Statistical Physics PDF

STATISTICAL PHYSICS Mr. P.R.S.Tissera Department of Physical Sciences Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka  HEAT: ENERGY IN TRANSIT  Heat is energy that flows between a system and its environment by virtue of a temperature difference...

STATISTICAL PHYSICS Mr. P.R.S.Tissera Department of Physical Sciences Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka  HEAT: ENERGY IN TRANSIT  Heat is energy that flows between a system and its environment by virtue of a temperature difference between them Figure 1 : (a) If the temperature TS of a system is less than the temperature TE of its environment, heat flows into the system until thermal equilibrium is established as in (b). (c) If the temperature of a system is greater than that of its environment, heat flows out of the system 2  Because heat is a form of energy, its units are those of energy, namely the joule (J), the calorie (cal), and the British thermal unit (Btu) 1 cal = 4.186 J and 1 Btu = 1055 J  Heat is similar to work in that both represent a means for the transfer of energy  Neither heat nor work is an intrinsic property of a system; that is we can not say that a system contains a certain amount of heat or work 3 HEAT CAPACITY AND SPECIFIC HEAT  We can change the state of a body by transferring energy to or from it in the form of heat, or by doing work on the body  One property of a body that may change in such a process is its temperature T  Let, the change in temperature T corresponds to the transfer of a particular quantity of heat energy Q, it is convenient to define the heat capacity C of a body as the ratio of the amount of heat Q supplied to a body in any process to its corresponding temperature change T; that is, Q C  T 4  The heat capacity per unit mass of a body, called specific heat capacity C Q c   m m T  The specific heat capacity ‘c’ depends on the temperature T  Therefore, the heat that must be given to a body of mass ‘m’ whose material has a specific heat ‘c’, to increase temperature from initial temperature ‘Ti’ to final temperature ‘Tf ’ is given by Tf Q  m  Ti c dT 5  ‘c’ may be a function of the temperature  At ordinary temperatures and over ordinary temperature intervals, specific heats can be considered to be constants  Under these conditions, we can therefore write Q  m c  Tf  Ti   If mass is very small, then Q  n c v  Tf  Ti  6 Table 1 : Heat capacities of some substances Substance Specific heat capacity Molar heat capacity (J kg1 K1) (J mol1 K1) Elemental solids Tungsten 135 24.8 Silver 236 25.5 Copper 387 24.6 Carbon 502 6.02 Aluminum 900 24.3 Other solids Brass 380 Granite 790 Ice (10C) 2220 Liquids Ethyl alcohol 2430 Seawater 3900 Water 4190 7  HEAT OF TRANSFORMATION  When heat enters a solid or liquid, the temperature of the sample does not necessarily rise  Instead, the sample may change from one phase or state (that is, solid, liquid, or gas) to another  The minimum amount of heat per unit mass transferred during a phase change is called the heat of transformation or latent heat (symbol L) for the process Q L  m Where ‘m’ is the mass of the sample Q  mL that changes phase 8   The heat transferred during melting or freezing is called the heat of fusion (symbol Lf), and the heat transferred during boiling or condensing is called the heat of vaporization (symbol Lv) Table 2 : Some heat of transformation Substance Melting Point Heat of Fusion Boiling Point Heat of Vaporization (K) (kJ kg1) (K) (kJ kg1) Hydrogen 14.0 58.6 20.3 452 Oxygen 54.8 13.8 90.2 213 Mercury 234 11.3 630 296 Water 273 333 373 2256 Lead 601 24.7 2013 858 Silver 1235 105 2485 2336 Copper 1356 205 2840 4730 9 HEAT CAPACITIES OF AN IDEAL GAS Heat Capacity at Constant Volume  Let us introduce a certain amount of energy as heat Q into a gas that is confined inside a cylinder fitted with a piston  The gas can either 1. Store the energy as the random kinetic energy of its molecules (internal energy) or 2. Use the energy to do work on the environment (such as by raising a weight on the piston) 10  Let us first consider the case in which the piston is fixed, so that the volume of the gas remains constant, and no external work is done  In this case all the