20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 Absolute Entropy Third Law of thermodynamics • Absolute Entropies Absolute entropy of an ideal gas Start with fundamental equation dU = TdS − pdV dU + pdV dS = T for ideal gas: nRT dU = C dT and p = V V CV dT nR dS = + dV T V At constant T, dT=0 pdV dS = T T For an ideal gas, pV = nRT nRdV dS = T V At constant T d()pV = d (nRT )= 0 pdV = −Vdp plugging into dST: nRdp dS =− T p This allows us to know how S(p) if T held constant. Integrate! 1 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 For an arbitrary pressure p, p § · = o − nRdp = o − p S(p,T) S( p ,T) ³ o S(p ,T) nRln¨ ¸ p p © p o ¹ where po is some reference pressure which we set at 1 bar. S(p,T) = So(T) – nR lnp (p in bar) = 0 − S( p,T ) S (T ) R ln p S ↓ as P↑ ° S p But to finish, we still need S o(T) ! Suppose we had S o (0K) (standard molar entropy at 0 Kelvin) dH = TdS +Vdp = dH C pdT for ideal gas = + C pdT TdS Vdp C V dS = p dT − dp T T 2 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 § ∂ · C S = p Then using © ∂ ¹ T p T o we should be able to get S (T). Integrating over dS eqn, assuming Cp constant over T range: T2 C p2 nR dS = ³³p dT − dp T p T1 p1 So then § · § · ∆ = ¨ T2 ¸ − ¨ p2 ¸ S Cp ln¨ ¸ nRln¨ ¸ © T1 ¹ © p1 ¹ § · for p = 1bar = ¨ T2 ¸ − C p ln¨ ¸ nRln p © T1 ¹ Given Cp, T1, p1, → T2, p2, can calculate ∆S. We will use T=0K as a reference point. Consider the following sequence of processes for the substance A: A(s,0K,1bar) = A(s,Tm,1bar) = A(l,Tm,1bar) = A(l,Tb,1bar) = A(g,Tb,1bar) = A(g,T,1bar) ∆ T m C (s)dT H T C ()dT ∆H T C (g)dT S (T,1bar) = S o (0K) + ³ p + fus + ³ b p + vap + ³ p 0 T T Tm T T T T m b b So(T) C dT ∆S = ³ p T ∆H vap Liquid boils, ∆S = T ∆Hfus Solid melts, ∆S = T 0 T 3 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 Since ∆S0 is positive for each of these processes, the entropy must have its smallest possible value at 0 K. If we take S o (0K) = zero for every pure substance in its crystalline solid state, then we could calculate the entropy at any other temperature. This leads us to the Third Law of Thermodynamics: • THIRD LAW: First expressed as Nernst's Heat Theorem: Nernst (1905): As T → 0 K , ∆S → 0 for all isothermal processes in condensed phases More general and useful formulation by M. Planck: Planck (1911): As T → 0 K , S → 0 for every chemically homogeneous substance in a perfect crystalline state Justification: ? It works! @ Statistical mechanics (5.62) allows us to calculate the entropy and indeed predicts S o (0K) = 0. This leads to the following interesting corollary: It is impossible to decrease the temperature of any system to T = 0 K in a finite number of steps. How can we rationalize this statement? Recall the fundamental equation, dU = T dS – p dV dU = Cv dT For 1 mole of ideal gas, P = RT/V so Cv dT = T dS – (RT/V) dV dS = Cv d (ln T) + R d (ln V) 4 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 For a spontaneous adiabatic process which takes the system from T1 to a lower temperature T2, ∆S = Cv ln (T2/T1) + R ln (V2/V1) ≥ 0 but if T2 = 0, Cv ln (T2/T1) equals minus infinity ! Therefore R ln (V2/V1) must be greater than plus infinity, which is impossible. Therefore no actual process can get you to T2 = 0 K. But you can get very very close! In W. Ketterle's experiments on "Bose Einstein Condensates" (recent MIT Nobel Prize in Physics), atoms are cooled to nanoKelvin temperatures (T = 10-9 K) … but not to 0 K ! _______________ Some apparent violations of the third law (but which are not !) Any disorder at T = 0 K gives rise to S > 0 • mixed crystals If have an unmixed crystal, N atoms in N sites: N! Ω = = 1 N! S = k ln1 = 0 But if mixed crystal: NA of A NB of B NA + NB = N Ω = N! N A!N B! N! S = k ln N A!N B! Use Stirling’s approx: 5 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field Lecture 16 5.60/20.110/2.772 ln N!= N ln N − N S = k()N ln N − N − N ln N + N − N ln N + N A A A B B B = − ()− + + + k N ln N N A ln N A N A N B ln N B Using mole fractions: NA =xAN, NB =xBN ∆ =− + Smix nR[ XA ln XA XB ln XB ] > 0 Always !!! Even at T=0K But a mixed crystal is not a pure substance, so the third law is not violated. • Any impurity or defect in a crystal also causes S > 0 at 0 K • Any orientational or conformational degeneracies such as in a molecular crystal causes S > 0 at 0 K, for example in a carbon monoxide crystal, two orientations are possible: C O C O C O C O C O C O C O C O C O C O C O C O O C C O C O C O C O O C C O C O C O C O C O C O C O C O C O C O 6 .