Second Law

Question 1

Heat Engine

Description

Answer 1

-absorbs heat Qh from a hot reservoir (heat source) at temperature Th, converts part of it to work W, and discards heat Qc to a cold reservoir (heat sink) at Tc

Question 2

Direction of Spontaneous Change

Answer 2

• any useful engine needs to work spontaneously (i.e. an engine that needs another engine to drive it is useless)
• heat travelling from a hot reservoir to a colder one is observed experimentally
• the opposite is never observed
• heat exchange only occurs spontaneously in one direction

Question 3

Second Law

Claussius Statement

Answer 3

-heat cannot by itself pass from a colder to a hotter body

Question 4

What does the second law tell us?

Answer 4

• which processes are accessible by way of a spontaneous change
• introduces a state function. entropy, which increases during a spontaneous change

Question 5

Second Law

Entropy Statement

Answer 5

-the entropy of an isolated system increases during a spontatneous change dS ≥ 0

Question 6

Can entropy ever decrease?

Answer 6

-there are no physical processes that lead to a decrease in entropy of the universe (system and surroundings)

Question 7

Entropy

Reversible and Irreversible Processes

Answer 7

• irreversible processes generate entropy (disorder) and thus occur spontaneously
• reversible processes do not generate entropy but may transfer it from one part of the system to another

Question 8

Thermodynamic Definition of Entropy

Answer 8

-you can generate work by moving heat from a hot reservoir to a cold one (but not the other way round)
-when you generate entropy, your ability to do useful work is lost
-more useful work is retained the smaller the difference between Th and Tc
-for a perfectly efficient (reversible) process:
dS = dQ/T

Question 9

Entropy as a State Function

Answer 9

∮dQ/T = 0
-entropy change in going from a state A to a state B
Sb - Sa = ∫dQ/T
-since entropy is a state function, the change in entropy between two states depends only on the initial and final states and not on the path taken

Question 10

When does the minimum possible entropy change occur?

Answer 10

-the minimum possible entropy change in any system occurs for a reversible process

Question 11

Irreversible Change

Answer 11

• many physical processes such as the expansion of gas into a vacuum or mining of two liquids through diffusion are irreversible
• and in real processes that may otherwise be reversible, heat leakage, friction etc. generate additional entropy

Question 12

The Clausius Inequality

Answer 12

-even ‘reversible’ or ‘quasistatic’ changes are idealised processes
-in real systems, heat, leakage, friction etc. generate additional entropy
dStotal = dSirr + dQ/T
-> the Clausius statement:
dStotal ≥ dQ/T

dStotal = entropy of system and exterior
dQ/T = entropy change of reversible process of interest
dSirr = entropy change due to irreversible processes

Question 13

The Carnot Engine

Description

Answer 13

• the most efficient engine will act reversibly i.e. no entropy change
• a carnot engine has maximum efficiency
• heat energy is taken from a hot reservior at Th, part of this energy is converted to work and the rest is discarded to a to a cold reservoir at Tc

Question 14

The Carnot Engine

Equations

Answer 14

W = Qh - Qc

Qh/Th = Qc/Tc

Question 15

The Carnot Engine

Derivation

Answer 15

-if an amount of heat Qh leaves a hot source:
ΔSh = -ΔQh/Th
-do some work:
W = Qh - Qc
-an amount of heat Qc enters the cold sink:
ΔSc = ΔQc/Tc
-total entropy change:
ΔS = Qc/Tc - Qh/Th
-if Qh=Qc (W=0) then ΔS is positve since Th>Tc, the transfer of energy will occur spontaneously but no work will be done
-we are free to convert some of Qh to work W as long as ΔS is positive
ΔS = Qc/Tc - Qh/Th ≥ 0
-maximum work the engine can do is obtained when Qc is minimum
Qcmin = Qh Tc/Th
Qh/Th = Qc/Tc
Wmax = Qh(1 - Tc/Th)

Question 16

The Carnot Engine

Efficiency

Answer 16

ε = max. work generated/energy supplied as heat
= Wmax/Qh = Qh-Qc/Qh

ε = 1 - Tc/Th

Question 17

Carnot’s Theorem

Answer 17

-the maximum efficiency of the engine depends on the temperature of the source and sink ONLY, and that the most efficient engine MUST be reversible

