Section 3: Entropy

Question 1

consider moving between two points i and f on a PV diagram, connected via reversible paths R1 and R2

for the complete cycle we have

Answer 1

∫δQR/T=0

Question 2

∫δQR/T=0 so

Answer 2

∫ from i to f δQR1/T + ∫ from f to i δQR2/T=0 =0

∫ from i to f δQR1/T = ∫ from i to f δQR2/T

Question 3

R2 is reversible and therefore

Answer 3

∫ from i to f δQR/T is path independent

leads to definition as func of state - entropy

Question 4

entropy - for a reversible path R, delta S=

Answer 4

Sf-Si = ∫ from i to f δQR/T

Question 5

entropy - for an infinitesimal, reversible process, dS=

Answer 5

δQR/T

Question 6

consider moving between two points i and f on a PV diagram, connected via reversible path R and irreversible path I

Claus ineq gives

Answer 6

∫ from i to f δQI/T + ∫ from f to i δQR/T < or =0

∫ from i to f δQI/T < or = -∫ from f to i δQR/T

∫ from i to f δQI/T < or = ∫ from i to f δQR/T

=∫ from i to f dS

Question 7

entropy - irreversible paths

for an infinitesimal part of the process

Answer 7

δQ/T < or =dS

equality holds for a reversible process

Question 8

for a thermally isolated system, δQ and ΔS

Answer 8

δQ=0
ΔS> or=0

Question 9

consider heat transfer between bodies A and B - A has

Answer 9

lost +δQ and gained -δQ

Question 10

consider heat transfer between bodies A and B - B has

Answer 10

lost -δQ and gained +δQ

Question 11

δQA<0 and δQB>0 with

Answer 11

δQA = - δQB

Question 12

for heat flow from A to B, the temp

Answer 12

Ta>Tb and hence

|δQ/TA| < |δQ/TB|

Question 13

the entropy of a thermally isolated system increases for

Answer 13

any irreversible process and is at best unaltered for a reversible process

Question 14

as the universe can be considered as thermally isolated, it follows that

Answer 14

the total entropy of the unvierse cannot decrease

Question 15

local decreases in S

Answer 15

are allowed

principle only applies to net entropy changes

Question 16

it is necessary to consider the surroundings for

Answer 16

systems that are not thermally isolated

Question 17

with the introduction of entropy, the inexact differential δQ in the first law can be rewritten

Answer 17

dU=TdS-PdV

‘central equation of thermodynamics’

Question 18

all differentials in dU=TdS-PdV

Answer 18

are exact

as U, S and V are functions of state

holds for all processes as integration is path-independent

Question 19

T and S

Answer 19

conjugate variables

S is extensive
T is intensive

Question 20

simple processes of TS diagrams

Answer 20

adiabatic are vertical lines
isothermals are horizontal lines

so carnot cycle is a simple rectangle

Question 21

carnot cycle is reversible and for any reversible process

Answer 21

Q= ∫ TdS

Question 22

so net heat absorbed in the area enclosed by a reversible path on a TS diagram is

Answer 22

the area enclosed by a reversible cycle, which is equal to the work done during the cycle

Question 23

simple entropy changes - heating a body

Answer 23

dSbody = δQ/T = C dT/T

integrate for ΔSbody

Question 24

simple entropy changes - adding heat to a reservoir at constant T

Answer 24

temp remains const

ΔSres = Q/T

Question 25

simple entropy changes - phase changes

Answer 25

phase changes are isothermal processes where the added heat Q=mL causes phase change without change in T ΔS = mL/T

Question 26

ΔS is positive for

Answer 26

bodies that absorb heat

Question 27

ΔS is negative for

Answer 27

bodies that supply or lose heat

Question 28

entropy of an ideal gas

Answer 28

start with Cv = (∂U/∂T)V = dU/dT rearrange central eqn of thermod insert PV=nRT and Cv=ncv