Laws of Thermodynamics and Thermal Processes

Transcription

1 Unit 1 Laws of Thermodynamics and Thermal Processes 1.1 Temperature scale 1. Thermodynamic temperature and absolute zero 1. Ideal gas laws 1.4 Kinetic theory 1.5 The internal energy of an ideal gas 1.6 Important terms used in thermodynamics 1.7 The zeroth law of thermodynamics 1.8 The first law of thermodynamics 1.9 Thermal Processes 1.10 Specific heats for an ideal gas 1.11 The second law of thermodynamics 1.1 The third law of thermodynamics 1.1 Temperature scale The Fahrenheit scale goes from o to 1 o (a total of 180 divisions). The Celsius scale goes from 0 o to 100 o (a total of 100 divisions). Fahrenheit o F Celsius o C Boiling of water Freezing of water 0 Now, we he 100 C o correspond to 180 F o, that is, 1C o corresponds to 180/100 F o, or 1C o corresponds to (9/5) F o. Here, we he a formula to relate C o and F o 9 T F = + T C. 5 Example Two thermometers, one marked in Fahrenheit the other in Celsius, are placed in a bath. At what temperature will both thermometers read the same? Answer The same reading will occur at T, where 5 T = ( T ) gives T = 40 o. 9 1

2 1. Thermodynamic temperature and absolute zero Pressure versus temperature for a gas at constant volume As the gas is cooled, its atoms lose some of their thermal energy; they trel more slowly and collide with the chamber walls less frequently and less forcefully, As a result, the pressure in the chamber drops. A plot of pressure versus temperature is shown below. All gases show straight lines of the same intercept at 7.15 o C, but different gases he different slopes. P T ( o C) Volume versus temperature for a gas at constant pressure As the temperature of a gas is lowered, its atoms are less energetic. To keep the pressure constant, the volume must decrease, which will make it easier for atoms to strike the chamber walls frequently, thereby sustaining the pressure. All gases shrink toward zero volume at T = 7.15 o C. V T ( o C) The 7.15 o C is referred as the absolute zero temperature. An absolute temperature is given a SI unit, Kelvin K, in which 0 K corresponds to 7.15 o C; and has the same size as Celsius; therefore 0 o C corresponds to 7.15 K. In general, the temperature in Kelvin relates the Celsius degree by the relation T = T C

3 1. Ideal gas laws The ideal gas It is a gas of small, hard, randomly moving atoms that bounce off each other and the walls of their container but otherwise do not interact. The attractive forces (intermolecular forces) between gas particles are neglected. Two conditions are met: (1) The density is low (i.e. the atoms occupy a volume much smaller than that of the container and the atoms are far apart from each other), and () The temperature is well above the condensation point. Remark: The ideal-gas model does not distinguish between a monatomic gas and a diatomic gas; both are considered as simply small and hard spheres. With the ideal- gas model, we obtain the following relations. Boyle s law Keeping the temperature constant, the volume of a gas varies inversely with the pressure, which is equivalent to saying that pressure times volume is constant. 1 V or PV = constant P Charles s law When the pressure is kept constant, the volume of a given amount of any gas varies directly with the temperature (in Kelvin). V V T or = constant T Gay-Lussac s law When the volume is kept constant, the absolute pressure of a given amount of any gas varies directly with the temperature (in Kelvin). P P T or = constant T The ideal gas law As the pressure increases with the number of moles of gas molecules linearly, we he P n, where n is the number of moles of gas molecules. From the Boyle s law and the Gay- Lussac s law, we he 1 P and P T respectively. Summing up the relations, we V conclude that P nt V, i.e. PV = n T constant. The proportional constant is nearly the same

