THERMO7

Question 1

2nd Law of Thermodynamics

Calussius

Answer 1

-no process is possible whose sole result is the transfer of heat from a colder body to a hotter body

Question 2

2nd Law of Thermodynamics

Kelvin

Answer 2

-no process is possible whose solo result is the complete conversion of heat into work

Question 3

Coefficient of Performance

Equation

Answer 3

useful product / necessary input

Question 4

In a complete cycle on a PV diagram…

Answer 4

… ΔEint = 0
… net work done = are enclosed
… clockwise, Won is -ve and Qin is +ve
… anticlockwise, Won is +ve and Qin is -ve

Question 5

2nd Law of Thermodynamics

Refrigerator Statement

Answer 5

-it is impossible for a refrigerator working in a cycle to produce only the effect of extracting heat from a cold object and reject the same amount of heat to a hot object

Question 6

Refrigerators / Air Conditioning

Answer 6

Qh = W + Qc
COP = Qc / W
Qh = heat removed to hot reservoir
Qc = heat added from cold reservoir
W = work put in

Question 7

Heat Pump

Answer 7

Qh = W + Qc
COP = Qc / W
Qh = heat removed
Qc = heat added
W = work put in

Question 8

Heat Engines

Answer 8

Qh = W + Qc
COP = W / Qh = (Qh-Qc)/Qh = 1 - Qc/Qh
Qh = heat added from hot reservoir
Qc = heat removed to cold reservoir
W = work output from engine

Question 9

2nd Law of Thermodynamics

Heat-Engine Statement

Answer 9

-it is impossible for a heat engine working in a cycle to produce only the effect of extracting heat from a single reservoir and performing an equivalent amount of work

Question 10

Theoretical Maximum Efficiency

Answer 10

ε = 1 - Tc/Th

Question 11

Entropy

Definition

Answer 11

• a measure of the disorder of a system
• hot -> cold
• order -> disorder

Question 12

Entropy

Statistical Mechanics Approach

Answer 12

-disorder isn’t more favourable than order, it is just that there are many more ways / arrangements of a system that are disordered rather than ordered
-as time passes the entropy of the universe increases
S = k log

Question 13

Entropy

Classical Thermodynamics Approach

Answer 13

-for a reversible process, dS = dQ/T
-entropy is a state variable so ΔS depends only on the initial and final states, not the path
ΔS = ∫ dS = ∫ dQ/T

Question 14

Adiabatic Expansion and Entropy

Answer 14

dS = dQ/T
-but for an adiabatic process, dQ = 0 so dS =0
ΔS = ∫ dS so there is no change in entropy
-there is an increase in disorder from the gas occupying a greater volume as there are more possible disordered arrangements of the system
-but this is exactly balanced out by the decrease in disorder due to the reduced molecular speeds at a lower temperature

Question 15

Free Expansion and Entropy

Answer 15

-insulated container so, Q = 0
-the gas is expanding into a vacuum so, W = 0
-from the 1st law, ΔEint = Q + W = 0
-so, ΔT = 0
-the free expansion is isothermal so Qin = -nRTln(Vi/Vf)
dS = dQ/T
ΔS = Q/T = -nRln(Vi/Vf)

Question 16

Tc and Th

Answer 16

Tc = temperature of the system
Th = temperature of the surroundings

Question 17

Entropy of the Universe

Answer 17

-universe = system + surroundings
Tc < Th
ΔSsys = Q/Tc and ΔSsur = - Q/Th
ΔSuni = Q/Tc + (- Q/Th) = Q/Tc - Q/Th
-as Tc < Th this means that ΔS≥0
-so ΔS can never be negative so the entropy of a universe always tends to a maximum

Question 18

Carnot Cycle

Definition

Answer 18

• a theoretical thermodynamic cycle with maximum efficieny

- reversible

Question 19

Carnot Cycle

Processes

Answer 19

-cycle between four points on a PV diagram
1 to 2: isothermal
2 to 3: adiabatic
3 to 4: isothermal 
4 to 1: adibatic

Question 20

Carnot Cycle

Equations and Explainations

Answer 20

1 to 2: Qh is absorbed so entropy increases by ΔS=Qh/Th
2 to 3: Qin = 0 so ΔS = 0
3 to 4: Qc is expelled so entropy decreases by ΔS=Qc/Tc
4 to 1: Qin = 0 so ΔS = 0

Question 21

Carnot Cycle

Entropy and Reversibility

Answer 21

• to be reversible net ΔS = 0 in either direction

- because is ΔS was greater than zero, it would be negative in the opposite direction and ΔS≥0 would be violated

Question 22

Carnot Cycle

Maximum Theoretical Efficiency

Answer 22

ε = W / Qin = 1 - Qc/Qh
-but for carnot cycle, Qc=TcΔS and Qh=ThΔS
ε = 1 - TcΔS/ThΔS = 1 - Tc/Th