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

Q_1-2 = U₂ –U₁ + W_1-2

for process (1 -a – b – 2)

Q_l-a-b-2 = u₂– u₁ + w_l-n-b-2

w_l-2 = wl-a-b-2, we can write

Q_l-2 = Q_l-a-b-2

= Q_1-a + O_n-b + Q_b-2 =0+Q_{a-b+ 0}

Or                                Q_l-2 = O_a-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 dQ_l 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

or

Thus

“The cyclic integral of dQ for a reversible cycle is equal to zero”. This is the Clausius

theorem.