heat energy goes into internal energy: Q  E int  Molar heat capacity at constant volume Q E int CV   n T n T 11 Monatomic Ideal Gas 3 E int  n R T 2 3 n R T 3 CV  2  R  12.5 J mol1 K 1 n T 2 Diatomic Ideal Gas 5 E int  n R T 2 5 n R T 5 CV  2  R  20.8 J mol1 K 1 n T 2 12 Polyatomic Ideal Gas E int  3 n R T 3 n R T CV   3 R  24.1 J mol 1 K 1 n T  Molar heat capacity at constant volume only depends on the number of atoms in the system 13  Heat Capacity at Constant Pressure Figure 2 : Two idealgas isotherms differing in temperature by T are connected by the constantvolume process ab and the constantpressure process ac 14  Already we have established that the internal energy of an ideal gas depends only on the temperature  For all paths connecting the two isotherms of above figure, the change in internal energy has the same value, because all paths correspond to the same change in temperature  In particular, the change in internal energy is the same for paths ab and ac E int, ab  E int, ac 15  There are two contributions to the change in internal energy along path ac  They are;  the heat Q transferred to the gas, and  the work W done on the gas E int, ac  Q  W  The work done and heat exchanged along path ac, respectively W   P V   n R T and Q  n C P T  And also E int, ab  n C V T 16  E int, ac  E int, ab n C V T  n C P T   n R T  CP  CV  R CP  5 R  20.8 J mol 1 K 1 monatomic gas  2 CP  7 R  29.1 J mol 1 K 1 diatomic gas  2 C P  4 R  33.3 J mol 1 K 1 polyatomic gas  17 Ratio of molar heat capacities  CP   CV   5  1.67 monatomic gas  3   7  1.40 diatomic gas  5   4  1.33 polyatomic gas  3 18 Table 3 : Molar heat capacities of gases Gas CP CV CP  CV  (J mol1 K1) (J mol1 K1) (J mol1 K1) Monatomic Ideal 20.8 12.5 8.3 1.67 He 20.8 12.5 8.3 1.66 Ar 20.8 12.5 8.3 1.67 Diatomic Ideal 29.1 20.8 8.3 1.40 H2 28.8 20.4 8.4 1.41 N2 29.1 20.8 8.3 1.40 O2 29.4 21.1 8.3 1.40 Polyatomic Ideal 33.3 24.1 8.3 1.33 CO2 37.0 28.5 8.5 1.30 NH3 36.8 27.8 9.0 1.31 19  Example A family enters a winter vacation cabin that has been unheated for such a long time that the interior temperature is the same as the outside temperature (0C). The cabin consists of a single room of floor area 6m by 4m and height 3m. The room contains one 2 kW electric heater. Assuming that the room is perfectly airtight and that all the heat from the electric heater is absorbed by the air, none escaping through the walls or being absorbed by the furniture, how long after the heater is turned on will the air temperature reach the comfort level of 21C? (Assume that the air in the room behaves like an ideal diatomic gas.) 20 ANSWER 21  THE FIRST LAW OF THERMODYNAMICS Figure 3 : Two gases separated by a diathermic (heat conducting) wall in an insulated container 22  After a sufficient time in thermal contact the system comes to equilibrium at some intermediate temperature T  By this time, The energy lost as The energy gained as heat by the hotter gas = heat by the cooler gas (Q1) (Q2) Q1  Q 2 23  Instead of taking the combination as our system, let us choose gas 2, which absorbs heat Q2  After this energy is absorbed, there is no change in the system other than an increase of its temperature from T2 to T  For an ideal gas, we can calculate the corresponding change in internal energy Eint, 2  The only source for this change in internal energy is the absorbed heat, and so E int, 2  Q 2 , both quantities being positive  Similar statement of conservation of energy applied to gas 1 E int, 1  Q1 , both quantities are negative  General equation E int  Q 24 Figure 4 : A gases in an insulated container has external work done on it by gravity 25  Suppose that the load on the piston is gradually increased, so that the weighted piston descends through a certain distance  Gravity does a certain amount of work W on the system  The temperature increases in this process, and the system therefore experiences a positive change in internal energy  Since no heat transfer is involved, the internal energy of the gas increases by the work done on it, or E int  W both quantities being positive 26 Figure 5 : (a) A system in an initial state in equilibrium with its surroundings (b) A thermodynamic process during which the system may