Topic 4

Question 1

What is Clausius’ theorem for a reversible closed cycle?

Answer 1

For the Carnot Cycle:
∮d𝑄/𝑇=0

• Isothermal path: dQ=dW=-nRln(V2/V1) since dU= 0
• Adiabatic path: dQ=0

Question 2

What is Clausius’ inequality for any cycle?

Answer 2

• General Cycle
• dQ for a series of Carnot engine, does work and does not output any heat

Kelvin statement leads to clausis inequality

∮d𝑄/𝑇≤0

Equality corresponds to a reversible process

Heat in - dQ= nRTln(V2/V1) and negative for heat out

Carnot cycle is reversible
Net work out = the modulus of the work

Question 3

What is entropy?

Answer 3

Defined by S

d𝑆=(đ𝑄_rev)/𝑇

It is a function of state
Units: JK^-1

Question 4

What is the entropy change along an adiabatic path?

Answer 4

For an adiabatic change of state đQ = 0
There is no change in entropy along a reversible adiabatic path
(not true of an irreversible adiabatic change)

Question 5

What is the entropy change along an reversible isothermal path?

Answer 5

For an isothermal change of state,

Δ𝑆=𝑄/𝑇

where Q is the heat transferred into the system along the path (if the system discards heat, Q < 0)

Question 6

What is greater reversible or irreversible?

Answer 6

d𝑆=(đ𝑄_rev)𝑇≥đ𝑄/𝑇

reversible > irreversible

Question 7

What is the equation for entropy for a thermally isolated system?

Answer 7

d𝑆≥0

since đQ = 0

Question 8

What is the entropy statement of the second law?

Answer 8

A process cannot occur if it would result in a decrease in the total entropy of the Universe

Question 9

How do you calculate entropy in irreversible processes?

Answer 9

Since entropy is a function of state
we can find a reversible path between the initial and final states

we can calculate the change in entropy of the system using that path

(change in entropy of the Universe will not be the same for the reversible and irreversible processes)

Question 10

What is the entropy change around a closed reversible cycle?

Answer 10

d𝑆=0

Question 11

What is the entropy changes for a reversible cycle (regarding heat engines)?

Answer 11

ΔS (system) = ΔS(surroundings) = 0

ΔS(Universe) = ΔS (system) + ΔS(surroundings) = 0

Question 12

What is the entropy changes for a reversible change of state (A → B)?

Answer 12

ΔS (system) = −ΔS(surroundings) [can be positive, negative or zero]
ΔS(Universe) = ΔS (system) + ΔS(surroundings) = 0

Question 13

What is the entropy changes for an irreversible cycle?

Answer 13

ΔS (system) = 0; ΔS(surroundings) > 0

ΔS(Universe) = ΔS (system) + ΔS(surroundings) > 0

Question 14

What is the entropy changes for an irreversible change of state (A → B)?

Answer 14

ΔS (system) ≠ −ΔS(surroundings)

ΔS(Universe) = ΔS (system) + ΔS(surroundings) > 0

Question 15

What is the central equation of thermodynamics?

Answer 15

The first law in a path independent form in terms of entropy changes:

dU=TdS -pdV
Leads to:
𝑇=(𝜕𝑈/𝜕𝑆)_𝑉 𝑝=−(𝜕𝑈/𝜕𝑉)_𝑆

Question 16

What is the equations used in irreversible joule expansion (irreversible adiabatic change of state)?

Answer 16

• Thermally isolated container (ideal gas pi and Vi)
• Rest is a vacuum
• Remove partition allowing it flow into the evacuated region

Entropy is a state function so we just need to find the reversible path between the initial and final state

dU=0 as only U depends on T for a gas at constant T so we need to find the entropy change using a reversible thermal path

dU=dW=dQ=0
Δ𝑆_surroundings=0
Δ𝑆_Universe=Δ𝑆_system=𝑛𝑅 ln⁡(𝑉_𝑓 /𝑉_𝑖 ) > 0

Question 17

What is the equations used in reversible expansion?

Answer 17

Δ𝑆_system=𝑛𝑅 ln⁡(𝑉_𝑓 /𝑉_𝑖 )
Δ𝑆_surroundings=−𝑛𝑅 ln⁡(𝑉_𝑓 /𝑉_𝑖 )
Δ𝑆_Universe=0
𝑑𝑈=0
𝑑𝑊=−𝑝𝑑𝑉
𝑑𝑄=−𝑑𝑊=𝑝𝑑𝑉

Question 18

What is the change in entropy for (MIXING) irreversible, adiabatic change of state?

Answer 18

• Thermally isolated container (ideal gas pi and Vi)
• Other part is filled with different ideal gas
• Open partition of the two and they mix
• Liked two combined joule expansion

Δ𝑆=−𝑛𝑅(𝑥 ln⁡𝑥+(1−𝑥) ln⁡(1−𝑥) )

Question 19

What is the maximum of the entropy of an isolated system?

Answer 19

Attains this maximum at equilibrium, this means that any change in entropy away from equilibrium would serve decrease entropy which is not allowed by the second law

Question 20

What is Boltzmann’s relation?

Answer 20

𝑆=𝑘_𝐵 ln⁡𝑊

• LHS Gives the macroscopic view
• RHS Gives the microscopic view

Question 21

What is the total number of combined ways?

Answer 21

W_t=W_1W_2

Question 22

Why does water not like to mix with non-polar molecules?

Answer 22

Due to entropy decrease

Question 23

What happens in the local changes in entropy?

Answer 23

The second law does not forbid local decrease of a system THAT IS NOT ISOLATED (changes of states e.g crystallisation, biological growth)

the entropy of the universe must not decrease (local decrease compensated by increased entropy of the system)