Thermodynamics Second Law
The second law denies the possibility of utilization of heat out of a single body. The definitions of the second law of thermodynamics are.
It is impossible to construct law of thermodynamics are: in a cycle will produce no effect other Thant the extraction of heat form a reservoir and the performance of an equivalent amount of work. ( Kelvin Planck statement)
Heats connote flow itself from a colder to a hotter body.
It is impossible to have a process in which the entropy of an isolated system is decreased.
Adiabatic -->thermally insulating
Diathermia ----> thermally conducting
Heat engine a heat engine takes a heat Q1 form the furnace and rejects Q2 to the heat sink and does a work
WQ = Q2 – Q1
Thus efficiency of an ending η = W /Q1 = (Q1 – Q2 )/ Q1
Or η = 1 – (Q2 / Q1 ) = 1 – (T2 / T1)
Entropy ds = dQ /T or S2 – S1 = ∫dQ/ T
Note that T is not differentiable. Entropy is a measure of randomness or disorder in a system.
Claudius inequality ?dQ/ T ≥ 0
Or ?s ≥ ∫dQ / T or dQ = Tds ≥dU + pdV
Relation between entropy and statistical weight O (thermodynamic property)
S = k logeO
Where k is Boltzmann’s constant.
Amount of heat required to form a unit area of the liquid surface layer during the isothermal increase of its
Surface H = - T (dσ/dT) where σ is surface tension.
Carnot engine the French scientist (auto engineer) NL. Said Carnot in 1824suggested an idealized engine called Carnot engine. It has a cylinder piston system. The walls and the piston are completely adiabatic (insulating) and the haze is diathermia (thermally conducting). It contains and ideal gas. Assuming it undergoes isothermal expansion, adiabatic expansion, and isothermal compression, adiabatic compression to complete the cycle Pv and ST plots for a Carnot cycle ar shown Carnot engine is a reversible engine.
Carnot’s theorem all reversible engines operating between the same tow temperatures have equal efficiency and no engine iterating between the same two temperatures can have an efficiency greater than this according to Carnot’s thermo, maximum efficiency
η = 1 – (T2 / T1)
Since T2 cannot be zero (as 0 K cannot be obtained), therefore, efficiency cannot be 1.
Refrigerator or heat pump a heat engine takes heat from a hot body. Converts part of it into work and rejects to cold body. The reverse operation is done by a refrigerator (or heat pump). It takes an amount Q2of heat form a cold body, an amount of work W is done unit by the surrounding and a total heat Q1 = Q2 + W is supplied to hot body as illustrated in
Q1 / Q2 = T1 / T2
Q2 + (W / Q2 )= T1 / T2
Or W = Q2 (T1 / T2 -1)
This leads to another statement of second law: it is not possible to design a refrigerator which works in a cyclic process and whose only result is to transfer heat form a body to a hotter body. This is the clauses statement of the second law of thermodynamics.
Coefficient of performance.
K = heat extracted / work done = Q2 / W = Q2 / (Q1 – Q2)
In a perfect refrigerator K = ∞
That is Q1 = Q2 or W = 0
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