Let a process be done reversibly from state 1 to state 2. Let reversible adiabatics 1-a and b- 2 be drawn as show in Fig. 8.2. Then a- b is drawn isothermally in such a manner that the area under 1- a- b- 2 is same as the area under 1 – 2. Applying 1st law for process 1 – 2, we have
Q1-2 = U2 –U1 + W1-2
for process (1 -a – b – 2)
Ql-a-b-2 = u2- u1 + wl-n-b-2
wl-2 = wl-a-b-2, we can write
Ql-2 = Ql-a-b-2
= Q1-a + On-b + Qb-2 =0+Qa-b+ 0
Or Ql-2 = Oa-b
Equation (8.3) states that heat transferred in the process 1 – 2 is equal to heat transferred in the isothermal process a – b.
So any reversible path can be replaced by a zig-zag path between the same end states. The zig-zag path consists of a reversible adiabatic followed by a reversible isotherm and then by a reversible adiabatic in such a way that will satisfy equation (8.3) i.e., the heat transferred during the isothermal process is the same as that transferred during the original process.
Fig. 8.3 shows an original reversible cycle which is divided by number of strips of reversible adiabatic lines. The strips are closed on tops and bottoms by means of reversible isotherms. Thus each strip consists of rev. adiabatic-rev. isotherm-rev. adiabatic-rev. isothern lines. So each strip represents Carnot’s cycle. If the no. of strips are very large, then the zig-zag lines will coincide with the original reversible cycle.
For abed Carnot cycle; aQ1 is added at contact temperature T1 and aQ2 is rejected at temperature T2. So we can write
ɳ c(rev) = 1 – dQ 2 / dQ1 = 1 – T2 / T1
dQ 2 / dQ1
dQ 2 / dQ1
dQ 2 /T1 = – dQ1/ T2 , He have dQl is positive and d”Q2 is negative
thus dQ 2 /T1 = – dQ1/ T2 (taking proper sign of heat)
dQ 2 /T1 + dQ1/ T2 =0
Similarly for the elemental Carnot cycle efgh we have
dQ 3/T3 = – Dq4/ T4 =0
Thus applying the above procedure for all the elemental Carnot’s cycles, we get the
equation for whole original cycle.
Thus dQ 2 /T1 + dQ1/ T2 + … = 0
“The cyclic integral of dQ for a reversible cycle is equal to zero”. This is the Clausius