Isothermal expansion, the internal energy of gas remains unchanged, and external work needs to absorb heat. The absorbed heat q is equal to the work done externally.
Q =-w =-∫ pdv = rt1∫ (1/v) dv =-rt1ln (V2/v 1) the upper and lower limits of the integral are v2 and v1respectively.
In which pV=RT 1 p=RT 1/V is used.
Entropy becomes s = q/t1=-rln (v2/v1).
(2) Suppose a process of isobaric expansion from A (p 1 V 1 T 1) to B(p 1 V2 T2) and then compression to C (p2 V2 T 1) under isobaric pressure.
According to the first law of thermodynamics, dU=pdV+dQ.
Calculate the entropy from a to b first. The internal energy dU=kdT k in an ideal gas is a proportional constant.
s(AB)=∫(dU-pdV)/T =∫(k/T)dT-∫p/TdV = ∫( k/T)dT-∫R/VdV
(pV=RT p/T=R/V) On the integral formula, the integral limits of T are T 1 and T2, and the integral limits of V are V 1 and V2.
s(AB)= kln(T2/t 1)-Rln(V2/v 1)
From b to c, the volume is constant, pdV=0.dQ=dU.
The lower limit of S(BC)=∫dQ/T=∫dU/T integral is T2 and the upper limit is T 1.
S(BC)=kln(T 1/T2)
So s = s (ab)+s (BC) =-rln (v2/v1).