Phase Transitions and Third Law Flashcards
Thermodynamic Potentials and States of Matter
-free energy seeks a minimum (most negative) at equilibrium
G = U - TS + PV
-fix P, change T and leave V free
-U is determined by the strength of intermolecular interactions (more negative means they are stronger)
-at low temperatures, U wins and you get solid
-at high temperatures, S wins and you get gases
Phase Diagrams
- summarises the conditions of temperature and pressure under which a substance exists in different phases e.g. solids/liquids/gases
- boundaries between the regions show values at pressure and temperature at which the two phases coexist in equilibrium
Triple Point
-temperature and pressure at which the solid, liquid and vapour phases coexist
Critical Temperature
- liquid/vapour boundary ends
- at temperatures greater than the critical temperature you cannot compress a vapour to form a gas
First Order Phase Transitions
- associated with a latent heat, L
- most common transitions are first order e.g. melting, boiling
- during a first order transitions, heat supplied to the system doesn’t increase temperature, but is absorbed in converting the substance from one phase to another
- this implies that there is a discontinuous change in the entropy and therefore a singularity (e.g. an infinite!) heat capacity at the transition
Entropy Change During Transition
Equation
ΔStrans = ΔHtrans/Ttrans = Ltrans/Ttrans
Entropy Change During Transition
Derivation
ΔStrans = ΔQtrans/Ttrans -generally changes such as melting/boiling occur at fixed pressure since: H = U + PV dH = dQ + dW + PdV + VdP dH = dQ - PdV + PdV +VdP dH = dQ + VdP -at constant pressure: dH = dQ -therefore: ΔQtrans = ΔHtrans = Ltrans -so change in entropy is given by: ΔStrans = ΔHtrans/Ttrans = Ltrans/Ttrans
Chemical Potential
Definition
-the chemical potential is the Gibb’s free energy per particle for a homogeneous one-component system
Chemical Potential
Derivation
-for phase transitions we need to generalise the thermodynamic potentials to include situations where N can change
-for phase transitions, its most common to work in terms of T and P as G is extensive
G(T,P,N) = N * g(T,P)
μ = g(T,P)
dG = -SdT + VdP = Ndg = dμ
Clapeyron Equation
Description
- describes the phase equilibrium curve
- describes how transition temperature changes with pressure
- derived by setting the chemical potential to be the same for both phases
Clapeyron Equation
Derivation
-consider an infinitesimal change in P and T that occurs in such a way that phases A and B remain in equilibrium dμa = dμb -for each phase separately: dGi = Nidgi = dμi = -SidT + VidP -setting dμ for each phase equal: -SadT + VadP = -SbdT + VbdP -rearrange for dP/dT (Vb - Va) dP = (Sb - Sa) dT dP/dT = (Sb - Sa) / (Vb - Va) dP / dT = ΔS / ΔV -and since ΔS = ΔQ/T = Lab/T = ΔHtrans/T -therefore: dP/dT = Lab/TΔV = ΔHtrans/TΔV
Clapeyron Equation
Equation
dP/dT = Lab/TΔV = ΔHtrans/TΔV
Clapeyron Equation
Solid-Liquid Boundary
dP/dT = ΔHfus / TΔVfus
-enthalpy of melting is almost always positive
-volume change is positive but small
=> the slope dP/dT is steep and usually positive
(water is an exception as it contracts on melting)
Solid to Liquid Transition - The Effect of Volume
- when a solid melts into a liquid, if there is an increase in volume (ΔV>0) then when pressure is increased (dP>0), melting temperature increases (dT>0)
- if there is a decease in molar volume on melting then increasing pressure (dP>0), then melting temperature will decrease (dT<0)
Clapeyron Equation
Liquid-Vapour Boundary
dP/dT = ΔHvap / TΔVvap
- enthalpy change on vaporisation is positive
- change in volume is always positive (and large) as gas volume > liquid volume
- dP/dT is always positive, but smaller than for the solid/liquid boundary since ΔV is greater
Claussius-Clapeyron Approximation
-for transition at the liquid vapour boundary: Vgas >> Vliquid => ΔV = Vgas - Vliquid = Vgas -this gives dP/dT = ΔHvap/TΔVvap = ΔHvap/TV -if the gas is ideal, V=nRT/P dP/dT = ΔHvap*P/RT² 1/P dP/dT = ΔHvap/RT² -integrate ln (P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Nernst Heat Theorem
-Nernst measured change in G & H for chemical reactions that started and finished at the same temperature
-at lower temperatures, ΔG and ΔH became closer together:
ΔH=TΔS+VΔP & ΔG=-SΔT+VΔP :
ΔG = ΔH - TΔS
-so Nernst observed that:
ΔH - ΔG -> 0 as T->0
-which implies TΔS->0 as T->0 , which is not surprising
BUT:
-Nernst also realised that ΔH-ΔG -> 0 at a faster than linear rate:
ΔH-ΔG/T -> 0 as T -> 0
-this is surprising as 1/T is increasing as T->0
-and since, ΔH-ΔG/T=ΔS :
ΔS -> 0 as T -> 0
Statistical Mechanics
Entropy
-the microscopic entropy is defined in statistical mechanics as:
S = k * ln(Ω)
-where Ω=’statistical weight’, the number of states that the system can be in
Statistical Mechanics
T=0
-at T=0, the system is in its unique ground state so:
at T=0, Ω=1 so S=0
Statistical Mechanics
The Third Law
-the entropy change in a process tends to zero as the temperature, T, tends to absolute zero
OR
-as the absolute zero of temperature is approached, the entropy of all bodies tends to zero
Heat Capacities, Classical Equipartition and the Third Law
C = ∂Q/∂T = T ∂S/∂T
-but the third law tells us ∂S/∂T->0 as T->0
-this implies that for heat capacity C:
C->0 as T->0
-but the classical equipartition theorem gives constant heat capacities
-therefore the third law of thermodynamics is in direct disagreement with classical equipartition
-classical equipartition breaks down at low temperatures due to quantum effects
The Carnot Cycle and Absolute Zero
-from the equations of the carnot cycle: Qh/Th = Qc/Tc and W=Qh-Qc -which gives: W/Qc = (Th-Tc )/ Tc -as Tc->0 , the work required to cool the system, W->∞
Unattainability of Absolute Zero
- an alternative definition of the third law is:
- it is impossible to reach T=0 in a finite number of steps
Chemical Potential and Gibbs Free Energy
G(T,P,N) = N * g(T,P)
μ = g(T,P)
-chemical potential is equal to the Gibbs free energy per particle for a homogeneous, one component system
dG = -SdT + VdP = N dg = dμ
