16.8 PT Clausius Clapeyron equation.pdf, Lecture notes of Physics

Download 16.8 PT Clausius Clapeyron equation.pdf and more Lecture notes Physics in PDF only on Docsity! Clausius-Clapeyron equation Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: November 27, 2017) The Clausius–Clapeyron relation, named after Rudolf Clausius and Benoît Paul Émile Clapeyron, is a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent. On a pressure–temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. Mathematically, VT L V S dT dP      The derivation of this equation was a remarkable early accomplishment of thermodynamics. Both sides of this equation are easily determined experimentally. The equation has been verified to high precision. 1. Phase diagram of water The P-V phase diagram of the water is shown below. Fig. The phase diagram of H2O showing the solid (ice), liquid (water) and gaseous (vapor) phases. The horizontal dashed line corresponds to atmospheric pressure, and the normally experienced freezing and boiling points of water are indicated by the open circles. Critical point: Tc = 647.1 K, Pc = 218 atm = 22.08 MPa Triple point: Ttr = 273.16 K Ptr = 4.58 Torr= 0.00602632 atm The phase diagram of H2O is divided into three regions, indicating the conditions under which ice, water, or stream, is the most stable phase. A phase diagram of water shows the solid, liquid and gaseous phases. The three phases coexist at the triple point. The solid-liquid phase boundary is very steep, reflecting the large change in entropy in going from liquid to solid and the very small change in volume. This phase boundary does not terminate, but continues indefinitely. By way of contrast, the phase boundary between liquid and gas terminates at the critical point. ((H2O)) Latent heat at gas-liquid transition         kJ/mol .7704J/g 76.2264 kJ/mol 73.44J/g 2485 0L C 100at C 0at 1 cal = 4.184 J, H2O = 18 g/mol L0 = 2264.76 J/g = 9.7228 kcal/mol = 540 cal/g Phase diagram of H2O Typical curves for the P-T phase diagram in the van der Waaks gas are sketched below. Two temperatures are taken to be close together, TTT 1 and TT 2 . Now we can carry out a Carnot cycle using these two neighboring isotherms. Starting at 1, where the system is just all liquid and the temperature is TTT 1 , the cycle is as follows. 1→2: isothermal expansion during which the liquid evaporates. The evaporation requires latent heat, so that the heat Q1 is absorbed in the system. 2→3: A small adiabatic expansion during which the temperature falls by the small amount dT; TT 2 3→4: An isothermal compression to the point where all the vapor is condensed. Latent heat 2Q is rejected. 4→1: A small adiabatic compression during which the temperature rises by T back to TTT 1 Fig. The gas-liquid co-existence in the P vs V curve for the van der Waals gas. Carnot cycle Path 1→2: isothermal process at the temperature TTT 1 Path 2→3: isentropic process Path 3→4: isothermal process at the temperature TT 2 Path 4→1: isentropic process The work done by the system is VPW  where V is the volume difference between the points 3 and 4, and the work corresponds to the area of closed rectangle path (1→2→3→4→1). The efficiency of the Carnot cycle is given by 1 2 1 21 1 1 T T Q QQ Q W    Thus we have 1 2 1 21 1 1 T T Q QQ Q VP      or 11 21 T T T TT L VP      where LQ 1 (the latent heat) Then we get the Clausius-Clapeyron equation; VT L dT dP T P     . 3. Thermal equilibrium: chemical potential Thermodynamic conditions for the co-existence of two phases are the conditions for the equilibrium of two systems that are in thermal, diffusive and mechanical contact. 21 TT  21   21 PP  For liquid and gas TTT gl  , gl   , PPP gl  . The chemical potential: ),(),( TPTP gl   At a general point in the P-T plane the two phases do not co-exist. If gl   , the liquid phase alone is stable. If lg   , the gas phase alone is stable. ((Note)) Metastable phases may occur, by supercooling or superheating. 4. Derivative of the co-existence curve, P vs T: ),(),( 0000 TPTP lg   This is a condition of co-existence. We start with an equation ),(),( 0000 dTTdPPdTTdPP lg   We make a series expansion dT T dP P TPdT T dP P TP P l T l l P g T g g                                      ),(),( 0000 In the limit as dP and dT approach zero, dT T dP P dT T dP P P l T l P g T g                                  This may be rearranged to give 12 12 vv ss dT dP    (Clausius-Clapeyron equation) 6. Clausius-Clapeyron equation ' ( )g ld Q T s s l   l : latent heat of vaporization/molecule lg vvv  Thus we have the Clausius-Clapeyron equation, dP l dT T v   Two approximations: (a) lg vv  g g glg N V vvvv  (b) TkNPV Bgg  (ideal gas law) or TkPv Bg  Using these two approximations, we have 2 g g B dP l l lP lP dT T v Tv TPv k T      Suppose that the latent heat l is independent of T; 2 B dP l dT P k T   or ln B l P k T   +constant or 