TITLE.PM5

114 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-1.pm5

The work done is given by the shaded area, and this area can be evaluated by integration.

i.e., Wpdv
v

v

=z

1

2

Therefore, since pvγ = constant, C, then

W = C

dv

v v

v

1 γ

2

z Q p

C

v

L =

NM

O

γQP

i.e., WC

dv

v

C v

v

v

v

v

==

z −+

−+

γ

γ

1 γ

2

1

1 2

1

= −

−

F

HG

I

KJ

= −

−

F

HG

I

KJ

−+ −+ −+ −+

C vv^2 C vv

1

1

1

1

1

2

1

11

γγ γ γ

γγ

The constant in this equation can be written as p 1 v 1 γ or as pv 22 γ. Hence,

W=pv v −p v v pv pv

−

= −

−

−+ −+

11 1

1

22 2

1

11 2 2

11

γγ γ γ

γγ

i.e., W

=pv−p v

−

11 2 2

γ 1 ...(4.32)

or W
=RT T−
−

() 12

γ 1 ...(4.33)

Relationship between T and v, and T and p :

By using equation pv = RT, the relationship between T and v, and T and p, may by derived

as follows :

i.e., pv = RT

∴ p=RTv

Putting this value in the equation pvγ = constant

RT

v .v

γ = constant

i.e., Tvγ–1 = constant ...(4.34)

Also v = RTp ; hence substituting in equation pvγ = constant

p RT

p

F

HG

I

KJ

γ

= constant

∴ T

p

γ

γ− 1 = constant

or
T

()p

γ

γ

− 1 = constant ...(4.35)

Therefore, for a reversible adiabatic process for a perfect gas between states 1 and 2, we can
write :