Thermodynamics
I. Temperature and Temperature Scales A. Definitions 1. Thermodynamics is the study of thermal energy and thermal motion. 2. Temperature is a direct measure of the average KE of atoms or molecules in a system. 3. Two or more objects are said to be in thermal contact if they have the ability to exchange heat with one another. This does not necessarily require physical contact! 4. Two or more systems in thermal contact are said to be in thermal equilibrium when they achieve the same temperature. Energy exchange between the two systems is constant so no heat is exchanged. 5. Heat is energy that is exchanged between two systems in thermal contact due exclusively to a temperature difference between the two systems. Note that heat and temperature are NOT the same things!!! B. Zeroth Law of Thermodynamics 1. If object A is in thermal equilibrium with object C, and if object B is in thermal equilibrium with object C, then object A is in thermal equilibrium with object B. 2. Objects A, B, and C are all at the same temperature 3. If they are not at the same temperature, then hot objects cool down and cool objects warm up until an equilibrium temperature between the extremes is achieved. C. Temperature Scales 1. Fahrenheit Scale a. water freezes at 32 °F at atmospheric pressure b. water boils at 212 °F at atmospheric pressure c. scale is based upon human body temperature of 100 °F, but the calibration was too high by 1.4 °F 2. Celsius Scale a. water freezes at 0 °C at atmospheric pressure b. water boils at 100 °C at atmospheric pressure c. scale is based upon properties of water 3. Kelvin Scale a. water freezes at 273.15 K at atmospheric pressure b. water boils at 373.15 K at atmospheric pressure c. absolute zero = 0 K is basis for Kelvin scale d. same size unit as Celsius degree but offset by 273.15 K 4. Conversions 9 ° a. TF 5 TC 32 F
b. TK TC 273.15 K 9 c. Celsius and Kelvin have same size unit; Fahrenheit unit is 5 the size of the Celsius and Kelvin unit. d. Temperature only doubles with a doubling of absolute temperature (e.g., 100 °C ≠ 2 * 50 °C, 100 °F ≠ 2 * 50 °F, but 100 K = 2 * 50 K). Therefore, absolute temperatures MUST be used in most thermodynamic calculations unless a temperature difference is specified. 5. Example 19.1 D. Thermometers 1. Thermometers as measuring devices are based upon the principle that some physical property of a system is a function only of temperature. a. volume of a liquid b. length of a solid c. gas pressure at constant volume d. gas volume at constant pressure e. electric resistance of a conductor f. color of an object 2. mercury and alcohol thermometers are based upon a fluid height at a given temperature a. because of differences in expansion properties, temperatures may be slightly inaccurate far from calibration points b. only a limited range of temperature can be covered before the liquid changes state at high or low temperatures 3. gas thermometers a. good agreement with gases as pressure is reduced, but vacuums are tough to approach b. must stay above condensation point or gas turns to liquid c. when lines on pressure-temperature graphs are extended, they all intersect at –273.15 °C = 0 K = absolute zero E. Definition of Absolute Zero 1. Absolute zero is the absence of all motion of individual particles and the KE of the system is zero 2. zero-point energy must exist (Heisenberg Uncertainty Principle) so 0 K can never be truly attained. It is a theoretical limit.
