Thermodynamic Equations of State • Thermodynamic equations of state will lead to an understanding of concepts such as surface tension, etc. • Leads to a knowledge of how to predict the physical property or at least relations between physical properties. Fundamentals of Thermodynamics: • Variables in the lab: P, V, T • First law: dE = dw + dq • Energy is state function; any combination of heat and work possible • Microscopic scale: energy is sum of rotational, vibrational, translational and electronic energy levels. • Remember w = -PdV; negatvie sign indicates system energy increases when work done on system. • Assume. only PdV work and defining entropy as dqrev/T º dS • Leads to first law: dE = -PdV + TdS • Gibbs Free Energy: G = H -TS • Helmholtz Energy: F = E - TS 1 Relationship between E and Volume in terms of P, V, T • Take partial of First Law: dE = -PdV + TdS with respect to V at constant T: æ ¶E ö æ ¶S ö ç ÷ = -P +T ç ÷ è ¶V øT è ¶V øT • Equations should be expressed in terms of P, V, T. • Helmholz free energy: F º E -TS • Differentiate: dF º dE - TdS - SdT • Substitute from first law. dF º -PdV +TdS -TdS - SdT = -PdV - SdT æ ¶F ö æ¶F ö • Total Differential of F: dF = ç ÷ dV + ç ÷ dT è¶V ø è ¶T ø T 2 V æ ¶F ö æ ¶ F ö æ ¶P ö • Comparing leads to: = -P Ù ç ÷ = - ç ÷ ç ÷ ç ÷ è ¶V øT è ¶V¶T ø è ¶T øV 2 æ ¶F ö æ ¶ F ö æ ¶S ö • And also: = -S Ùç ÷ = - ç ÷ ç ÷ ç ÷ è ¶T øV è ¶T¶V ø è ¶V øT æ ¶2F ö æ ¶2F ö • But: ç ÷ = ç ÷ ç ¶T¶V ÷ ç ¶V¶T ÷ è ø è ø æ ¶S ö æ ¶P ö • Substituting from previous two equations:ç ÷ = ç ÷ è ¶V øT è ¶T øV æ ¶E ö æ ¶P ö • Which is leads to: ç ÷ = -P +Tç ÷ è ¶V øT è ¶T øV 2 Pressure Dependence of Enthalpy • Enthalpy is defined in terms of energy, pressure and volume: H ºE + PV. • Differentiating: dH = dE + PdV + VdP. • From the first law: dE = - PdV + TdS; • Substituting: dH = PdV+ VdP- PdV+ TdS = VdP+ TdS æ ¶H ö æ ¶S ö • Divide by dP and hold T constant: ç ÷ = V + Tç ÷ è ¶P øT è ¶P øT • Use Gibbs Free Energy: G = H - TS or dG = dH - TdS - SdT • Substitute for dH: dG=VdP+TdS-TdS-SdT=VdP-SdT • Write total differential for free energy, G(T,P): æ ¶G ö æ ¶G ö dG = ç ÷ dP+ ç ÷ dT è ¶P øT è ¶T øP æ ¶G ö æ ¶G ö • By inspection: ç ÷ = V Ù ç ÷ = -S è ¶P øT è ¶T øP 2 2 æ ¶ G ö æ¶V ö æ ¶ G ö æ ¶S ö • Second derivative:ç ÷ = Ù ç ÷ = - ç ÷ ç ÷ ç ÷ ç ÷ è ¶P¶T ø è ¶T øP è ¶T¶Pø è ¶P øT æ¶V ö æ ¶S ö • Or: ç ÷ = -ç ÷ è ¶T øP è¶P øT æ ¶H ö æ ¶V ö • Substitute for enthalpy equation: ç ÷ = V - Tç ÷ è ¶P øT è ¶T øP 3 Equations of State:Temperature Dependence • Recall 3 variables to be used are P, V, T. • Knowledge of two of the variables allows determination of the energy state of the system. • Two polynomial equations of state used here. – V = f(P) and – P = f(V) 2 • V = Vo[1 + ao(T) – a1(T)P + a2(T)P + …] where – coefficients aj are functions of temperature. – Vo = volume at absolute zero. • Differentiate with respect to T and neglect higher terms: æ ¶V ö æ ¶ao ö ç ÷ = Voç ÷ è ¶T øP è ¶T øP • When using only the first two terms of the series expansion, we have: V = Vo[1 + ao(T)]. • Substitute for Vo: Vo = V/[1+ao(T)]. 1 æ ¶V ö 1 æ ¶a ö ç ÷ = ç o ÷ V è ¶T øP [1 + ao (T)]è ¶T øP • When ao << 1, the equations reduces to: æ ¶a ö 1 æ ¶V ö ç o ÷ = ç ÷ = a è ¶T øP V è ¶T øP where a = volume expansivity, relative change in volume with temperature; related to temperature variation of ao. 4 Equations of State: Pressure Dependence • Found by taking derivative of equation of state with respect to pressure: æ ¶V ö 1 æ ¶V ö Vo ç ÷ = -Voa1 Ú - ç ÷ = a1 è ¶P øT V è ¶P øT V where higher pressure terms are ignored (only first order considered significant). • But V = Vo[1+ ao]; so that 1 æ ¶V ö a1 c = - ç ÷ = » a1 V è ¶P øT 1+ ao