Corolaries of 2nd Law of Thermodynamics

Corollary 1: The clausius statement of second law of thermodynamics is the first corollary i. e. ‘It is impossible to construct a device operating in a closed cycle that performs no effect other than the transfer of heat from a cooler body to a hotter body. ’ The mathematical statement of which is given by; Q1? Q2=W Q1 W Q2 Corollary 2: It is impossible to construct an engine operating between only two heat reservoirs, which will have a higher efficiency than a reversible heat engine operating between the same two reservoirs. Hot reservoir Cold reservoir Heat engine Q1 Q1 WX WR Q1? WX Q1?

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WR Suppose these conditions are true, Let? X>? R If the reversible engine also receives heat Q1 from the source, it will do work WR. Let us convert that heat engine to a heat pump, then we have, that R receives Q1? WR & receives WR from the surroundings & rejects Q1 heat to high temperature reservoir. Coupling it to heat engine the net work done now becomes; (Q1? WR) – (Q1? WX) =WX? WR Thus of corollary 2 is true, then we have we have that following condition is also possible. X R Q1 Q1 WX? WR Q1? WX Corollary 3: All reversible heat engines operating between the same two reservoirs have same efficiency.

Since the second corollary has been proved true then it is impossible that one has higher efficiency. Thus according to the Carnot theorem both heat engines have same efficiency. i. e. ?= (Q1+Q2)/Q1 = (T1? T2)/T1 Corollary 4: A scale of temperature can be defined which is independent of any particular thermometric substance and which provides an absolute zero of temperature. ?????????????????????? X R Q W Q0 Let us suppose a heat engine as shown in figure above Thus from corollary 3, the efficiency of that heat engine depends only on the temperature of the reservoirs. = 1–Q/Q0 Let the temperature of the heat reservoir be T0, Now, if T be the temperature of cold reservoir, Then since?? (1 – Q / Q1) Also? = 1 – T / T0 {Carnot’s efficiency} Hot reservoir Heat engine Cold reservoir Q0a Q0b Heat pump Q0n (reversible) Q1b Q1a Q1n Q1a Q1n W Q2z Q2p Q2p Q2z Q2q Heat pump (reversible) Q0z Q0q Q0p Q2q Z p q Heat engine b a n Q1b T0 T0 Total work done by original engine; n z ? Q1 – ? Q2 a p The work delivered in auxilliary engines; z ? (Q2 –Q0 ) p And work applied to heat pump; n ? (Q1 –Q0 ) a Net work done; n z z n ? Q1 ?? Q2 + ? (Q2 ? Q0 ) ?? Q1 ? Q0 ) ? 0 a p p a Thus,n z ? Q0 ?? Q0 ? 0 a p Thus,n z ? Q1T0/T1 ?? Q2T0/T2 ? 0 a p Then; T=T0Q/Q0 Thus, we have defined a temperature scale called as thermodynamic scale & since it begins from T=0, it is an absolute scale of temperature. Corollary 5: The efficiency of any reversible engine operating between more than two reservoirs must be less than that of that of a reversible engine operating between two reservoirs which have temperatures equal to the highest and lowest temperature of the reservoirs in original engine Corollary 6: Whenever a system under goes a cycle,? dQ/T) is zero if the cycle is reversible and negative if irreversible i. e. in general ? (dQ/T) < 0 . It is also called as Clausius inequality. Proof: Let us consider an engine operating between a finite numbers of reservoirs Q1a Q1n Q2z Q2p Q2q Z p q Heat engine b a n Q1b Let quantity of heat Q1a , Q0b?????????? Q1n be supplied from finite number of reservoirs at temperatures T1a?????????? T1n. Let the heat rejected to sinks T2z?????????? T2qetc. Now let the sources be supplied with their quantities of heat by a series of reversible heat pumps receiving quantities of heat Q0a,Q0b?????????? Q0n from a single reservoir T0

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