Lecture 11: Fundamental Equations, Absolute Entropy, and The Third Law

Lecture #11 page 1 Fundamental Equations, Absolute Entropy, and The Third Law • Fundamental Equations relate functions of state to each other using 1st and 2nd Laws 1st law with expansion work: dU = đq - pextdV need to express đq in terms of state variables because đq is path dependent Use 2nd law: đqrev = TdS For a reversible process pext = p and đq = đqrev =TdS So…… ** dU = TdS – pdV ** This fundamental equation only contains state variables Even though this equation was demonstrated for a reversible process, the equation is always correct and valid for a closed (no mass transfer) system, even in the presence of an irreversible process. This is because U, T, S, p, and V are all functions of state and independent of path. AND The “best” or “natural” variables for U are S and V, ** U(S,V) ** Lecture #11 page ** U(S,V) ** ⎛ ∂U ⎞ From dU = TdS – pdV ⇒ ** ⎜ ⎟ =T ⎝ ∂S ⎠V ; ⎛ ∂U ⎞ ⎜ ⎟ = −p ** ⎝ ∂V ⎠S We can write similar equations for enthalpy H = U + pV ⇒ dH = dU + d(pV) = dU + pdV + Vdp inserting dU = TdS – pdV ⇒ ** dH = TdS + Vdp ** The natural variables for H are then S and p ** H(S,p) ** ⎛ ∂H ⎞ From dH = TdS + Vdp ⇒ ** ⎜ ⎟ =T ⎝ ∂S ⎠ p ; ⎛ ∂H ⎞ ⎜ ⎟ = V ** ∂p ⎝ ⎠S _______________ We can use these equations to find how S depends on T. From dU = TdS – pdV ⇒ ⎛ ∂S ⎞ = 1 ⎛ ∂U ⎞ = CV ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠V T ⎝ ∂T ⎠V T From dH = TdS + Vdp ⇒ ⎛ ∂S ⎞ = 1 ⎛ ∂H ⎞ = Cp ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ p T ⎝ ∂T ⎠ p T 2 Lecture #11 • page Absolute Entropies Absolute entropy of an ideal gas dU + pdV T pdV At constant T, dU=0 ⇒ dST = T nRdV For an ideal gas, pV = nRT ⇒ dST = V At constant T d(pV) = d(nRT) = 0 ⇒ pdV = -Vdp From dU = TdS – pdV ⇒ dS = dST = − So nRdp p For an arbitrary pressure p, p S(p, T) = S(p o , T) − ∫po ⎛ p⎞ nRdp = S(p o , T) − nRln⎜⎜ o ⎟⎟ p ⎝p ⎠ where po is some reference pressure which we set at 1 bar. ⇒ S(p,T) = So(T) – nR lnp (p in bar) S (p, T) = S o (T) − Rln p (p in bar) But to finish, we still need S o (T) ! Suppose we had S o (0K) (standard molar entropy at 0 Kelvin) Cp ∂S ⎞ Then using ⎛⎜ we should be able to get S o (T) ⎟ = ⎝ ∂T ⎠ p T 3 Lecture #11 page 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) Tm S (T,1bar) = S o (OK) + ∫ Cp (s)dT T 0 So(T) ΔS = ∫ + ΔHfus Tm Tb Cp (A)dT Tm T +∫ + ΔHvap Tb T Cp (g)dT Tb T +∫ CpdT T ΔH vap Liquid boils, ΔS = T ΔH fus Solid melts, ΔS = T 0 T Since ΔS0 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 !!! THIRD LAW: First expressed as Nernst's Heat Theorem: Nernst (1905): As T → 0 K , ΔS → 0 for all isothermal processes in condensed phases • 4 Lecture #11 page 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: c It works! d 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 so For 1 mole of ideal gas, P = RT/V Cv dT = T dS – (RT/V) dV dS = Cv d (ln T) + R d (ln V) For a spontaneous adiabatic process which takes the system from T1 to a lower temperature T2, ΔS = 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! 5 Lecture #11 page 6 In Prof. W. Ketterle's experiments on "Bose Einstein Condensates" (MIT Nobel Prize), atoms are cooled to nanoKelvin temperatures (T = 10-9 K) … but not to 0 K ! Another consequence of the Third Law is that It is impossible to have T=0K. How can we rationalize the alternate statement? Consider the calculation of S starting at T=0K S(s, T,1bar) = ∫ T Cp (s)dT 0 T to prevent a singularity at T=0, Cp → 0 as T → 0 K in fact, experimentally Cp = γT + AT 3 + ... That is, the heat capacity of a pure substance goes to zero as T goes to zero Kelvin and this is experimentally observed. Combining the above with dT = đqp/Cp , at T=0 any infinitesimally small amount of heat would result in a finite temperature rise. In other words, because Cp → 0 as T → 0 K, the heat đqp needed to achieve a temperature rise dT, (đqp=CpdT) also goes to zero at 0 K. If you somehow manage to make it to 0 K, you will not be able to maintain that temperature because any stray heat from a warmer object nearby will raise the temperature above zero, unless you have perfect thermal insulation, which is impossible. Lecture #11 • page 7 Some apparent violations of the third law (but which are not !) Any disorder at T = 0 K gives rise to S > 0 • For example in mixed crystals ΔSmix = −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 0K • Any orientational or conformational degeneracies such is in a molecular crystal causes S > 0 at 0 K, for example in a carbon monoxide crystal, two orientations are possible: C C C C O O O O C C C C O O O O C C C C O O O O CO CO O C CO C C C C O O O O CO O C CO CO C C C C O O O O