TITLE.PM5

BASIC CONCEPTS OF THERMODYNAMICS 45

dharm
M-therm/th2-2.pm5

Equilibrium

states

1

2

p

V

patm

pgas

170 mm

Liquid

(Sp. gravity = 0.8)

X X

Solution. Equating pressure on both arms above the

line XX (Fig. 2.27), we get

pgas + pliquid = patm. ...(i)

Now, pliquid = ρ.g.h

= (0.8 × 1000) × 9.81 × 1000170

= 1334.16 N/m^2

= 0.0133416 bar

patm. = 760 mm of Hg = 1.01325 bar

Substituting these value is eqn. (i) above, we have

pgas + 0.0133416 = 1.01325

∴ pgas = 0.9999 bar. (Ans.)

Example 2.9. Estimate the mass of a piston that can be supported by a gas entrapped

under the piston in a 200 mm diameter vertical cylinder when a manometer indicates a difference

of 117 mm of Hg column for the gas pressure. (Poona University, May 1996)

Solution. Refer Fig. 2.28.

Let m = mass of the piston, kg.

p = pressure of the gas

= 117 mm of Hg column

Dia. of vertical cylinder, d = 200 mm

Now, downward force = m.g ...(i)

and upward force = p × π/4 d^2 ...(ii)

Equating eqns. (i) and (ii), we get

m.g = p × π/4 d^2

m × 9.81 = 136 1000 981

117

..××× 1000

F

H

I

K×

π

4

200

1000

2

×FH IK (Q p = ρgh)

∴ m = 49.989 kg.(Ans.)

2.18. REVERSIBLE AND IRREVERSIBLE PROCESSES

Reversible process. A reversible process (also sometimes

known as quasi-static process) is one which can be stopped at any

stage and reversed so that the system and surroundings are exactly

restored to their initial states.

This process has the following characteristics :

1. It must pass through the same states on the reversed path
   as were initially visited on the forward path.


2. This process when undone will leave no history of events in
   the surroundings.


3. It must pass through a continuous series of equilibrium
   states.
   No real process is truely reversible but some processes may approach reversibility, to close
   approximation.

Fig. 2.28

Fig. 2.29. Reversible process.

Fig. 2.27