4 for all gases at low pressures and is referred as the universal gas constant R, where R = J/(mol K). The ideal gas law is given by PV = n RT (The equation of state for an ideal gas). The Avogadro s number N A = is the number of molecules per mole, and so nn A equals N molecules present in a sample of gas. Now, the ideal gas law can be rewritten as N R PV = n RT = RT = N( ) T = N k BT, N N A where k B is the Boltzmann s constant and k B = J/K. A Example A cylindrical flask of cross-sectional area A is fitted with an airtight piston that is free to slide up and down. Contained within the flask is an ideal gas. Initially the pressure applied by the piston is 10 kpa and the height of the piston above the base of the flask is 5 cm. When additional mass is added to the piston, the pressure increases to 170 kpa. Assuming the system is always at the temperature 90 K, find the new height of the piston. Answer Piston Piston h 1 P 1 h P The system is basically isotherms, that is, the temperature is kept constant. The Boyle s law is applied here as PV = constant, or P 1 V 1 = P V. (10 10 )(0.5 A) = ( )(h A) gives h = 19 cm. Example A person s lungs might hold 6.0 L of air at body temperature (10K) and atmospheric pressure (101 kpa). Given that the air is 1 % oxygen, find the number of oxygen molecules in the lungs. 4

5 Answer Assume that the air in the lung is ideal gas. The formula PV = N k T gives B PV N = = = kt B 5 ( )( ) ( )(10) (0.1)( ) oxygen molecules in the lung. air molecules in the lung. Hence, there are 1.4 Kinetic theory Pressure and temperature of a gas are macroscopic quantities. However, the macroscopic quantities connect the microscopic quantities, such as the position or velocity of an individual molecule. The kinetic theory of gas states the connection. We imagine a gas to be made up of a collection of molecules moving about inside a container of volume V. In particular, we assume the following: The container holds a very large number N of identical molecules. Each molecule has a mass m, and behes as point particle. The molecules move about the container in a random manner. They obey Newton s laws of motion at all times. When molecules hit the walls of the container or collide with one another, they bounce elastically. Other than these collisions, the molecules he no interactions. (a) The origin of pressure Imagine a container that is a cube of length L on a side. Its volume, then, is V = L. In addition, consider a given molecule of mass m that moves in the negative x direction toward a wall. If its velocity is v x and then v x after rebound from the container s wall. The change of momentum of the molecule is given by p = p p mv ( mv ) mv. x f, x i, x = x x = x L As the time required for this round trip of length L is t = L /, v x the erage force exerted by the wall on the molecule is p mvx mvx F = = =. t L / v L x v x v x Round trip = L The erage pressure exerted by the wall is F F mvx / L mvx mvx P = = = = =. A L L L V 5

6 Since a gas consists of molecules of different velocities, the relation above should be replaced ( m v by P ) x =, where represents the erage value. Now, the gas molecules moves in V a three dimensional space, we he 1 mv v = ( vx ) + ( v y ) + ( vz ) = ( vx ) which yields P = N( ) or V 1 =, where N P ρ v is the number of molecules in the container and ρ is the density of the gas. Rearrange the equation, we finally obtain an equation in kinetic theory of gas 1 PV = N mv. The L.H.S. of the above equation can be expressed in terms of the kinetic energy of individual molecule, e.g. PV PV = N k T, we then he a most important result K B = k BT. 1 mv = N mv = N( ) = N K. But, as we know that (b) RMS speed of a gas molecule The square root of ( v ) is given a special name the root mean square (rms) speed. As K = 1 m( v ) = k BT, we obtain k BT ( v ) =, hence m v rms k BT =. m The rms speed of the gas molecules can be expressed in terms of the molecular mass M, e.g. RT v rms =, since k B = R/N A. M (c) The speed distribution of molecules The distribution of speeds of molecules is governed by the Maxwell speed distribution. The following figure shows the distribution of speeds of oxygen gas at the temperatures T = 00 K, and T = 1100 K. Note that the most probable speed increases with increasing temperature. 6

7 1.5 The internal energy of an ideal gas The internal energy U of a substance is the sum of its potential energy and kinetic energy. In an ideal gas, as there are no interactions between gas molecules, the potential energy is then zero. As a result, the internal energy U of an ideal gas is represented by its kinetic energy. Since U U = N k BT. = N k B T = ( n N A ) k BT and R = N A k B, we obtain a similar relation, but this time in terms of moles U = nrt. Example A basketball at 90 K holds 0.95 mol of air molecules. What is the internal energy of the air in the ball? Answer Applying the relationu = nrt, we he U = (0.95)(8.1)(90) = 44 J. 7

