18 Thermodynamics & Kinetics-Fundamental Equation, Absolute Entropy, and The Third Law, Lecture notes of Chemistry

Download 18 Thermodynamics & Kinetics-Fundamental Equation, Absolute Entropy, and The Third Law and more Lecture notes Chemistry in PDF only on Docsity! 5.60 Spring 2007 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) ** 5.60 Spring 2007 Lecture #11 page 2 ** U(S,V) ** From dU = TdS – pdV ⇒ ** T S U V =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ; p V U 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) ** From dH = TdS + Vdp ⇒ ** T S H p =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ; ⎛ ⎞∂ =⎜ ⎟∂⎝ ⎠S H V p ** _______________ We can use these equations to find how S depends on T. From dU = TdS – pdV ⇒ T C T U T 1 T S V VV =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ From dH = TdS + Vdp ⇒ T C T H T 1 T S p pp =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ 5.60 Spring 2007 Lecture #11 page 5 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 )K0(S o = 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) 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! 5.60 Spring 2007 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 ∫= T 0 p T dT)s(C )bar1,T,s(S to prevent a singularity at T=0, Cp → 0 as T → 0 K in fact, experimentally ...ATTC 3 p ++γ= 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. 5.60 Spring 2007 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 [ BBAAmix XlnXXlnXnRS +−=∆ ] > 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 is 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