II. Heat and Internal Energy A. Definitions 1. The internal energy of a system is all of the energy associated with its atoms and molecules when viewed from a reference frame at rest with respect to the system 2. Heat is the transfer of energy to or from a system and its surroundings due to a temperature difference between them 3. Units of Heat a. A calorie is the amount of energy required to raise the temperature of 1 gram of water by 1 °C. The common Calorie (capital C) listed in food products is really 1000 calories (lowercase c) = 1 kilocalorie. b. A British Thermal Unit (BTU) is the amount of energy required to raise the temperature of 1 pound of water by 1 °F. This is a VERY large unit! c. A Joule is the SI unit of energy. 1 calorie = 4.186 Joules B. Work, Heat, Energy 1. Work is the change in KE due to external forces acting on the system as a whole. 2. Heat is the amount of energy transferred to or from a system due to temperature differences. 3. Internal energy is defined by the total average energy of all atoms and molecules in the system. Internal energy is directly related to temperature. 4. work, heat, and internal energy all have the same units, but represent three fundamentally different, but related physical concepts C. Mechanical Equivalent of Heat 1. Mechanical energy lost as friction is stored as internal energy in another system or as heat 2. Joule showed that the loss in mechanical energy is proportional to the change in the temperature of water, and the constant of proportionality is 4.186 J/cal. D. Heat Capacity and Specific Heat 1. When heat is added to a system or work is done, the temperature of the system usually rises (a phase transition is the only exception) 2. the temperature rise varies with the ability of a substance to absorb or emit heat 3. the heat capacity of a substance is the energy per unit mass per unit temperature required to raise 1 gram of a substance by 1 °C 4. the specific heat (capacity) of substance is heat capacity per unit mass
5. Q = mcpT Q = heat gained or lost from the substance m = mass of the substance cp = specific heat capacity at constant pressure T = change in temperature of the substance 6. specific heats for various substances are given in Table 19.3 7. specific heat at constant volume (cv) differs by a few percent from cp . These differences will be discussed later. 8. specific heat actually varies as a function of temperature but we will not be concerned with that at this level 9. water has a very high specific heat leading coastal areas to maintain more moderate temperatures than inland areas 10. cheese releases more heat than pizza crust at the same temperature so hot cheese is more likely to burn you than hot pizza crust at the same temperature E. Calorimetry 1. measures specific heat 2. sample at Tx > Twater is placed in known mass of water 3. little mechanical work is done and little energy is lost to environment, so most energy is converted to internal energy of the water 4. Derivation Qwater = -Qx At equilibrium temperature Tf
mwcw (T f Tw ) mx cx (T f Tx )