8 Example Two identical containers A and B are connected by a tap S that is initially closed. A contains an ideal gas at a pressure P 1 and temperature T 1. B contains the same gas at a pressure P and a temperature T. The taps is then opened. If the temperatures of containers A and B are maintained and remain unchanged, find the molar ratio of the gas in the two containers. Find also the pressure of the gas mixture. Answer Before opening S, the number of moles of gas in containers A and B: n PV 1 1 = and RT1 n PV =, RT A P 1 S P B where V is the volume of containers A and B. T 1 T After opening S, the number of moles of gas in containers A (or B): PV ' PV ' n1 ' = and n ' =. RT RT 1 PV ' Therefore, n ' RT T PV PV PV ' PV ' 1 = 1 =. As n1 + n = n1' + n ', we he 1 + = + n ' ' PV T1 RT1 RT RT1 RT, RT where P is the common pressure in the containers. After simplification, we obtain P1 P P' P' + = +. Rearrange the expression T T T T 1 1 P1 P + T1 T PT 1 + PT 1 P ' = =. 1 1 T1 + T + T T Important terms used in thermodynamics Heat: Heat is the energy transferred between objects because of a temperature difference Thermal equilibrium: Objects that are in thermal contact, but he no heat exchange between them, are said to be in thermal equilibrium. Temperature: Temperature is the quantity that determines whether or not two objects will be in thermal equilibrium. 8

9 1.7 The zeroth law of thermodynamics If object A is in thermal equilibrium with object C, and object B is separately in thermal equilibrium with object C, then objects A and B will be in thermal equilibrium if they are placed in thermal contact. 1.8 The first law of thermodynamics The change in a system s internal energy, U, is related to the heat Q and the work W as follows: U = Q+ W, where W = PdV. Q is positive Q is negative W is positive W is negative System gains heat System loses heat Work done on system Work done by system It is noted that if work is done on an insulated system (i.e. Q = 0) its internal energy increases. The work done W has a positive value when the external world does work on the system (e.g. compression), and it has a negative value when work is done by the system (e.g. expansion). U = U U = W f i On the other hand, if the system does work, and / or heat is removed, its internal energy decreases. 1.9 Thermal Processes All processes discussed in this section are assumed to be quasi-static, which means the processes occur so slowly that at any given time the system and its surroundings are essentially in equilibrium. Thus, in a quasi-static process, the pressure and temperature are always uniform throughout the system. If friction or dissipative forces are neglected, the process is considered to be reversible. For a reversible process, it must be possible to return both the system and its surroundings to exactly the same states they were in before the process began. 9

10 (a) Isothermal processes Constant-temperature heat bath Piston Piston Q Consider the following reversible processes at constant temperature, the piston moves down without friction. Applying the first law of thermodynamics, U = Q+ W, and making use the property of ideal gas which has its internal energy depends only on the temperature, we he U = 0 and Q = W, since the process is carried out at constant temperature. The work done, W, is positive as work is done on the system. Hence we he a negative Q, that is, heat is given out by the system to the external world. The reversed process in the figure below performs the same idea, but this time the work W is negative, as work is done by the system. Since U = 0 again, the heat Q is then positive or we conclude that heat is gained by the system from the external world. Constant-temperature heat bath Piston Piston Q 10

11 Remarks: PV plots The PV plot of one mole of an ideal gas at the temperatures 00 K, 500 K, 700 K and 900 K. Notice that each isotherm has the shape of a hyperbola. As the temperature is increased, however, the isotherms move farther from the origin. Thus, the pressure corresponding to a given volume increases with temperature, as one would expect. In an isothermal expansion from the volume V i to the volume V f, the absolute value of the work done is equal to the shaded area. For n moles of an ideal gas at the temperature T, the work done on the gas is Vf Vf nrt Vf dv V V f f W = P dv = dv nrt nrt ln V V nrt ln V = i i Vi V = = Vi V Vi Example A cylinder contains 0.50 mol of an ideal gas at a temperature of 10 K. As the gas expands isothermally from an initial volume of 0.1 m to a final volume of 0.45 m, find the amount of heat that must be added to the gas in order to maintain a constant temperature. Answer V f V i T = constant Q Since the process is isothermal, the internal energy of an ideal gas has no change, i.e. U = 0. The relation U = Q+ W gives Q = W, where W is negative (work is done by the system). 11

