THERMO7 Flashcards
2nd Law of Thermodynamics
Calussius
-no process is possible whose sole result is the transfer of heat from a colder body to a hotter body
2nd Law of Thermodynamics
Kelvin
-no process is possible whose solo result is the complete conversion of heat into work
Coefficient of Performance
Equation
useful product / necessary input
In a complete cycle on a PV diagram…
… ΔEint = 0
… net work done = are enclosed
… clockwise, Won is -ve and Qin is +ve
… anticlockwise, Won is +ve and Qin is -ve
2nd Law of Thermodynamics
Refrigerator Statement
-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
Refrigerators / Air Conditioning
Qh = W + Qc COP = Qc / W Qh = heat removed to hot reservoir Qc = heat added from cold reservoir W = work put in
Heat Pump
Qh = W + Qc COP = Qc / W Qh = heat removed Qc = heat added W = work put in
Heat Engines
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
2nd Law of Thermodynamics
Heat-Engine Statement
-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
Theoretical Maximum Efficiency
ε = 1 - Tc/Th
Entropy
Definition
- a measure of the disorder of a system
- hot -> cold
- order -> disorder
Entropy
Statistical Mechanics Approach
-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
Entropy
Classical Thermodynamics Approach
-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
Adiabatic Expansion and Entropy
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
Free Expansion and Entropy
-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)
Tc and Th
Tc = temperature of the system Th = temperature of the surroundings
Entropy of the Universe
-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
Carnot Cycle
Definition
- a theoretical thermodynamic cycle with maximum efficieny
- reversible
Carnot Cycle
Processes
-cycle between four points on a PV diagram 1 to 2: isothermal 2 to 3: adiabatic 3 to 4: isothermal 4 to 1: adibatic
Carnot Cycle
Equations and Explainations
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
Carnot Cycle
Entropy and Reversibility
- 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
Carnot Cycle
Maximum Theoretical Efficiency
ε = W / Qin = 1 - Qc/Qh
-but for carnot cycle, Qc=TcΔS and Qh=ThΔS
ε = 1 - TcΔS/ThΔS = 1 - Tc/Th