mwcw (T f Tw ) cx mx (Tx T f ) 5. Example 19.3 in class F. Latent Heat 1. Phase changes involve energy transfer but no change in temperature a. solid liquid (melting, freezing) b. liquid gas (vaporizing, condensing) c. solid gas (sublimating, condensing) 2. Latent heat is the energy required to change the phase of a substance a. Lf = latent heat of fusion (melting) b. Lv = latent heat of vaporization c. Ls = latent heat of sublimation (rarely used) d. Table 19.4 latent heats for various substances 3. Q = mL represents heat energy required to change state. Therefore L = Q/m 4. Example: What is the energy required to convert 1 gram of ice at –30 °C to steam at 120 °C. The specific heats of ice, water, and steam are respectively ci = 2090 J/kg °C, cw = 4186 J/kg °C, cs = 2010 J/kg °C. The latent heats of fusion and vaporization are respectively Lf = 3.33 5 6 x 10 J/kg and Lv = 2.26 x 10 J/kg. a. Qtot = Qi + Qf + Qw + Qv + Qs
b. Qi = mciTi -3 Qi = (1.0 x 10 kg)(2090 J/kg °C)(0 °C-(-30 °C)) Qi = 62.7 J (raises ice to melting point) c. Qf = mLf -3 5 Qf = (1.0 x 10 kg)(3.33 x 10 J/kg) Qf = 333 J (melts ice to water)
d. Qw = mcwTw -3 Qw = (1.0 x 10 kg)(4186 J/kg °C)(100 °C - 0 °C) Qw = 419 J (raises water to boiling point) e. Qv = mLv -3 6 Qv = (1.0 x 10 kg)(2.26 x 10 J/kg) Qv = 2260 J (vaporizes water to steam)
f. Qs = mcsTs -3 Qs = (1.0 x 10 kg)(2010 J/kg °C)(120 °C - 100 °C) Qs = 40.2 J (raises steam to 120 °C) g. Qtot = 62.7 J + 333 J + 419 J + 2260 J + 40.2 J Qtot = 3114.9 J G. Example 19.4 in class H. Note that melting/freezing and boiling/condensation points depend upon pressure and composition 1. higher elevations are at lower pressure, so the boiling point of water is lower 2. salt + water lowers freezing point so roads are salted in cold weather
III. Work in Thermodynamic Processes A. Equation of State 1. Pressure 2. Internal energy 3. Volume 4. Temperature 5. Density B. Equation of state relates two or more of these variables when system is in thermal equilibrium C. Work on a System 1. quasi-static (slow changes) are required to maintain thermodynamic equilibrium 2. dW = F dx = PA dx = P dV 3. if system expands, dV > 0 and dW > 0; if system contracts, dV < 0 and dW < 0 4. work done BY gas is negative; work done ON gas is positive 5. + work = transfer of energy out of the system; - work = transfer of energy into the system. This is the opposite of mechanics due to an unfortunate historical convention. Vf 6. dW = P dV so W P(V)dV Vi e r u s s e r 7. Work is area under P-V curve and it is P path dependent! 8. Consider several different paths as shown V olu me below. Case (b) represents the greatest amount of work, case (c) represents the least amount of work, and case (a) represents an intermediate situation. e e r r e u r u s u s s s s e s e r r e P r P P
Volume Vo lu me Vo lu me (a) (b) (c)
9. Work done by/on a system depends upon a. initial state (i) b. final state (f) c. path followed from i to f 10. Energy transfer as heat is also path dependent
IV. First Law of Thermodynamics A. Two methods for exchanging heat with Surroundings 1. Work is mechanical intervention 2. Heat is energy transfer controlled by temperature differences B. Path Dependence 1. Work and heat depend upon the path taken from the initial to the final state 2. Difference between work and heat (Q – W) is independent of path!
