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Flashcards in Second Law Deck (32)
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1
Q

Heat Engine

Description

A

-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

2
Q

Direction of Spontaneous Change

A
  • 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
3
Q

Second Law

Claussius Statement

A

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

4
Q

What does the second law tell us?

A
  • which processes are accessible by way of a spontaneous change
  • introduces a state function. entropy, which increases during a spontaneous change
5
Q

Second Law

Entropy Statement

A

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

6
Q

Can entropy ever decrease?

A

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

7
Q

Entropy

Reversible and Irreversible Processes

A
  • 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
8
Q

Thermodynamic Definition of Entropy

A

-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

9
Q

Entropy as a State Function

A

∮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

10
Q

When does the minimum possible entropy change occur?

A

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

11
Q

Irreversible Change

A
  • 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
12
Q

The Clausius Inequality

A

-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
13
Q

The Carnot Engine

Description

A
  • 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
14
Q

The Carnot Engine

Equations

A

W = Qh - Qc

Qh/Th = Qc/Tc

15
Q

The Carnot Engine

Derivation

A

-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)

16
Q

The Carnot Engine

Efficiency

A

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

ε = 1 - Tc/Th

17
Q

Carnot’s Theorem

A

-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

18
Q

Efficiency of Real Engines

A
  • 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
19
Q

Perpetual Motion of the Second Kind

A
  • 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
20
Q

Claussius Statement of the First Two Laws

A

1) the energy of the universe is constant

2) the entropy of the universe approaches a maximum

21
Q

Absolute Scale of Temperature

A
  • 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
22
Q

Thermodynamics of Refrigeration

Description

A
  • 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
23
Q

What needs to happen for a fridge to work?

A
  • 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
24
Q

Refrigerator

Equations

A

Qh = Qc + W

Qh/Th = Qc/Tc

25
Q

Refrigerator

Derivation

A

-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

26
Q

Refrigerator

Coefficient of Performance

A

η = Qc/W = Qc / (Qh-Qc) = Tc/(Th-Tc)

27
Q

Reversible Energy Changes in an Ideal Gas

Heat Expression

A
  • 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
28
Q

Reversible Energy Changes in an Ideal Gas

Isothermal Change in Volume

A
-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)
29
Q

Reversible Energy Changes in an Ideal Gas

Isothermal Change in Pressure

A

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

30
Q

Reversible Energy Changes in an Ideal Gas

Heating at Constant Volume

A
-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)
31
Q

Reversible Energy Changes in an Ideal Gas

Heating at Constant Pressure

A

-by analogy with heating at constant volume:

ΔS = Cp ln(Tf/Ti)

32
Q

Units and Heat Capacity

A
  • 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