exchange heat Q or work W with its environment. (c) A final equilibrium state reached as a result of the process 27  The system starts out in an initial equilibrium state ‘i’ in which the thermodynamic variables have certain values  Finally, the system is in a final equilibrium state ‘f’  Now take the system from state ‘i’ to state ‘f’ along a variety of Figure 6 : A system in an initial equilibrium state different paths ‘i’ is brought to a final equilibrium state ‘f’ along three different paths 28  Both W and Q depend on the path  Guided by our previous discussion of internal energy, we evaluate the quantity Q + W for each path  We find that in every case the quantity Q + W has the same value  Even though Q and W individually depends on the path taken, the quantity Q + W is independent of the path, depending only on the initial and final equilibrium states ‘i’ and ‘f’ of the gas  This brings us to the first law of thermodynamics 29  First law of thermodynamics In any thermodynamic process between equilibrium states ‘i’ and ‘f’, the quantity Q + W has the same value for any path between ‘i’ and ‘f’. This quantity is equal to the change in the value of a state function called the internal energy  Mathematically, the first law is E int  Q  W  If the system undergoes only an infinitesimal change in state, only an infinitesimal amount of heat dQ is absorbed and only an infinitesimal amount of work dW is done. In such a case the first law is written in differential form dE int  dQ  dW 30 APPLICATION OF THE FIRST LAW Adiabatic Process  In an adiabatic process, the system is well insulated so that no heat enters or leaves, in which case dQ = 0 dE int  dQ  dW dE int  0  dW dE int  dW adiabatic process  We know that dE int  n CV dT 31 Thus P dV   dW   dE int   n C V dT P dV   n C V dT A  d PV  d n R T P dV  V dP  n R dT V dP  n R dT  P dV  n R dT  n C V dT  n dT R  C V  V dP  n dT C P B 32 B V dP   n dT C P C  P   A  P dV n dT C V CV dP dV    P V Pf Vf dP dV  pi P     Vi V Pf Vf ln    ln Pi Vi Pi Vi  Pf Vf P V   constant 33  We can rearrange these results in terms of temperature, using the ideal gas equation of state P V  V   1  constant k T  V   1  constant T V   1  constant  Also, we have already shown W  1  Pf Vf  Pi Vi  (adiabatic process)   1  34  Isothermal Process  In an isothermal process, the temperature remains constant  If the system is an ideal gas, then the internal energy must therefore remains constant E int  0 Q  W  0 Q  W  If an amount of positive work W is done on the gas, an equivalent amount of heat Q = W is released by the gas to the environment  None of the work done on the gas remains with the gas as stored internal energy 35  Constant Volume Process  If the volume of the gas remains constant, it can do no work  Thus W 0 and E int  Q constant volume process   In this case all the heat that enters the gas is stored as internal energy 36  Cyclical Process  In a cyclical process, we carry out a sequence of operations that eventually restores the system to its initial state Figure 7 : A gas undergoes a cyclical process starting at point A 37  Because the process starts and finishes at the point A, the internal energy change for the cycle is zero  Thus, Q  W  0 cyclical process  Where Q and W represent the totals for the cycle  The work done W0 Path 1 W0 Path 2 W0 Path 3 W0 cyclical process 38  Thus, the heat exchange Q  W cyclical process  Q0 cyclical process   For any cycle that is done in a counterclockwise direction, W  0 and Q  0  For any cycle that is done in a clockwise direction, W  0 and Q  0 39  Free Expansion Figure 8 : Free expansion. Opening the stopcock allows gas to flow from one side of the insulating container to the other. No work is done, and no heat is transferred to the environment 40  No work is done in this process  The process is adiabatic  Hence, with W = 0 and Q = 0, the first law gives E int  0 Free expansion   Thus the internal energy of an ideal gas undergoing a free expansion remains constant  Because the internal energy of an ideal gas depends only on the temperature, its temperature must similarly remain constant 41 Table 4 : Application of the First Law 42 EXAMPLE 43 Answer 44 45 46 47 END OF THE CHAPTER - 6 48