3. Eint = Q – W is the First Law of Thermodynamics (conservation of energy), and it depends only upon the initial and final states of the system. The path is unimportant! 4. Q > 0 when energy enters the system and W > 0 when system does work. Typically, if Q > 0, then W > 0 unless outside forces are added. Note that Q and W are not properties of the system; they are simply agents that affect the system. C. Special Cases of the First Law of Thermodynamics 1. dEint = dQ – dW (state function) 2. isolated system a. no energy transfer occurs b. Q = W = 0
c. Eint = Q – W d. the internal energy of an isolated system is constant 3. cyclic processes a. begin and end process at the same state
b. Eint = Q – W = 0 c. Q = W d. In a cyclic process, the net work done by the system per cycle equals the area enclosed by the path on a P-V diagram 4. no work done (see isovolumetric processes) a. W = 0
b. Eint = Q
c. If energy enters the system, Eint increases; if energy leaves the
system, then Eint decreases 5. Adiabatic processes (no heat is exchanged) a. Q = 0 because no heat is exchanged b. can achieve adiabatic behavior if system is either well- insulated or if change occurs too rapidly for heat transfer
c. Eint = - W
1. If system expands, W > 0 and Eint decreases
2. If system is compressed, W < 0 and Eint increases d. adiabatic free expansion 1. insulated so Q = 0 2. expansion into vacuum so W = 0
3. Eint = 0 and no temperature change occurs 6. Isobaric processes (pressure is constant) a. constant pressure b. neither Q nor W equals zero
c. W = P V = P(Vf – Vi) 7. Isovolumetric processes (volume is constant) a. constant volume implies W = 0 because V = 0
b. Eint = Q c. all changes in the system occur as heat transfer d. this is a specific case of a “no work” process e. can of spray paint tossed into a fire 1. energy enters system from fire 2. P and T rise, but V = 0 3. eventually P exceeds yield strength of can and can explodes 8. Isothermal processes (constant temperature) a. P vs. V at constant T yields a curve called an isotherm
b. Eint = Eint(T) for ideal gases so Eint = 0 in an isothermal process c. Q = W is similar to a cyclic process D. Example 19.5
V. Energy Transfer mechanisms (3 types) A. Conduction 1. two systems at different temperatures T1 > T2 are in direct physical contact 2. energy is transferred when atoms of T1 collide with atoms of T2 increasing the energy in T2 and decreasing the energy in T1 3. heat transfer ends when T1 = T2 = Tequilibrium Q T dT 4. A A t x dx Q P power t dT P kA dx 5. k = thermal conductivity in Watts/m °C dT temperature gradient dx 6. good thermal conductors (metals) have large k 7. good thermal insulators (non-metals) have small k (home insulation) B. Convection 1. heat energy is transferred by the movement of a heated substance 2. two systems not necessarily in direct contact will come to an equilibrium temperature if the medium between them is convecting 3. examples of natural convection a. heated air around a fire b. mixing of cool and warm air at the beach c. water evaporated by sunlight 4. examples of forced convection a. fans cool air b. heat pumps c. boiling water on stove C. Radiation 1. energy released as E-M waves 2. does not require direct physical contact between objects (Sun and Earth) 4 3. Stefan’s law: P AT a. = 5.6696 x 10-8 W/m2 K4 (Stefan-Boltzmann constant) b. = emissivity (0 ≤ ≤ 1) equals the fraction of incoming radiation that is absorbed c. T = temperature 2 4. Isolar = 1340 W/m from sun on average 5. At night, Earth re-radiates most (99.98%) of energy it absorbs from the Sun during the day 6. A(T 4 T 4 ) where T is the object’s temperature and T is the P net 0 0 temperature of the surroundings 7. = 1 is an ideal absorber and emitter (also called a blackbody) 8. = 0 is an ideal reflector D. Dewar Flasks 1. one example is a Thermos bottle 2. all Dewar flasks minimize all three types of heat transfer E. Examples 19.6 and 19.7
