Thermodynamics Flashcards
making bonds is ___
breaking bonds is ___
make = exothermic
break = endothermic
Second Law of Thermodynamics
In a spontaneous process, the entropy of the Universe increases
molecular interpretation of entropy
randomness = how do molecules distribute themselves amongst available energy levels
–> statistical analysis
in statistical analysis, what does the weight of a distribution refer to?
the total number of ways a particular configuration can be achieved
what is the most probable distribution?
the one configuration with a larger value W than all the others
= the Boltzmann Distribution
Boltzmann Distribution eqn for population of level i, n(i), with energy ε(i)
+ interpret
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
Eqn relating entropy to number of ways W
S = k lnW
k: Boltzmann’s constant
W: weight, ways
molecular energy level distribution when a system is heated
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
molecular energy level distribution when a system expands
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
how does the size of entropy change depend on the T of a system?
the increase in entropy resulting from a certain amount of heat being supplied is GREATER the LOWER the temperature of the system
eqn classical definition of entropy
differential form
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!
eqn calculate entropy change of surroundings
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
relate spontaneity of a process to ∆S(univ)
+ve = spontaneous (entropy of universe increases)
-ve = not spontaneous
= 0 is equilibrium, no tendency for further change
The First Law of Thermodynamics + eqn
energy cannot be created or destroyed but is transformed from one form into another
∆U = q + w
what are the different “forms” of energy?
heat, work and internal energy
key different between internal energy and heat/work
internal energy is a STATE function
heat/work are PATH functions
note that the symbol ‘∆’ is only appropriate for state functions
derive eqn for work done on an ideal gas by expansion
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
when a gas expands, what happens to the internal energy?
work is done so Internal energy falls
derive expression for expansion work done against CONSTANT external pressure
integrate p(ext) dV for an expansion between Vi and Vf
p(ext) constant, move out of integral
w = - p(ext) [Vf-Vi]
what must p(ext) be to do maximum expansion work for a gas in a cylinder?
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)
define reversible reaction
a process whose direction can be changed by an infinitesimal change in some variable
reversible vs spontaneous reactions
reversible:
infinitely slow
at equilibrium
do max work
irreversible = spontaneous:
finite rate
not at equilibrium
less than max work
how to find the expression for work done in a finite expansion for an ideal gas expanding reversibly
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)
what is an indicator diagram?
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
what can be said about the magnitude of heat in a reversible process?
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
internal energy of an ideal gas
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
expression for heat for an isothermal, reversible expansion of ideal gas
q(rev) = w’ = nRT ln(Vf/Vi)
expression for entropy change for any isothermal expansion of an ideal gas from Vi to Vf
∆S = nR ln(Vf/Vi)
confirming entropy increases as volume increases
from q(rev) = nRT ln(Vf/Vi) and ∆S=q(rev)/T
first law for infinitesimal changes
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
first law for infinitesimal changes and expansion work only
dU = 𝛿q - p(ext)dV
expression for heat absorbed under conditions of constant volume
dU = 𝛿q(const vol)
because p(ext)dV = 0, no expansion
so to find ∆U, measure the heat under constant volume conditions
relationship between heat supplied and temperature change
q = c∆T
where c is the heat capacity
q = nC(m) ∆T
where C(m) is molar heat capacity and n is moles
definition of heat capacity at constant volume
C(V,m) = (∂U(m) / ∂T)|V
what is enthalpy? give eqn
H = U + pV
a state function whose value is equal to the heat supplied at constant pressure (sum of internal energy and pV work)
complete differential of the enthalpy equation
dH = dU + pdV + Vdp
from H=U+pV
relate enthalpy change to heat
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
definition of heat capacity at constant pressure
C(p,m) = (∂H(m) / ∂T) |p
equation for converting enthalpy at one temperature to another
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
how to express entropy S in terms of heat capacity (something you can actually measure!)
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
justification for entropy at absolute zero = 0
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.
absolute entropy S(T*) expression
= integral of Cp,m(T)/T wrt T + sum of entropy of phase changes
convert entropy S at T1 to T2
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
define Gibbs energy state function
∆G = ∆H - T∆S
all for the SYSTEM
relate change in G to spontaneity
dG < 0 = spontaneous
at equilibrium G reaches a minimum, dG =0
the first thermodynamic Master Equation
dU = TdS - pdV
aka “First and Second Laws combined”
dU = 𝛿q-𝛿W for pV work, and 𝛿q=TdS
the second thermodynamic Master Equation
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
how does Gibbs energy vary with pressure (constant T)?
for ideal gas
dG = Vdp (dT=0)
sub V = nRT/p and integrate
–> Gm(p) = Gm˚ + RT ln(p/p˚)
how does Gibbs vary with volume at constant T?
V inversely proportional to p
p2/p1 = V1/V2
sub in expression for G(p)
G(V2) - G(V1) = nRT ln(V1/V2)
Gibbs-Helmholtz equation
d/dT (G/T) = -H/T^2
at constant pressure
general method: how to find Gibbs energy of mixtures
using partial pressures of gases (for ideal gases)
partial pressure defintion
p(i) = x(i) p(tot)
where x(i) is the mole fraction of I
equation for Gibbs energy of a mixture of ideal gases
G = sum of n * Gm(p)
n is number of moles, Gm(p) is molar gibbs energy for partial pressure of a species
what is the chemical potential for ideal gases?
µ is the same as molar gibbs energy
so that G = sum of nµ
what is the chemical potential for solutions?
µ(c) = µ˚ + RT ln(c/c˚)
same as molar gibbs energy eqn except using conc instead of pressure
for ideal solutions
what is the chemical potential of a solid or liquid?
solids and pure liquids are always in standard state so = µ˚
definition of standard state
standard state of a substance =. in pure form at one bar and specified temp
∆rH˚ =
sum of standard enthalpies of formation * stoic coefficients
