Sackur-Tetrode equation Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: October, 02, 2017) The Sackur-Tetrode equation is important historically, because it suggests quantization (and a value for Planck’s constant h) based solely on experimental work in thermodynamics. The result provides a basis for quantum theory that is independent of Planck’s first conception of it, which he developed in the theory of blackbody radiation, and from Einstein’s conception of quantization based on the photoelectric effect. The Sackur–Tetrode equation is an expression for the entropy of a monatomic classical ideal gas which incorporates quantum considerations which give a more detailed description of its regime of validity. The Sackur–Tetrode equation is named for Hugo Martin Tetrode (1895– 1931) and Otto Sackur (1880–1914), who developed it independently as a solution of Boltzmann's gas statistics and entropy equations, at about the same time in 1912. 1. Calculation of ZC1( ) Now we calculate the partition function (canonical ensemble) using the density of state for the one particle system ZC1( ) exp( k ) k V dk exp( ) 2( )3 k V ℏ k 22 4k 2dk exp( ) 3 2( ) 0 2m 2/3 V 1 m 2 3 2 2 ℏ mk T V B 2ℏ2 2. N particle system (identical case) The difficulty encountered in the Gibbs paradox is due to the treatment of the gas molecules as individually distinguishable. The interchanging positions of two identical molecules would lead to a physically distinct state of the gas. This is not true. In fact, if we treat the gas by quantum mechanics, the molecules would have to be regarded as completely indistinguishable. The distinguishability of the molecules must be taken into account explicitly in calculating the partition function. This can be done by dividing the classical partition function by the permutation N!. 1 Z ( ) Z ( ) N C N! C1 N 3/ 2 V 1 m N [ 2 2 ] N! 2 2 ℏ 3N / 2 V N m 2 N! 2ℏ 3N / 2 V N mk T B N! 2ℏ2 3N / 2 V N 2mk T B N! h2 Note that 1 1 1 h , ℏ 2 2 2 8 3 (2 )3 / 2 2 The Helmholtz free energy: F kBT lnZ C(,, T V ) 3N 3N 2mk k T[N lnV ln N! lnT ln( B )] B 2 2 h2 3N 3N 2mk k T[N lnV N ln N N lnT ln( B )] B 2 2 h2 V 3 3 2mkB NkBT[ln lnT 1 ln( )] N 2 2 h2 or V 3 3 2mkB F NkBT[ln lnT 1 ln( )] N 2 2 h2 lnZ C(,, T V ) is obtained as V 3 3 2mkB lnZ C(,, T V ) N[ln lnT 1 ln( )] N 2 2 h2 The average energy F U T 2 T T V 2 kBT ln Z T V ln Z V 3 Nk T 2 B or 3 U Nk T 2 B The heat capacity at constant volume: U 3 CV NkB T V 2 3 For N N , we have C R , where R is the gas constant. The heat capacity at constant A V 2 pressure is 5 C Cv R R , (Mayer’s relation) P 2 The entropy S is E F S T 3 V 3 3 2mkB NkB NkB[ln lnT 1 ln( )] 2 N 2 2 h2 V 3 2mkBT 5 NkB[ln ln( 2 ) ] N 2 h 2 or V 3 2mk BT 5 S Nk B[ln ln( ) ] (Sackur-Tetrode equation) N 2 h2 2 The pressure P is F Nk T P B V T V or PV Nk BT 3. Gibbs paradox We need to use the Sackur–Tetrode equation for the identical gases V 3 S Nk B[ln ln( T ) C] N 2 where C is a constant, and N nN A . The entropy of the system indeed becomes a true extensive quantity. Thus the very root of the trouble has been eliminated by the recipe of Gibbs. Fig. The initial state before mixing and the final state after mixing. The entropy in the initial state before the mixing V 3 1 Si 1Rn [ln ln( T) C] n1N A 2 V 3 2 n2 R[ln ln( T ) C] n2 N A 2 RT 3 1Rn [ln ln( T ) C] PN A 2 RT 3 n2 R[ln ln( T) C] PN A 2 kBT 3 (n1 n2 )R[ln ln( T) C] P 2 since PV 1 n1RT and PV 2 n2RT V 3 1 Si 1Rn [ln ln( T )] C n1N A 2 The entropy in the final state after the mixing is V V 3 1 2 S f (n1 n2 )R[ln ln(T) C] (n1 n2 )N A 2 RT 3 (n1 n2 )R[ln ln(T) C] PN A 2 kBT 3 (n1 n2 )R[ln ln(T ) C] P 2 since P(V1 V2 ) (n1 n2 )RT . So we have Si S f . The Gibbs paradox is no