VI. Second Law of Thermodynamics A. Heat always flows from hot to cold B. Mechanical energy cannot be completely converted to other useful forms of energy without some loss of heat C. Some processes are irreversible which means that the probability of a process occurring is vastly greater in one direction than the reverse direction 1. oscillating pendulum slows down due to collisions with air molecules; they never speed up 2. breaking a triangle of pool balls is easy, but striking a ball in such a way as to restore the triangle after the break is virtually impossible 3. death is irreversible 4. time flows in the forward direction only D. All closed (isolated) systems proceed from order to disorder. If a part of the system becomes more ordered, it does so at the expense of greater disorder to the system as a whole 1. easier to make a mess than to clean it up 2. easier to grow older than younger 3. life on Earth becomes more ordered (humans vs. bacteria) but only because Earth receives a tremendous energy input from the sun 4. entropy is a measure of disorder a. entropy is a state function and is therefore independent of thermodynamic path b. entropy of the universe increases in all real processes c. in reversible processes, entropy always increases d. in reversible processes (extremely rare but not impossible) the entropy remains zero e. entropy of a non-closed system can be decreased, but only at the expense of a greater entropy in the combined system dQ Q f. dS S T T E. Examples of Entropy Change 1. Thermal Conduction (T2 > T1) a. –Qlost = Qgained = Q Qlost Q b. S2 (entropy decreases) T2 T2
Qgained Q S2 (entropy increases) T1 T1
c. Snet S1 S2 Q Q 1 1 S Q T1 T2 T1 T2 d. Define S = 0 before heat transfer begins
S S net 0 Snet 1 1 Snet Q T1 T2 1 1 Since T2 T1 Snet 0 T2 T1 S 0 2. Free Expansion (T = constant) dQ f dQ 1 f a. dS dQ T i T T i b. If T = constant then dQ = dW = P dV k c. Suppose P (k 0) V 1 f k f dV dS PdV T T V i i k V dS ln( f ) T Vi
d. Since k > 0, T > 0, and Vf > Vi in expansion, dS > 0 3. Calorimetric Processes (T2 > Tf > T1) dQ a. dS T b. Qgained = -Qlost c. Q = mc T
d. Qgained m1c1 (T f T1 )
Qlost m2c2 (T f T2 ) m c (T T ) m c (T T ) 1 1 f 1 2 2 f 2 (m1c1T1 m2c2T2 ) T f (m1c1 m2c2 )
e. Snet = S1 +S2
f. Sinitial = 0 so S = Snet – 0 = Snet
T f T f dQ m c dT T f g. S 1 1 m c ln( ) 1 T T 1 1 T T1 T1 1 T f T f dQ m c dT T f S 2 2 m c ln( ) 2 T T 2 2 T T2 T2 2
Tf Tf h. Snet m1c1 ln( ) m2c2 ln( ) T1 T2
if T1 = T2 = Tf then Snet = 0
if Tf T1 then m2c2 0 and Snet > 0 Tf if Tf T2 then m1c1 0 but ln( ) 0 and Snet > 0 T2 4. Examples 21.1 and 21.2
VII. Heat Engines A. Heat engines convert internal energy to mechanical energy 1. Power plants produce electricity by burning coal, nuclear fuel, etc., which heats water to steam. Steam turns a turbine that drives an electric generator. 2. General operation of a heat engine a. energy is absorbed from a high-temperature reservoir b. work is done by the engine c. energy is expelled by the engine into a low-temperature reservoir
d. the process is cyclic so Eint = 0
e. Eint = Q – W = 0 W = Qnet W = Qh – Qc (Qh > 0, Qc > 0) 3. Net work done in a cyclic process is the area enclosed by the process on a P-V diagram 4. Efficiency = e W e Q Q Q e h c Qh Q e 1 c Qh 5. Efficiency of real engines is always < 100% 6. 2nd Law statements a. cannot get more energy out of a cyclic process than the amount of energy put into the process b. cannot break even because Qc = 0 in that case, and the machine would be a perpetual motion machine. Furthermore the cold reservoir would not remain at 0 K after that. B. Refrigerators and Heat Pumps 1. refrigerators and heat pumps are heat engines running in reverse 2. energy is absorbed from Qc and deposited to Qh 3. requires work to be done on the engine because heat flows from hot to cold according to the 2nd law of thermodynamics 4. since no engine is perfect (100%) efficiency, it takes more work to remove heat than the amount of heat that is removed. 5. cooling a room with a refrigerator does not work because the compressor expels more heat than the refrigerator can remove C. Carnot Engine 1. heat engine that operates through a theoretical, ideal, reversible cycle (Carnot cycle) is theoretically the most efficient engine possible 2. represents theoretical upper limit on efficiency, because real engines do not operate through a reversible cycle and also suffer from friction and other energy losses Q 3. e 1 c Qh
Qc = mc(Tc- 0 K) = mcTc (heat of cold reservoir relative to 0 K) Qh = mc(Th- 0 K) = mcTh (heat of hot reservoir relative to 0 K) Q T c c Q T h h T e 1 c Th