Question 18

Efficiency of Real Engines

Answer 18

• for real engines other deficiencies such as friction, less than perfect fuel conversion etc. further reduce the Carnot efficiency
• high efficiencies may be achieved, ε->1 when Tc->0 or Th->∞
• thus engines are designed to run at high temperatures
• even modern heat engines are only ~38% efficient

Question 19

Perpetual Motion of the Second Kind

Answer 19

• it is not possible to construct an engine which will work in a complete cycle and convert ll of the heat it absorbs from a reservoir into work
• such a non-existent engine is known as perpetual motion of the second kind
• it is impossible and would violate the Claussius statement

Question 20

Claussius Statement of the First Two Laws

Answer 20

1) the energy of the universe is constant

2) the entropy of the universe approaches a maximum

Question 21

Absolute Scale of Temperature

Answer 21

• the fact that engine maximum efficiency is independent of the physical/chemical nature of the engine, but depends only on the temperature of the hot and cold reservoirs provides an absolute scale of temperature
• in the Kelvin scale an efficiency of ε=1 defines the absolute zero of temperature

Question 22

Thermodynamics of Refrigeration

Description

Answer 22

• a refrigerator performs the opposite function to a heat engine it uses work to transfer heat from a cold to a hot system
• to go against the natural tendency of change (hot to cold) you need to perform work
• the thermodynamic concept is similar to the heat engine operated in reverse

Question 23

What needs to happen for a fridge to work?

Answer 23

• if an amount of heat Q is removed from a cold body and placed in a hotter one, entropy change is negative
• that means that this process is impossible spontaneously
• to make a fridge work we need to add enough work to cancel this entropy reduction

Question 24

Refrigerator

Equations

Answer 24

Qh = Qc + W

Qh/Th = Qc/Tc

Question 25

Refrigerator

Derivation

Answer 25

-entropy change of cold reservoir when heat is removed:
ΔSc = -Qc/Tc
-entropy change of the hot reservoir when heat is added:
ΔSh = Qh/Th
-for the fridge to function with the minimum amount of work the total entropy change must be zero:
ΔS = Qh/Th - Qc/Tc = 0
so:
Qh/Th = Q/Tc

Question 26

Refrigerator

Coefficient of Performance

Answer 26

η = Qc/W = Qc / (Qh-Qc) = Tc/(Th-Tc)

Question 27

Reversible Energy Changes in an Ideal Gas

Heat Expression

Answer 27

• first law: dU = dQ + dW
• second law: dS = dQ/T

dU = dQ + dW
= dQ - PdV
dQ = dU + PdV
= ∂U/∂T|V dT + ∂U/∂V|T dV + PdV
-for an ideal gas ∂U/∂V is zero
dQ = ∂U/∂T|V dT + PdV
= Cv dT + P dV

Question 28

Reversible Energy Changes in an Ideal Gas

Isothermal Change in Volume

Answer 28

-starting from the heat expression
dQ = Cv dT + P dV
-for an isothermal change dT=0
dQ = PdV
dS = dQ/T = PdV/T
ΔS = ∫PdV/T
-substituting P=nRT/V
ΔS = ∫nRdV/V
ΔS = nR ln(Vf/Vi)

Question 29

Reversible Energy Changes in an Ideal Gas

Isothermal Change in Pressure

Answer 29

ΔS = ∫dQ/T = -∫VdP/T = -∫nR/P dP = nR ln(Pi/Pf)

Question 30

Reversible Energy Changes in an Ideal Gas

Heating at Constant Volume

Answer 30

-starting from the heat expression
dQ = Cv dT + P dV
-at constant volume dV=0
dQ = Cv dT
ΔS = ∫CvdT/T 
ΔS = Cv ln(Tf/Ti)

Question 31

Reversible Energy Changes in an Ideal Gas

Heating at Constant Pressure

Answer 31

-by analogy with heating at constant volume:

ΔS = Cp ln(Tf/Ti)

Question 32

Units and Heat Capacity

Answer 32

• For nCv:
• if heat capacity is given in J/K then n should be given in moles
• of heat capacity is given in J/(kgK) then n should be given in kg