since a1<<a0<<1 § c = isothermal compressibility. § Pressure as a function of T, V: 2 éVo - V ù éVo - V ù P = Po(T) + P1(T)ê ú + P2 (T)ê ú + .... ë Vo û ë Vo û § Pi = material dependent coefficients; determined experimentally. § Po = pressure required to decrease the volume of the solid at higher temperature to what it would be at 0 K and no pressure. § P0<< P1 or P2 5 Relationship between Pi and ai 2 • Recall: V = Vo[1 + ao(T) – a1(T)P + a2(T)P + …] V 2 • Solve for Volume: = 1+ ao(T) - a1(T)P + a2(T)P Vo V 2 - 1 = ao(T) - a1(T)P + a2(T)P Vo Vo - V 2 = -ao(T) + a1(T)P - a2(T)P Vo • We use this in the earlier equation: 2 éVo - V ù éVo - V ù P = Po(T) + P1(T)ê ú + P2 (T)ê ú + .... ë Vo û ë Vo û P = P (T) + P (T )(- a (T) + a P - a P + ...)+ • To get o 1 o 1 2 2 P (T) - a (T ) + a P - a P +... 2 +.... 2 ( o 1 2 2 ) • Expand the second term • Neglect squared and higher terms in terms of ao and P . P = P + P - a + a P - a P2 + ... + o o 1 ( o 1 2 ) 2 P - 2a a P + a a P2 + a2P2 + ... +.... 2 ( o 1 o 2 1 ) • This can only be true when the following happens: Po - P1ao = 0, P1a1 - 2P2aoa1 =1, 2 6 - P1a2 + 2P2aoa2 + P2a1 = 0 Relationship between Pi and ai (cont.) P - P a = 0, • Solve for ai from o 1 o P1a1 - 2P2aoa1 =1, 2 - P1a2 + 2P2aoa2 + P2a1 = 0 • Gives: Po ao = , P1 1 1 æ 2P P ö a = = ç1+ o 2 ÷, 1 P - 2P a P ç 2 ÷ 1 2 o 1 è P1 ø P æ 6P P ö a = 2 ç1 + o 2 ÷ 2 3 ç 2 ÷ P1 è P1 ø • Pi can also be expressed in terms of ai, • Figure shows that compressibility increases with atomic number. • Slopes at high pressures are similar for all. • Coefficients of expansion nearly constant at absolute zero (see figure), but increase at higher temperatures . 7 Pressure Dependence of Heat Capacity • Recall the definition of heat capacity: æ ¶H ö CP = ç ÷ è ¶T øP • Take its derivative with respect to P at constant T æ ¶C ö ¶ éæ ¶H ö ù ¶ éæ ¶H ö ù ç P ÷ = ç ÷ = ç ÷ ¶P ¶P ê ¶T ú ¶T ê ¶P ú è øT ëè øP ûT ëè øT ûP æ ¶H ö æ ¶V ö • But earlier we showed: ç ÷ = V - Tç ÷ è ¶P øT è ¶T øP æ ¶C ö ¶ éæ ¶H ö ù ¶ é æ ¶V ö ù • Substitute: ç P ÷ = ç ÷ = V - Tç ÷ ¶P ¶P ê ¶T ú ¶T ê ¶T ú è øT ëè øP ûT ë è øP ûP 2 æ ¶V ö æ ¶T ö æ ¶V ö æ ¶ V ö = ç ÷ - ç ÷ ç ÷ - Tç ÷ è ¶T ø è ¶T ø è ¶T ø ç 2 ÷ P P P è ¶T øP 2 æ ¶C ö æ ¶ V ö ç P ÷ = -Tç ÷ è ¶P ø ç 2 ÷ T è ¶T øP • We can now use the equation of state to determine an equation for calculating heat capacity under various conditions. 2 • Recall: V = Vo[1 + ao(T) – a1(T)P + a2(T)P + …] • Substitute: æ 2 2 ö æ ¶CP ö ç ¶ [1 + ao(T) + a1(T )P + a2(T)P ]÷ ç ÷ = -TVo è ¶P ø ç 2 ÷ T è ¶T øP æ 2 2 2 ö æ ¶CP ö ç ¶ ao(T) ¶ a1(T ) ¶ a2(T ) 2 ÷ ç ÷ = -TVo + P + P è ¶P ø ç 2 2 2 ÷ 8 T è ¶T ¶T ¶T øP CP Vs P(cont) • Integration gives the pressure dependence of CP. æ ¶2a (T ) ¶2a (T ) ¶2a (T ) ö CP¶C = -TV Pç o + 1 P + 2 P2 ÷dP ò o P oò0 ç 2 2 2 ÷ CP è ¶T ¶T ¶T ø æ ¶2a (T ) 1 ¶2a (T ) 1 ¶2a (T ) ö C = C o -TV ç o P + 1 P2 + 2 P3÷ P P ç 2 2 2 ÷ è ¶T 2 ¶T 3 ¶T ø o • C P = heat capacity at zero pressure. • The first second derivative term is dominant at high temperature and heat capacity is expected to decrease with increasing pressure in this temperature regime (see negative sign in equation). æ ¶a ö • Recall that alpha is the volume expansitivity: a = ç o ÷ è ¶T øP • The first term is the temperature coefficient of thermal expansion. • This term nearly linear at high temperatures. 9 Cv vs P æ ¶E ö æ ¶P ö • Earlier we showed: ç ÷ = -P +T ç ÷ è ¶V øT è ¶T øV • Take derivative of both sides with respect to T: ¶2E ¶2E ¶P ¶T ¶P ¶2P = = - + + T ¶T¶V ¶V¶T ¶T ¶T ¶T ¶T 2 æ ¶E ö • But: CV = ç ÷ è ¶T øV 2 æ ¶C ö ¶ P • Substitute: V = T ç ÷ 2 è ¶V øT ¶T • Earlier we used the equation of state: 2 éVo - V ù éVo - V ù P = Po(T) + P1(T)ê ú + P2(T)ê ú +...