4. Note that e = 100% at Tc = 0 K; however, the addition of energy to a reservoir initially at 0 K raises its temperature instantaneously and therefore lowers the efficiency of the engine. 5. Since 0 K is neither attainable nor sustainable, and because Carnot engines represent the theoretical upper limit of efficiency, no real engines can ever attain efficiencies of 100% 6. Examples 21.3 and 21.4 D. Coefficients of Performance energy from high temp Q 1. COP (heating mode) h work done by pump W a. pumping heat into hot reservoir requires external work b. Qh is generally greater than W so COP > 1 c. higher COP is desirable because it means more heat is transferred for a given amount of work d. at 25 °F, COP ~ 4 e. for temperatures below ~ 15 °F COP < 1 and use of heat pumps for heating houses is inappropriate Q Q T f. COP (heating mode) = h h h W Q h Q c Th Tc energy removed low temp Q 2. COP (cooling mode) c work done by pump W a. pumping heat out of cold reservoir requires external work b. higher COP is desirable because it means more heat is removed for a given amount of work Q Q T c. COP (cooling mode) = c c c W Q h Q c Th Tc d. As Th Tc 0, COP , but practically COP < 10. This arises because it is more difficult to oppose the second law of thermodynamics with greater temperature differences
VIII. Ideal Gases at the Macroscopic Level A. Assumptions 1. T must not be too high or too low 2. P must be relatively low 3. molecular volume << volume of container B. Equation of State total mass M 1. # moles = n = molecular mass m 2. PV = nRT = NkbT 23 3. # molecules = N; NA = Avogadro’s # = 6.02 x 10 molecules/mol -23 4. kb = Boltzmann’s constant = 1.38 x 10 J/K 5. R = gas constant = 8.315 J/mol K = 0.08214 L atm/mol K PV C. Ideal gases always have nT cnst at all pressures D. P, V, T are thermodynamic variables E. Examples 20.1 and 20.2 F. Special Cases of Ideal Gas behavior 1. Isothermal expansion V V f f nRT a. W PdV dV V Vi Vi
V f b. W nRT ln( ) Vi 2. Quasi-static and Reversible Process a. cv = specific heat capacity at constant volume b. dQ = dEint + P dV c. dEint = ncv dT nRT d. dQ nc dT dV v V dQ dT dV e. dS nc nR T v T V S f f dT f dV dS nc nR v T V Si i i
T f V f S ncv ln( ) nR ln( ) Ti Vi
IX. Ideal Gases at the Microscopic Level A. Kinetic Theory of Gases 1. individual molecules are considered to be hard solid spheres 2. spheres move randomly 3. spheres collide elastically with each other and container walls 4. spheres do not deform 5. works for monatomic gases and must be modified for diatomic, triatomic, and other molecules B. Microscopic Model 1. # of molecules is large 2. volume of molecules << container volume 3. molecules obey Newton’s laws of motion 4. molecules move in random directions with a distribution of speeds 5. all collisions are elastic 6. forces between molecules are negligible 7. molecules interact with each other only during collisions 8. all molecules are identical 9. molecular rotations and vibrations do not affect the average motions of particles C. Derivation of pressure on Container Walls 1. consider the collision of 1 molecule moving initially in the +x direction with the container wall
2. px p f pi mvx (mvx ) 2mvx
px F1 t 2mvx (F1 force on molecule) 3. t = time between collisions with the same wall 2d t vx 2mv 2mv F x x 4. 1 t 2d vx mv 2 F x 1 d rd 5. F1w F1 (Newton's 3 Law) mv 2 F x 1w d
m 2 2 2 6. Fnet Fiw (v1x v2x ⋯ vix ) i d 1 7. v 2 (v 2 v 2 ⋯ v 2 ) x N 1x 2x Nx m 8. F (Nv 2 ) net d x 2 2 2 2 9. v vx v y vz 2 2 2 vx v y vz if motion is random 2 2 2 2 2 v vx v y vz 3vx m Nm v 2 10. F (Nv 2 ) ( ) net d x d 3 F F 11. P A d 2 1 Nm v 2 Nmv 2 P ( ) d 2 d 3 3d 3 Nmv 2 P 3V 2 N 1 P ( mv 2 ) 3 V 2 12. Pressure is proportional to the number density of molecules and the average KE of the molecules 13. You can increase the pressure in a container by either increasing the # of molecules or by increasing their average KE D. Ideal Gas law 1. Standard Temperature and Pressure (STP) are 0 °C and 1 atm 2 N 1 1 2. P ( )( mv 2 ) N( mv 2 ) 3 V 2 3 1 3. PV Nk T N( mv 2 ) B 3 1 k T mv 2 B 3
