TITLE.PM5

394 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th8-1.pm5

(i)What mass of original gas must have escaped if the dimensions of the balloon is not
changed?
(ii)Find the amount of heat to be removed to cause the same drop in pressure at constant
volume.
Solution. Diameter of the spherical balloon = 6 m
Pressure of hydrogen gas, p 1 = 1 bar abs.
Temperature of hydrogen gas, T 1 = 20°C or 293 K
At a later time pressure of the gas, p 2 = 0.94p 1 at 293 K.
(i)Mass of original gas escaped :
∆m = m 1 – m 2
[where m 1 and m 2 are the initial and final masses of the gas]

=

pV

RT

11

1

=

pV

RT

22

2

=

V

RT

1

1

(p 1 – p 2 ) [Q V 1 = V 2 , T 1 = T 2 and p 2 = 0.94p 1 ]

=

V

RT

1

1

(p 1 – 0.94p 1 ) =

pV

RT

11

1

(1 – 0.94)

∴ %age mass escaped = ∆m

m 1

× 100

=

pV

RT

pV

RT

11

1

11

1

(.) 1094 −

= 6%. (Ans.)

(ii)Amount of heat to be removed :

Using the gas equation,

pV

T

11

1

=

pV

T

22

2

or

p

T

1

1

=

(^0941)
2

. p
T
(Q V 1 = V 2 and p 2 = 0.94p 1 )

∴ T 2 = 0.94T 1 = 0.94 × 293 = 275.4 K or 2.42°C

The heat to be removed is given by

Q = mcv(T 1 – T 2 )

where m =

pV

RT

11

1

=

110 4

3

3

8314

2

293

×××^53

×

π

= 9.28 kg

Q MR

R

MH

=

∴=

=

L

N

M

M

M

M

O

Q

P

P

P

P

8314
8314
2
as for 2 2
cv = 10400 J/kg K for H 2
∴ Q (heat to be removed) = 9.28 × 10400 (293 – 275.4) = 1.69 MJ. (Ans.)
Example 8.4. A vessel of capacity 3 m^3 contains 1 kg mole of N 2 at 90°C.
(i)Calculate pressure and the specific volume of the gas.
(ii)If the ratio of specific heats is 1.4, evaluate the values of cp and cv.
(iii)Subsequently, the gas cools to the atmospheric temperature of 20°C ; evaluate the final
pressure of gas.