TITLE.PM5

FIRST LAW OF THERMODYNAMICS 145

dharm

/M-therm/Th4-3.pm5

∴ Change in internal energy during constant volume process 2–3,

U 3 – U 2 = mcv(T 3 – T 2 )

= 0.00813 × 10.2 (300 – 544.5) (Q T 3 = T 1 )

= – 20.27 kJ. (Ans.)

(– ve sign means decrease in internal energy)

● During constant volume cooling process, temperature and hence internal energy is

reduced. This decrease in internal energy equals to heat flow to surroundings since

work done is zero.

(iii)Net work done and heat transferred during the cycle :

W1–2 = pV^11 pV^22 mR T T^12

11

−

−

= −

γγ−

()

=

0 00813 4 1 300 544 5

402 1

..(.)×−

1. −
   = – 20.27 kJ
   W2–3 = 0 ... since volume remains constant

W3–1 = p 3 V 3 loge

V

V

1

3

F

HG

I

KJ

= p 1 V 1 loge

p

p

3

1

F

HG

I

KJ

(Q p 3 V 3 = p 1 V 1 )

= (1 × 10^5 ) × 0.1 × loge

44

1

F.

HG

I
KJ
= 14816 Nm (or J) or 14.82 kJ
∴ Net work done = W1–2 + W2–3 + W3–1
= (– 20.27) + 0 + 14.82 = – 5.45 kJ
–ve sign indicates that work has been done on the system. (Ans.)
For a cyclic process : zzδδQW=
∴ Heat transferred during the complete cycle = – 5.45 kJ
–ve sign means heat has been rejected i.e., lost from the system. (Ans.)
Example 4.32. 0.15 m^3 of an ideal gas at a pressure of 15 bar and 550 K is expanded
isothermally to 4 times the initial volume. It is then cooled to 290 K at constant volume and then
compressed back polytropically to its initial state.
Calculate the net work done and heat transferred during the cycle.

Solution. Given : V 1 = 0.15 m^3 ; p 1 = 15 bar ; T 1 = T 2 = 550 K ; V

V

2

1

= 4 ; T 3 = 290 K

Refer to Fig. 4.25.

Considering the isothermal process 1–2, we have

p 1 V 1 = p 2 V 2 or p 2 =

pV

V

11

2

or, p 2 = 15 0 15

4015

×

×

.

(.)

= 3.75 bar

Work done, W1–2 = p 1 V 1 loge

V

V

2

1

F

HG

I

KJ