entropy

Question 1

what is entropy, is it a path or state function

Answer 1

the flow of heat resulting in the system losing the capacity to do work,

state function

Question 2

define entropy in equation form

Answer 2

deltaS = Sb - Sa =

intergral of change of heat over Temp

Question 3

how does the flow of heat affect the capacity of doing work

Answer 3

the flow of heat reduces the capacity to do work as dU = dbarQ + dw

Question 4

what is dbarQrev at constant temp

Answer 4

dbarQrev = TdS

Question 5

what about entropy is an equilibrium condition

Answer 5

dS = dbar q / T because deltaS= integral dqrev / T only holds for reversible path

Question 6

what are the conditions for a spontaneous process

Answer 6

dbarq dbarq/T for all observable processes

Question 7

what is general definition of equilibrium

Answer 7

dS = dbarq/T

Question 8

what are the conditions of an obseravable change

Answer 8

dS>0 as dS>dbarq/T

Question 9

what happens when a spontaneous change occurs in an isolated system (second law of thermodynamics)

Answer 9

this will produce a states of higher entropy until reaches a max value at which the system will be in equilibrium and entropy will remain constant at max value.

Question 10

how do you prove entropy is a state function

Answer 10

consider perfect gas where dU=dbarqrev - PdV if system only does PV work

for one mole of perfect gas: dU = CdT and as P=RT/V

so dS=dbarqrev/T = Cdt/T + RdV/V which intergrates tp:

Sb - Sa = Cln(Tb/Ta) + Rln(Vb/Va)
showing dbarqrev/T is exact differential and that S is state function for a perfect gas

Question 11

how to show entropy change of reversible expansion of perfect gas

Answer 11

deltaU = q + w,
perfect gas U independent of volume deltaU = 0 so
q = -w
gas expands pressure drops, external pressure continuously adjusted so that p = pex + dP expansion carries out reversibly doing max work and
-wrev = integral PdV
p=nRT/V
-wrev = nRT integral dV/V = nRTln(Vb/Va)
deltaS = qrev/T
= nRTln(Vb/Va)
for reversibly deltaSoverall=0
= qrev/T - qrev/T

Question 12

what is entropy of an irreversible expansion of a perfect gas

Answer 12

irreversible expansion of a perfect gas into a vacuum no work done so deltaSoverall no longer 0

deltaS=nRln(Vb/Va) ,
gas does so work q = -w = 0 so no heat absorbed
deltaS = nRln(Vb/Va) + 0/T

Vb>Va total entropy of system and surroundings will have increased,
entropy change in gas is same for reversible and irreversible expansion so investigate surroundings before determining if rev or not.

Question 13

how is entropy change when heat flows from Th to Tc

Answer 13

dS = (-dbarq/Th) + (dbarq/Tc) 
dS = dbarq((Th-Tc)/ThTc)

Question 14

what are the fundamental thermodynamic equations

Answer 14

when volume can change, tendency to maximise S(V) predicts expansion of gases

particle numbers change max S(N) predict change in composition

energy can change, max S(U) predict tendency for heat flow

if all can change fundamental TD equation for entropy S = S(U,V,N)

Question 15

what are the 2 factors that drive chemical systems to equilibrium

Answer 15

tendency to minimise their energy

tendency to maximise entropy

Question 16

how do you show entropy as a function of pressure and temperature

Answer 16

S = Sb - Sa = nRln(Vb/Va)
P inversely proportional to V of perfect gas under isothermal conditions
S=Sb-Sa = -nRln(Pb/Pa)

S0 = entropy of perfect gas at 1atm
S=S0 - Rln(P/P0) =
=S0 - Rln(P/1) = S0 - Rln(P)

shows that pressure increases as entropy decreases

Question 17

show Cv and Cp in terms on dbar-q-rev

express the entropy terms

Answer 17

Cv = (dbar-q-rev/dT)v
Cp = (dbar-q-rev/dT)p

as dS= (dbar-q-rev / T)
dS=(Cv/T)dT
dS=(Cp/T)dT

integrate to
deltaS = C ln(Tb/Ta)

if So is entropy at 0 K entropy at T:
S(T) = So + integral Cp/T dT

Question 18

what is the third law of thermodynamics

what is common for most pure substances?

Answer 18

the entropy of all perfect crystals is zero at the absolute zero of temperature

form perfect crystals at low temp so assume So = 0

Question 19

describe the entropy for the states of matter

Answer 19

solids - molecules / atoms tightly bound together so entropy is low

gases - molecules free to move in large volumes so entropy is high

liquids - intermediate in their properties

Question 20

how does entropy change during vaporisation

Answer 20

molecules leave state of modest freedom to one of high freedom so positive entropy change,
deltavapS = qrev / T
= deltavapH/T

Question 21

when got box A and box B where box A is half-filled while B is full-filled how is the entropy change per mole between the states found

which definition of entropy does this lead to

Answer 21

probA/probB = (Va/Vb)^M
ln(probB/probA) = Mln(Vb/Va)
1mol gas Sb-Sa = Rln(Vb/Va)
M = Na 
Sb-Sa = (R/Na)ln(ProbB/ProbA) 

S = Kb ln(omega) 
Kb = R/Na 
omega = num microstates

Question 22

when counting events and objects are indistinguishable how can number of permutations (omega) be calculated

Answer 22

(N!/n1!n2!…nt!) = omega

N = number of trails
n = results
N!/n!(N-n)!

Question 23

what is a macrostate

Answer 23

macroscopic observable or controllable which can perform experiment in which we vary or measure volume or density N/V of the system

Question 24

what is a microstate

Answer 24

individual snapshots that are counted and equally probable which macrostates are made up of. probability of observing a particular macrostate is proportional to how many microstates it has

Question 25

how does the value of omega link to diffusion and the probable layouts of energy distribution

Answer 25

Diffusion: larger omega predicts diffusion as S=Kbln(omega) and link entropy in terms of set of probabilities

energy: having particles spread over more energy levels is most probable as has largest omega and corresponds to max number of microstates

Question 26

what is deltaS for energy distributions when in thermodynamic equilibrium

Answer 26

deltaS = -K ln(ni/n0)
or = thermalEi/T
ni/n0 = e^-thermalEi/kT
(Boltzmann distribution)