TITLE.PM5

FIRST LAW OF THERMODYNAMICS 149

dharm

/M-therm/Th4-3.pm5

Vacuum

dy

Air

Q

y

y 1

y 0

0 0

2

1

c

p (Pressure)

V (Volume)

Fig. 4.28 Fig. 4.29

Now, substituting the values of p 1 , V 1 , A in eqn. (vi), we get

p =^150

4 8

2

π^2

FH × IK

VC+

or p = 0.0594 V + C ...(vii)
where p is in N/cm^2 and V is in cm^3.
∴ p 1 = 0.0594 V 1 + C
30 = 0.0594 × 45 + C
∴ C = 27.33
Hence, p-V relationship for the process is,
p = 0.0594 V + 27.33 ...(viii)
During the process the piston is moved by a distance of 3.5 cm.
This increases the volume of gas by

3.5 × A^2 = 3.5 ×

π

4 8

F × 2

H

I

K = 175.9 cm^3

Hence, the final volume of air,

V 2 = 45 + 175.9 = 220.9 cm^3

Substituting this value in equation (viii), we get

p( = p 2 ) = 0.0594 × 220.9 + 27.33 = 40.45 N/cm^2

The work done W during the process is given by

W = zpdV =

A

p S pdp

p 2

1

2

z