TITLE.PM5

FIRST LAW OF THERMODYNAMICS 113

dharm

M-therm/th4-1.pm5

Dividing both sides by T, we get

cv dTT + Rdvv = 0

Integrating

cv loge T + R loge v = constant

Substituting T = pvR

cv loge pvR + R loge v = constant

Dividing throughout both sides by cv

loge pvR + R

cv

. loge v = constant

Again c

R

v=()γ− 1 or

R

cv = γ – 1

Hence substituting

loge pvR + (γ – 1) loge v = constant

∴ loge pvR + loge vγ−^1 = constant

loge pv v×R

−γ 1

= constant

i.e., loge pvR

γ

= constant

i.e., pvR

γ

= econstant = constant

or pvγ = constant ...(4.31)
Expression for work W :
A reversible adiabatic process for a perfect gas is shown on a p-v diagram in Fig. 4.8 (b).

(a) (b)

p. v = Constantγ

2

v 1 v 2

p 2

p 1

p

1

v

Gas

Piston

Insulated

system

Fig. 4.8. Reversible adiabatic process.