2 1 T ( mv 2 ) 3k B 2 4. T is therefore a direct measure of average molecular KE 1 3 5. mv 2 k T 2 2 B 1 6. v 2 v 2 x 3 1 1 1 1 mv 2 k T mv 2 mv 2 2 x 2 B 2 y 2 z 7. Each direction is a single degree of freedom and contains some of the molecular energy 8. Equipartition of Energy Theorem: Each degree of freedom 1 contributes 2 k BT to the energy of the system 1 2 3 3 9. Etrans N( 2 mv ) 2 Nk BT 2 nRT 3k T 3RT 10. v v 2 B rms m M 1 11. vrms m E. Molar Specific Heat of an Ideal Gas
1. Energy required to raise Ti to Tf depends upon the path because Eint = Q – W is independent of path, but Q and W alone are not 2. Molar Specific Heats
a. Qv ncv T (constant volume)
b. Q p nc p T (constant pressure) c. c p cv because at constant pressure, both internal energy AND work are increased; in constant volume processes, no work is done 3. Monatomic Ideal Gas 3 3 a. Eint 2 Nk BT 2 nRT b. at constant volume, dV = 0 so dW = PdV = 0
Eint = Qv – W = Qv = ncv T c. dEint = ncv dT Eint = ncv T 3 d. Eint 2 nRT dE int 3 nR nc dT 2 v 3 cv 2 R (monatomic gases only) e. At constant pressure
Eint = Qp – W = Qp = ncp T - PV
Eint = ncv T PV = nR T
ncv T = ncp T - nR T cv = cp - nR for all ideal gases f. monatomic gases 3 1. cv 2 R
cv c p R 5 c p 2 R
c p 2. Define (ratio of specific heats) cv 5R 2 5 1.667 3R 3 2 3. = 1.667 agrees very well with experimental results for monatomic gases, but disagrees with experiment for more complex gases g. complex gases 1. add a degree of freedom for the rotation and vibration of each atom in a molecule 1 2. each degree of freedom contributes 2 k BT to the total internal energy of the system 3. diatomic gas example a. a second molecule adds two degrees of freedom, 3 1 5 so Eint 2 k BT 2( 2 k BT ) 2 k BT 5 b. cv 2 R
cv c p R 7 c p 2 R 7R 2 7 c. 1.4 5R 5 2 4. for more than 2 atoms per molecule, different vibration patterns make the results more complex h. important note for the future is that energy is quantized in steps 1 of 2 k BT which forbids non-integer steps in ideal gases i. Brief Aside for Solids 1. if an atom in a solid is slightly displaced, it can rotate and vibrate in x, y, z, but it cannot translate (it is stuck in place) 2. rotation and vibration provide 6 additional degrees of 1 freedom so Eint 0 6( 2 k BT ) 3k BT
3. cv 3R is the Dulong-Petit Law for solids 4. at high temperatures, all solids appear to obey the
Dulong-Petit Law and cv 3R 25 J/mol K 4. Examples 20.7 and 20.8 F. Adiabatic Processes for an Ideal Gas 1. adiabatic requires that Q = 0
2. Eint = Q – W = -W and dEint = -dW 3. examples include the gasoline engine and slow expansion of thermally-insulated gas 4. PV = nRT so d(PV) = PdV + VdP = nRdT 5. Derivation
dEint dW
ncv dT PdV PdV VdP nRdT PdV dT ncv PdV R PdV VdP nR( ) PdV ncv cv
R c p cv c c PdV VdP ( v p )PdV (1 )PdV cv PdV VdP (1 )PdV Divide by PV dV dP dV (1 ) V P V dV dP 0 V P lnV ln P ln(cnst)
PV cnst 6. PiVi Pf V f (adiabatic) nRT 7. Substitute P V 1 1 TiVi T f V f 8. Example 20.9 G. Gasoline and Diesel Engines 1. Otto cycle a. intake stroke (O to A): piston moves down drawing in air and fuel mixture at atmospheric e r u s s e r pressure. Volume P increases from V1 to V2. Energy is carried into the system by mass transfer (fuel) and is represented by Volume Qh. b. compression stroke (A to B): system is compressed adiabatically from V2 to V1; temperature rises from TA to TB; work done is negative. c. spark plug fires (B to C): pressure and temperature rise, but volume remains constant; no work is done d. power stroke (C to D): gas expands adiabatically from V1 to V2; positive work is done on the piston e. exhaust valve opens (D to A): energy is released and pressure suddenly drops; no work is done f. exhaust stroke (A to O): piston moves upward while exhaust valve remains open and volume decreases from V2 to V1. 2. Efficiency of engine if air-fuel mixture is ideal gas V 1 e 1 ( 1 ) 1 1 a. V 1 V2 2 V1 V b. 2 is called the compression ratio V1 c. ideal gas engines attain e = 56% d. real engines attain e = 15% - 20% 3. Diesel engines operate on the Otto cycle but do not use a spark plug. Instead, the compressed gas is itself at a temperature high enough for self-ignition H. Entropy on a Microscopic Scale 1. Gas molecules move randomly so there is an infinitesimally tiny chance that they may all end up in an ordered state on one side of the container. This is just one possible state.
