Thermodynamics

Question 1

making bonds is ___
breaking bonds is ___

Answer 1

make = exothermic
break = endothermic

Question 2

Second Law of Thermodynamics

Answer 2

In a spontaneous process, the entropy of the Universe increases

Question 3

molecular interpretation of entropy

Answer 3

randomness = how do molecules distribute themselves amongst available energy levels
–> statistical analysis

Question 4

in statistical analysis, what does the weight of a distribution refer to?

Answer 4

the total number of ways a particular configuration can be achieved

Question 5

what is the most probable distribution?

Answer 5

the one configuration with a larger value W than all the others
= the Boltzmann Distribution

Question 6

Boltzmann Distribution eqn for population of level i, n(i), with energy ε(i)
+ interpret

Answer 6

n(i) = n(0) exp (-ε(i) / kT)

where k is Boltzmann’s constant
T is temperature

ie. as the energy of a level increases, its population decreases
as Temp raised, molecules move to higher energy levels

Question 7

Eqn relating entropy to number of ways W

Answer 7

S = k lnW

k: Boltzmann’s constant
W: weight, ways

Question 8

molecular energy level distribution when a system is heated

Answer 8

energy is supplied, some molecules move to higher energy levels
–> more spread out amongst energy levels so more ways of achieving the distribution, hence entropy increases

Question 9

molecular energy level distribution when a system expands

Answer 9

when a system expands, the spacing of translational energy levels DECREASES
hence more energy levels available to the molecules
distributed over more levels, W increases, S increases

Question 10

how does the size of entropy change depend on the T of a system?

Answer 10

the increase in entropy resulting from a certain amount of heat being supplied is GREATER the LOWER the temperature of the system

Question 11

eqn classical definition of entropy

differential form

Answer 11

dS = 𝛿q(rev) / T

NOTE: even if the process itself is not reversible, must work out what the heat would be if we were to go by a reversible path!

Question 12

eqn calculate entropy change of surroundings

Answer 12

surroundings are so large that all heat transfers are seen as reversible

∆S(surr) = q(surr) / T(surr) = -q(sys) / T(sys)
for a finite change

Question 13

relate spontaneity of a process to ∆S(univ)

Answer 13

+ve = spontaneous (entropy of universe increases)
-ve = not spontaneous
= 0 is equilibrium, no tendency for further change

Question 14

The First Law of Thermodynamics + eqn

Answer 14

energy cannot be created or destroyed but is transformed from one form into another

∆U = q + w

Question 15

what are the different “forms” of energy?

Answer 15

heat, work and internal energy

Question 16

key different between internal energy and heat/work

Answer 16

internal energy is a STATE function
heat/work are PATH functions

note that the symbol ‘∆’ is only appropriate for state functions

Question 17

derive eqn for work done on an ideal gas by expansion

Answer 17

piston with area A experiences p(external)
force on the piston due to external pressure is p(ext) * A

small amount of work done BY the gas is force x distance dx (piston moved out)
δw = p(ext) * A dx = p(ext) dV

work done ON the gas: δw = - p(ext) dV

Question 18

when a gas expands, what happens to the internal energy?

Answer 18

work is done so Internal energy falls

Question 19

derive expression for expansion work done against CONSTANT external pressure

Answer 19

integrate p(ext) dV for an expansion between Vi and Vf
p(ext) constant, move out of integral

w = - p(ext) [Vf-Vi]

Question 20

what must p(ext) be to do maximum expansion work for a gas in a cylinder?

Answer 20

to maximise work done, external pressure needs to be as high as possible but cannot exceed internal pressure (or else compression would occur, not expansion)
–> p(ext) infinitesimally smaller than p(int)

Question 21

define reversible reaction

Answer 21

a process whose direction can be changed by an infinitesimal change in some variable

Question 22

reversible vs spontaneous reactions

Answer 22

reversible:
infinitely slow
at equilibrium
do max work

irreversible = spontaneous:
finite rate
not at equilibrium
less than max work

Question 23

how to find the expression for work done in a finite expansion for an ideal gas expanding reversibly

Answer 23

essentially, p(ext) = p(int)
work = integral of p(int) dV between Vi and Vf

for ideal gas, p(int)V=nRT –> sub for p(int)
Assume isothermal (constant T)

w(rev) = - nRT ln(Vf/Vi)

Question 24

what is an indicator diagram?

Answer 24

essentially a graph of p(ext) vs V
work done in a finite expansion is the integral of p(ext)dV between Vi and Vf, so work done is the area under the curve

Question 25

what can be said about the magnitude of heat in a reversible process?

