Phase Transitions and Third Law

Question 1

Thermodynamic Potentials and States of Matter

Answer 1

-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

Question 2

Phase Diagrams

Answer 2

• 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

Question 3

Triple Point

Answer 3

-temperature and pressure at which the solid, liquid and vapour phases coexist

Question 4

Critical Temperature

Answer 4

• liquid/vapour boundary ends

- at temperatures greater than the critical temperature you cannot compress a vapour to form a gas

Question 5

First Order Phase Transitions

Answer 5

• 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

Question 6

Entropy Change During Transition

Equation

Answer 6

ΔStrans = ΔHtrans/Ttrans = Ltrans/Ttrans

Question 7

Entropy Change During Transition

Derivation

Answer 7

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

Question 8

Chemical Potential

Definition

Answer 8

-the chemical potential is the Gibb’s free energy per particle for a homogeneous one-component system

Question 9

Chemical Potential

Derivation

Answer 9

-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

Question 10

Clapeyron Equation

Description

Answer 10

• 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

Question 11

Clapeyron Equation

Derivation

Answer 11

-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

Question 12

Clapeyron Equation

Equation

Answer 12

dP/dT = Lab/TΔV = ΔHtrans/TΔV

Question 13

Clapeyron Equation

Solid-Liquid Boundary

Answer 13

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)

Question 14

Solid to Liquid Transition - The Effect of Volume

Answer 14

• 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)

Question 15

Clapeyron Equation

Liquid-Vapour Boundary

Answer 15

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

Question 16

Claussius-Clapeyron Approximation

Answer 16

-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)

Question 17

Nernst Heat Theorem

Answer 17

-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

Question 18

Statistical Mechanics

Entropy

Answer 18

-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

Question 19

Statistical Mechanics

T=0

Answer 19

-at T=0, the system is in its unique ground state so:

at T=0, Ω=1 so S=0

Question 20

Statistical Mechanics

The Third Law

Answer 20

-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

Question 21

Heat Capacities, Classical Equipartition and the Third Law

Answer 21

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

Question 22

The Carnot Cycle and Absolute Zero

Answer 22

-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->∞

Question 23

Unattainability of Absolute Zero

Answer 23

• an alternative definition of the third law is:

- it is impossible to reach T=0 in a finite number of steps

Question 24

Chemical Potential and Gibbs Free Energy

Answer 24

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