longer a paradox. No entropy changes 4. Comment by Huang Gibbs resolved the paradox in an empirical fashion by postulating that we have made an error in calculating the partition function. Gibbs assumed that correct answer is N! times smaller than N we thought it was. It is not possible to understand classically why we must divide ZC1 by N! to obtain the correct counting of states. The reason is inherently indistinguishable in the following sense. A state of the gas is described by an N-particle wave function, which is either symmetric or antisymmetric with respect to the interchange of any two particles. A permutation of the particles can at most change the wave function by a sig, and it does not produce a new N state of the system, From this fact it seems reasonable that we should divide ZC1 by N!. 5. Calculation of entropy using the Sackur-Tetrode equation Calculation of entropy for argon, neon and helium gas at T = 300 K and P = 1 atm. We use the Sackur0Tetrode equation. S(Kr) = 164.106 J/(K mol) for krypton gas S(Ar) 154.867 J/(K mol) for argon gas S(Ne) 146 35. J/(K mol) for neon gas S(He) 126.174 J/(K mol) for helium gas Clear "Global` " ; rule1 cal 4.19, atm 1.013 10 5, NA 6.02214179 10 23 , R 8.314472, me 9.1093821545 10 31 , u 1.660538782 10 27 , eV 1.602176487 10 19 , qe 1.602176487 10 19 , kB 1.3806504 10 23 , c 2.99792458 10 8, h 6.62606896 10 34 , 1.05457162853 10 34 , mNe 20.1797 u, mAr 39.948 u, mHe 4.002602 u, P1 1 atm , T1 300 ; kB T1 3 2 m kB T1 5 S1 NA kB Log Log ; P1 2 h2 2 S1 . m mAr . rule1 154.867 S1 . m mNe . rule1 146.35 S1 . m mHe . rule1 126.174 6. Entropy of neon at 27.2 K (Kittel) We consider the phase transion on Ne at P = 1 arm. (i) Solid (for 0 K<T<24.55 K) Ssolid 29.14 J/(mol K) . (ii) Melting point T= 24.55 K. The entropy of melting is 335 S 64.13 J/(mol K) melting 24.55 (iii) Liquid (24.55<T<27.2 K) Sliquid 85.3 J/(mol K) (iv) Vaporization at T= 27.2 K. The entropy of vaporization is 1761 S 62.64 J/(mol K) vaporization 27.2 So the experimental value of entropy of neon gas at 27.2 K at 1 atm is Sgas Ssolid Smelting Sliquid Svaporization = 96.40 J/(mol K) (experiment) The calculate value of the entropy of neon at 1 atm at 27.2 K is Sgas 96.4515 J/(mol K) from Sackur-Tetrode equation) There is an excellent agreement between the experiment and theory (( Mathematica )) Clear "Global` " ; rule1 cal 4.19, atm 1.013 10 5, NA 6.02214179 10 23 , R 8.314472, me 9.1093821545 10 31 , u 1.660538782 10 27 , eV 1.602176487 10 19 , qe 1.602176487 10 19 , kB 1.3806504 10 23 , c 2.99792458 10 8, h 6.62606896 10 34 , 1.05457162853 10 34 , mKr 83.798 u, mNe 20.1797 u, mAr 39.948 u, mHe 4.002602 u, P1 1 atm ; kB T1 3 2 m kB T1 5 S1 NA kB Log Log ; P1 2 h2 2 S1 . m mNe . T1 27.2 . rule1 96.4515 7. Entropy of Kr From Francisco Jos ́e Pa ̃nos Exp ́osito The Sackur-Tetrode equation and the measure of entropy http://diposit.ub.edu/dspace/bitstream/2445/59903/1/TFG-Pa%C3%B1os- Exp%C3%B3sito-FranciscoJos%C3%A9.pdf T(solid-liquid)= 115.77 K (Kr) T(liquid=gas) = 119.93 K )Kr) Fig. Heat capacity of krypton as a function of temperature. Fig. Entropies as a function of temperature for Ne, Ar, Kr, and Hg. S 1( K) .0 002 J/(mol K) S 1( K 12K) .3 518 J/(mol K) S(12K 115.77K) 49.712 J/(mol K) 1640 S 14.166 J/(mol K) melting 115.775 S(115.77K 119.81K) .1 500 J/(mol K) 9080 S 75.787 J/(mol K) vaporization 119.81 So the experimental value of entropy of krypton gas at 119.81 K at 1 atm is Sgas (Kr) Ssolid Smelting Sliquid Svaporization = 144.683 J/(mol K) The calculate value of the entropy of krypton at 1 atm at 119.81 K is Sgas 145.048 J/(mol K) from Sackur-Tetrode equation) There is an excellent agreement between the experiment and theory.