Answer 25

the magnitude of heat is a MAXIMUM

because for any particular initial and final states, ∆U is constant
as ∆U = q+w, if a path is changed such that w increases, q must change by the same amount to keep ∆U fixed hence q must also be at a maximum for such a path

Question 26

internal energy of an ideal gas

Answer 26

depends ONLY on temperature
because particles do not interact, if heat added it is only stored in Ek

hence for isothermal expansion, ∆U=0
and w’=q

Question 27

expression for heat for an isothermal, reversible expansion of ideal gas

Answer 27

q(rev) = w’ = nRT ln(Vf/Vi)

Question 28

expression for entropy change for any isothermal expansion of an ideal gas from Vi to Vf

Answer 28

∆S = nR ln(Vf/Vi)

confirming entropy increases as volume increases

from q(rev) = nRT ln(Vf/Vi) and ∆S=q(rev)/T

Question 29

first law for infinitesimal changes

Answer 29

dU = 𝛿q + 𝛿w

note: 𝛿 not d because d implies “a small change in” but heat and work are state functions hence must talk about “a small amount of” heat/work

Question 30

first law for infinitesimal changes and expansion work only

Answer 30

dU = 𝛿q - p(ext)dV

Question 31

expression for heat absorbed under conditions of constant volume

Answer 31

dU = 𝛿q(const vol)

because p(ext)dV = 0, no expansion
so to find ∆U, measure the heat under constant volume conditions

Question 32

relationship between heat supplied and temperature change

Answer 32

q = c∆T
where c is the heat capacity

q = nC(m) ∆T
where C(m) is molar heat capacity and n is moles

Question 33

definition of heat capacity at constant volume

Answer 33

C(V,m) = (∂U(m) / ∂T)|V

Question 34

what is enthalpy? give eqn

Answer 34

H = U + pV

a state function whose value is equal to the heat supplied at constant pressure (sum of internal energy and pV work)

Question 35

complete differential of the enthalpy equation

Answer 35

dH = dU + pdV + Vdp

from H=U+pV

Question 36

relate enthalpy change to heat

Answer 36

dH = 𝛿q(const.press)
enthalpy change is equal to heat measured under conditions of constant pressure

by subbing dU=𝛿q-pdV and taking Vdp=0, const press

Question 37

definition of heat capacity at constant pressure

Answer 37

C(p,m) = (∂H(m) / ∂T) |p

Question 38

equation for converting enthalpy at one temperature to another

Answer 38

Hm(T2) = Hm(T1) + C(p,m) [T2-T1]

from integrating the definition of heat cap at constant pressure, assuming constant C

note C(p,m) is a molar quantity and as such gives change in molar enthalpy

Question 39

how to express entropy S in terms of heat capacity (something you can actually measure!)

Answer 39

using dS = dH/T
and dH(m) = C(p,m) dT

over wide temp range, CANNOT assume constant C
integrate C(T)/T wrt T over some wide temp range

Question 40

justification for entropy at absolute zero = 0

Answer 40

at absolute 0, kT goes to 0 so all molecules in lowest energy level (ground state), only one way to arrange so W=1 and S=klnW=0.

Question 41

absolute entropy S(T*) expression

Answer 41

= integral of Cp,m(T)/T wrt T + sum of entropy of phase changes

Question 42

convert entropy S at T1 to T2

Answer 42

Sm(T2) = Sm(T1) + C(p,m) ln(T2/T1)

from integrating Cp,m/T wrt dT assuming constant C over small T1 to T2

Question 43

define Gibbs energy state function

Answer 43

∆G = ∆H - T∆S

all for the SYSTEM

Question 44

relate change in G to spontaneity

Answer 44

dG < 0 = spontaneous
at equilibrium G reaches a minimum, dG =0

Question 45

the first thermodynamic Master Equation

Answer 45

dU = TdS - pdV

aka “First and Second Laws combined”
dU = 𝛿q-𝛿W for pV work, and 𝛿q=TdS

Question 46

the second thermodynamic Master Equation

Answer 46

dG = Vdp - SdT

from the differential form of H=U+pV
and sub in first master eqn
then sub dH expression into differential form of G=H-TS

Question 47

how does Gibbs energy vary with pressure (constant T)?
for ideal gas

Answer 47

dG = Vdp (dT=0)
sub V = nRT/p and integrate

–> Gm(p) = Gm˚ + RT ln(p/p˚)

Question 48

how does Gibbs vary with volume at constant T?

Answer 48

V inversely proportional to p
p2/p1 = V1/V2

sub in expression for G(p)
G(V2) - G(V1) = nRT ln(V1/V2)

Question 49

Gibbs-Helmholtz equation

Answer 49

d/dT (G/T) = -H/T^2

at constant pressure

Question 50

general method: how to find Gibbs energy of mixtures

Answer 50

using partial pressures of gases (for ideal gases)

Question 51

partial pressure defintion

Answer 51

p(i) = x(i) p(tot)

where x(i) is the mole fraction of I

Question 52

equation for Gibbs energy of a mixture of ideal gases

Answer 52

G = sum of n * Gm(p)
n is number of moles, Gm(p) is molar gibbs energy for partial pressure of a species

Question 53

what is the chemical potential for ideal gases?

Answer 53

µ is the same as molar gibbs energy

so that G = sum of nµ

Question 54

what is the chemical potential for solutions?

Answer 54

µ(c) = µ˚ + RT ln(c/c˚)

same as molar gibbs energy eqn except using conc instead of pressure

for ideal solutions

Question 55

what is the chemical potential of a solid or liquid?

Answer 55

solids and pure liquids are always in standard state so = µ˚

Question 56

definition of standard state

Answer 56

standard state of a substance =. in pure form at one bar and specified temp

Question 57

∆rH˚ =

Answer 57

sum of standard enthalpies of formation * stoic coefficients