V = 3 R and c 2 p = 5 R are true for all 2 monatomic ideal gas The ratio of the molar specific heat capacity is introduced: γ = c p c V = 5 3 3 Isothermal process (constant temperature) In this process U = 0 Therefore, from equation, Q = W All of heat is converted into work 4 Adiabatic process (no heat) In this process, Q = 0, and hence U = W In this process, it can be shown that p 1 V 1 γ = p 2 V 2 γ, T 1 V 1 γ 1 = T 2 V 2 γ 1, p 1 1 γ T 1 γ = p 2 1 γ T 2 γ The p-v diagrams for thermodynamic processes are illustrated in Figure 3
Example 1 For diatomic gas at room temperature, show that the ratio of molar specific heat capacity 3 γ = 7 5 Example 2 Two moles of ideal monatomic gas expand isobarically from V 1 = 003 m 3 to V 2 = 007 m 3 The pressure throughout is p = 15 10 5 Pa a) Calculate T 1 and T 2 b) Calculate change in internal energy U c) Calculate work done by gas W d) Calculate heat absorbed by the gas Q Example 3 Two moles of an ideal monatomic gas expand adiabatically from T 1 = 400 K and p 1 = 10 6 Pa to final pressure p 2 = 10 5 Pa a) Calculate final temperature of the gas b) Find the work done by gas Example 4 One mole of diatomic gas at T 1 = 300 K is compressed adiabatically from V 1 = 20 10 3 m 3 to V 2 = 08 10 3 m 3 a) Calculate the final temperature T 2 b) Find work done by gas
Example 5 One mole of a monatomic ideal gas is taken through a cycle 1 2 3 1 as shown in the Figure The process 1 2 is adiabatic The temperatures at 1, 2 and 3 are 600, 450 and 300 K respectively 4 a) Calculate the amount of heat entering Q in and the amount of heat leaving Q out in one cycle b) How much work is done by the gas in one cycle? Second law of thermodynamics Consider a system absorbing heat by amount dq at temperature T The change in entropy ds of the system is defined as ds = dq T S = 1 2 dq T The integral runs from state 1 to state 2 The second law of the thermodynamics states that entropy of an isolated system never decreases This means that S 0 The entropy change S = 0 when the process is reversible In irreversible processes, S > 0 A reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings An irreversible process cannot be reversed without altering the surroundings A process that is allowed by the first law may not be allowed by the second law; such process never takes place For example, if an ice cube is put into hot coffee We know that the ice melts and the coffee gets colder By conservation of energy, it is allowed that the coffee gets hotter by acquiring energy from the ice and the ice becomes colder However, such process will never take
place It violates the second law Whether or not a process will takes place is determined by the second law of thermodynamics The universe evolves in such the way that its entropy is increasing The entropy is said to be the arrow of time Example 6 Two moles of ideal diatomic gas undergo an isochoric process in which its temperature is changed from T 1 = 300 K to T 2 = 400 K Calculate the change in entropy of the gas 5 Heat engine A heat engine works between two heat reservoirs with different temperatures The hotter reservoir (source) is at temperature T h while the colder reservoir (sink) at temperature T c as shown in Figure 4 The engine extracts amount of heat from the source, converts part of it into useful work W, and discards an amount Q c into the sink By conservation of energy, = W + Q c W = Q c The efficiency e of the engine is defined as Figure 4: Heat engine e = W e = Q c = 1 Q c It must be noted that to have e = 1, Q c must be zero That means the engine converts all the heat input into useful work Such ideal engine does not exist However, the engine with the efficiency closest to the ideal engine is the Carnot engine whose efficiency is theoretically maximum In Figure 5, a Carnot cycle is illustrated; the cycle consists of reversible alternating isothermal and adiabatic processes During AB, the change in entropy is T h and during CD, the entropy change is Q c T c (negative because heat is leaving) There is not heat entering or leaving during BC and DA because of they are adiabatic processes Thus, the total entropy change is S = T h Q c T c Since the Carnot cycle is reversible, S = 0, yielding Figure 5: Carnot cycle Q c = T c T h
6 Hence, the efficiency of the Carnot engine e = 1 Q c e = 1 T c T h Refrigerator (heat pump) works in the opposite way to the heat engine In Figure 6, the pump extracts heat Q c from the cold reservoir expels heat to the hot reservoir provided that work W is done on the system: W + Q c = W = Q c The coefficient of performance of the refrigerator is defined as Figure 6: Refrigerator K = Q c W K = Q c Q c Example 7 A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K a) How much heat does it discard? b) How much work does it do? c) What is the efficiency? Example 8 A power plant generates 13 10 9 W of electric power The plant burns 462 10 7 kg of coal per day; the heat of combustion of coal is about 62 10 6 J kg -1 What is the overall efficiency of the plant? Example 9 A freezer has a coefficient of performance of 240 The freezer is to convert 180 kg of water at 25 C to 180 kg of ice at 0 C in 1 hour a) How much heat is removed from the water at 25 C to convert it to ice at 0 C? b) How much electrical energy is consumed during this hour? c) How much wasted heat is rejected to the room where the freezer sits?