0( ) exp( ) B l P T P k T   where L0 is the latent heat of vaporization of one molecule. If 0L refers instead to one mole, then A B A lN L k N R  0( ) exp( ) L P T P RT   . where R = 8.3144598 [J/(K mol)]. AL lN . L = 2256 x 18.01528 = 40.642 (kJ/mol): latent heat of vaporization/mol for water: 7. Phase diagram of H2O and Clausius-Clapeyron equation p (bar) 10° 10 10 solid 200 triple point 300 l 400 11 500 T (K) critical point 1 600 1 700 800 sl ss  In summary, we have slg sss  8. Phase diagram of liquid 4He 100- Solid (bec Liquid He-| 10F . -lin Critical 8 point a 4b. sssags : Gas: Liquid oe He-ll (superfluid 0.1+ . : Triple pare 7 a 0.01 fi | iE L 1 2 3 4 =~ 5: 2.17 4.22 5.20 (7) (7) (T) Temperature, T/K  melting curve s A-line Q ™~s Q He-I vapor pressur LL vapor pressure 00 10 20 30 40 50 6.0 T/K 6 Normal Liquid Pressure (MPa) co G 0 T T T —== T as 0 1 2 3 4 5 6 Temperature (K) Fig. Phase diagram of ‘He. 40h | Spin-ordered solid : Spin-disordered solid N 30f N N A phase Melting curve B phase N fe 20 Y (atm) Liquid 10 Critical point an 0 f : a! Vapor, 5 10-4 lo-3 10-2 107! 1 33 10 T(K) mm | 8 3He A, Solid 34+ Liquid 1+ ts + }—k _ 3.2 33 T (K) 40 P 35 (atm) 30 25 po po a L 2.3 87 10 100 1000 T (millikelvin) Fig. Phase diagram of *He.  Fig. Phase diagram of 3He. The Clausius-Clapeyron equation, 0    sl sl vv ss dT dP Since sl vv  , we have sl ss  . The entropy of liquid is less than the entropy of the solid. When we cross the co-existence curve ( lg   ) VPE NVPE STL lg    )(  The enthalpy H is defined by PVEH  VdPPdVdEdH  At constant pressure ( 0dP ) lg HHHVPESTL  Values of H are tabulated, PPPP P T H T V P T E T T S TC                                   dTCH P 11. Example-1 (Kittel 10-3) Calculation of dPdT ./ Calculate from the vapor pressure equation the value of dT/dP near P = 1 atm for the liquid- vapor equilibrium of water. The heat of vaporization at 100 C is 2260 J/g. Expain the result in K/atm. ((Solution)) Clausius-Clapeyron equation: 2)( RT PL PVT PL TV L V S dT dP gg     where R is the gas constant, R=8.3144598 J/(mol K) = 1.9872036 cal/(mol K) L is the latent heat of vaporization /mol. 7.40764/005.9743/705.2264  molcalgJL J/mol For T = 373 K and P = 1 atm, we get )/( 0352397.0 2 Katm RT PL dT dP  or  dP dT 28.3771 K/atm. 12. Example-2 (Kittel 10-3) Heat of vaporization of ice The pressure of water vapor over ice is 3.88 mmHg at -2 C and 4.58 mmHg at 0 C. Estimate in J/mol the heat of vaporization of ice at -1 C. ((Solution)) Clausius-Clapeyron equation 2)( RT PL PVT PL TV L V S dT dP gg     leading to RT L P ln +const (1) For the two points ),( 11 TP and ),( 212 TP on the curve given by Eq.(1), ) 11 (ln 212 1 TTR L P P       or ) 11 ( ln 22 2 1 TT P P R L         = 6135.55 or J/mol9.51013L @ InP Data 2 4:26:22 PM 11/30/2016 10 T 6 s 5 =a a = 0) Prien coiennabien 5 0 1 2 3 4 5 Fig.2 Plot of -dinP/d(1/T) vs T  The latent heat can be obtained from the value of )/1(/ln TdPd at each temperature. T = 0.2 K slope = 3.0 l 8.15 x 107 (erg/g) T = 0.4 K slope = 3.4 l 9.24 x 107 (erg/g) T = 0.6 K slope = 3.9 l 10.60 x 107 (erg/g) T = 0.8 K slope = 4.4 l 11.95 x 107 (erg/g) T = 1.0 K slope = 4.8 l 13.04 x 107 (erg/g) T = 1.2 K slope = 5.2 l 14.13 x 107 (erg/g) 14 Example 4 (Blundell-Blundell 28-6) It is sometimes stated that the weight of a skater pressing down on their thin skates is enough to melt ice, so that the skater can glide around on a thin film of liquid water. Assuming an ice rink at -5° C, do some estimates and show that this mechanism will not work. (In fact, frictional heating of ice is much more important, see S.C. Colbeck, Am. J. Phys. 63, 888 (1995) and S.C. Colbeck, L. Najarian, and H.B. Smith Am. J. Phys. 65, 488 (1997).) ((Solution)) 31000.1 l kg/m3, 310917.0 s kg/m3, 31000.1 1  l lV  m3/kg, 310091.1 sV m3/kg, 334fL kJ/kg (latent heat of fusion for water) 7 3 5 1034.1 10091.0 1034.3 273 11           sl f sl sl VV L TVV SS dT dP Pa/K = -132.2 atm/K. where 1 atm = 1.01315 x 105 Pa. For a very heavy skater (100 kg), only making the contact with ice over an area 10 cm x 1 mm = 10-4 m2. 410 8.9100   P = 9.8 x 106 Pa = 96.7 atm Then APPENDIX-1 Derivation of Clausius-Clapeyron equation from the Maxwell’s relation. Using the Maxwell’s relation we have TV V S T P                 which leads to the Clausius-Clapeyron equation VT L V S dT dP      . APPENDIX-II Maxwell relation (ii) Maxwell's relation VT T P V S                 Here we consider the Maxwell's relation VT T P V S                 For TV S         , in the Born diagram, we draw the lines along the vectors SV and VT . The resulting vector is VTSVST  (the direction of sun light) For VT P         , in the Born diagram, we draw the lines along the vectors PT and TV . The resulting vector is TVPTPV  (the direction of water flow). Then we have the positive sign in front of VT P         such that VT T P V S                 T V P F S T V P F S