12 Now, we know that the gas gains heat from the surrounding (positive Q). As the work done W is given by V f Vi PdV, we obtain W as Vf Vf nrt Vf dv V V f f W = PdV = dv = nrt = nrt lnv = nrt ln( ) V V i Vi Vi i V V V. Substituting n = 0.50 mol, R = 8.1 J/(mol K), V f = 0.45 m and V i = 0.1 m, then we obtain Q = W = 480 J. i (b) Adiabatic processes A process is said to be adiabatic if there are no heat flows into or out of the system. We can he such process if the system is well insulated. The figure below is an adiabatic compression, the gas heats up and its pressure increases. Insulation Piston P increases T increases Similarly, for an adiabatic expansion, the gas cools down and the pressure decrease. Example Describe and explain what happen when a rod is pushed downward rapidly through the cylinder as shown in the right figure. Answer When a rod that fits snugly into a cylinder is pushed downward rapidly, the temperature of the gas increases before there is time for heat to flow out of the system. Thus, the process is essentially adiabatic. As a result, the temperature of the gas can increase enough to ignite bits of paper in the cylinder. In a diesel engine the same principle is used to ignite an air-gasoline mixture. 1

13 Example An amount of gas is compressed from volume A isothermally and the PV plot is obtained as shown in curve (ii). If the same gas is compressed adiabatically, which curve would show the correct curve? Curve (i) or curve (iii)? Answer In an adiabatic compression, no heat change occurs in the process. According to the first law of thermodynamics, U = Q+ W = W, U is positive as W is positive during compression. The increasing temperature gives an increasing pressure for a given volume of gas. Hence, curve (iii) is the correct curve Specific heats for an ideal gas (a) Constant volume A heat Q flows into a constant-volume container, which is filled with n mole ideal gas, the temperature of the gas rises by the amount T, and its pressure increases as well. The heat Q is given by Q V = n C V T, where C V is the molar specific heat of a gas at constant volume. In other words, we can write QV CV =. n T (b) Constant pressure Similarly, the whole process can be carried out at constant pressure, but this time with the temperature and volume varies. If a heat Q is added to the gas then its temperature and volume increase. The molar specific heat of a gas at constant pressure is related by Q P = n C P T, where C P is the molar specific heat of a gas at constant pressure. And, we he QP CP =. n T According to the first law of thermodynamics U = Q+ W, we he in process (a) the internal energy U = Q, since W = P V = P(0) = 0. But, U = nr T, hence we he C V nr T QV = = = R. n T n T 1

14 For process (b), it is a constant-pressure case, as heat is added to the container, the increasing volume involves a work done W by the gas, wherew = P V = nr T. Making using the relation U = Q+ W again and we obtain 5 Q = U W = nr T + nr T = nr T. Plug it into the expression of C P, we obtain C P 5 nr T QP 5 = = = R. n T n T Now, we conclude that C P is greater than C V, as extra work is required for expansion while increasing the temperature. C C R. P V = 1.11 The second law of thermodynamics When objects of different temperatures are brought into thermal contact, the spontaneous flow of heat that results is always from the high temperature object to the low temperature object. Spontaneous heat flow never proceeds in the reverse direction. The second law of thermodynamics is more restrictive than the first law; it says that of all processes that conserve energy, only those that proceed in a certain direction actually occur. In a sense, the second law enforces a certain directionality on the behior of nature. 1.1 The third law of thermodynamics The third law of thermodynamics states that there is no temperature lower than absolute zero, and that absolute zero is unattainable. It is possible to cool an object to temperatures arbitrarily close to absolute zero experiments he reached temperatures as low as K but no object can ever be cooled to precisely 0 K. To cool an object, you can place it in thermal contact with an object that is colder. Heat transfer will occur, with your object ending up cooler and the other object ending up warmer. In particular, suppose you had a collection of objects at 0 K to use for cooling. You put your object in contact with one of the 0-K objects and your object cools, while the 0-K object warms slightly. You continue this process, each time throwing away the warmed up 0-K object and using a new one. Each time you cool your object it gets closer to 0 K, without ever actually getting there. 14