Vi 2. # states for a single molecule is wi where Vi is the container Vm
volume and Vm is the volume of a single molecule N 3. For N molecules, the number of possible states is W* w Vi N V f N 4. W*i ( ) and W* f ( ) Vm Vm
W* f V f 5. ( ) N W*i Vi
W* f V f V f ln( ) N ln( ) nN A ln( ) W*i Vi Vi
W* f V f V f k B ln( ) k B nN A ln( ) nR ln( ) W*i Vi Vi 6. Entropy for free expansion of a gas V S nR ln( f ) Vi
W* f S k B ln( ) W*i 7. Entropy is a measure of microscopic disorder. 8. Entropy is described by statistical mechanics. The probability of finding all molecules in 1 mol of a gas in one corner of a container is
1 23 P(W* ) 1.610 N A 9. Examples 21.5 and 21.6
X. Maxwell-Boltzmann Distribution A. Molecules in a gas do not all move at the same speed. B. Distribution of molecular speeds depends upon the molecular mass and the temperature of the gas and is called the Maxwell-Boltzmann Distribution 1. If N = # molecules in a gas, then # molecules with speeds between v and v + dv is dN = Nvdv mv2 m 3 2 2 2kBT 2. N v 4N( ) v e 2 k BT
3. vmp v vrms where vmp is the most probable speed 3k T 4. v B root mean square speed rms m 8k T v B average speed m 2k T v B most probable speed mp m
5. Function produces a bell-shaped curve with most molecules near vmp, but with outliers moving extremely fast or extremely slow 6. The difference in speeds allows some molecules to escape Earth, fewer to escape Jupiter, and more to escape the Moon. This accounts in large part for the differences in atmospheric compositions between planets of various sizes. 7. Example 20.3 C. Boltzmann Distribution 1. Probability of finding molecules in a particular energy state depends E upon k T where E is the energy of the state e B E
2. kBT nV (E) n0e 3. Examples 20.5 and 20.6 D. Mean Free Path 1. The mean free path of an atom or molecule is the average distance between collisions with other gas particles 2. In time t, a particle of diameter d sweeps out a cylinder whose length is vt vt 1 l 3. The # of collisions in time t is therefore n where d 2 vtn v 4 v
nv is the number density of gas molecules in the container and is the cross-sectional area of the molecule. 1 v 4. The collision frequency is f n v t l v 1 1 5. The mean free time is tmf f nv v 6. actual calculations including the motion of molecules in the cylinder shows that more exactly 1 v l and f 2nv v 2nv l 7. Example 20.4 E. Exponential Atmospheres
1. PV NkBT (assumes atmosphere is isothermal) N 2. n v V
3. P nvkBT 4. Consider air of thickness dy
weight mgnvV mgnv Ady Fy 0 (P dP)A nv Ady mg PA
dP nv mg dy
P nv k BT
dP k BT dnv
k BT dnv nv mg dy
dn mg v dy nv k BT mgy E kBT kBT nv (y) n0e n0e mgy kBT P P0e
where P0 = n0kBT 5. Law of Atmospheres mgy y a. k BT H nv (y) n0e n0e mgy y b. k BT H P P0 e e k T thermal energy c. H scale height B mg weight
XI. Thermal Expansion of Solids and Liquids A. Thermometers are based upon the principles that: 1. as T increases, V increases 2. as T decreases, V decreases 3. water is a weird exception because as T falls toward 4 °C, water contracts. Below 4 °C, water expands and density decreases. As a result, ice floats on water and water freezes from the top down instead of the bottom up. B. Engineering Applications 1. Metal and concrete expand when heated, so joints are used on roads, bridges, and railroad tracks to allow expansion when heated by the sun. 2. Plastic containers enclosing liquids will expand or contract with temperature differences, and the material must withstand those stresses (e.g., water bottle partially filled sits in sun all day, and then when placed in refrigerator the plastic collapses C. Linear Expansion of Solids 1. Assume expansion is small relative to original size. L linear strain L 2. i cnst if T is small change in temp T
3. L Li T L f Li 4. has units of °C –1 or K-1 5. Table 19.2 lists coefficients of linear expansion 6. Note that if is negative, the substance contracts as temperature increases D. Volume Expansion of Solids and Liquids 1. linear expansion in 3 dimensions so define = 3 2. derivation
Vi V (l l)(w w)(h h)
Vi V (l l T )(w w T )(h h T) 3 Vi V lwh (1 T) 3 Vi V Vi (1 T) 2 3 Vi V V i [1 3 T 3( T ) ( T ) ] 2 3 V V i [3 T 3( T ) ( T) ] If T 1 then high orders vanish rapidly
V 3 T V i T V i
3. You can also show that A 2 T V i for area expansion 4. Example 19.2 E. Bimetallic strips 1. Different metals have different coefficients of expansion. 2. Place 2 long strips of different metals together at room temperature 3. raise temperature and metal with higher bends more than metal with lower ; so bimetallic strip bends into an arc with higher on the outside. 4. lower temperature cause strip to bend into an arc with lower on the outside 5. thermostats make use of this principle