Mechanical Engineering Principles

264 MECHANICAL ENGINEERING PRINCIPLES

25 °C. Air is released from the cylinder until

the pressure falls to 300 kPa and the

temperature is 15°C. Determine (a) the mass

of air released from the container, and

(b) the volume it would occupy at STP.

Assume the characteristic gas constant for air

to be 287 J/(kg K).

V 1 = 1 .2m^3 (=V 2 ),p 1 =1MPa= 106 Pa,

T 1 =( 25 + 273 )K=298 K,

T 2 =( 15 + 273 )K=288 K,

p 2 =300 kPa= 300 × 103 Pa

and R=287 J/(kg K).

(a) Using the characteristic gas equation,

p 1 V 1 =m 1 RT 1 , to find the initial mass of air

in the cylinder gives:

( 106 )( 1. 2 )=m 1 ( 287 )( 298 )

from which, massm 1 =

( 106 )( 1. 2 )

( 287 )( 298 )

= 14 .03 kg

Similarly, usingp 2 V 2 =m 2 RT 2 to find the

final mass of air in the cylinder gives:

( 300 × 103 )( 1. 2 )=m 2 ( 287 )( 288 )

from which, massm 2 =

( 300 × 103 )( 1. 2 )

( 287 )( 288 )

= 4 .36 kg

Mass of air released from cylinder

=m 1 −m 2 = 14. 03 − 4. 36 = 9 .67 kg

(b) At STP,T=273 K andp= 101 .325 kPa.
Using the characteristic gas equation

pV=mRT

volume,V=

mRT

p

=

( 9. 67 )( 287 )( 273 )

101325

= 7 .48 m^3

Problem 16. A vesselXcontains gas at a

pressure of 750 kPa at a temperature of

27 °C. It is connected via a valve to vesselY

that is filled with a similar gas at a pressure

of 1.2 MPa and a temperature of 27°C. The

volume of vesselXis 2.0 m^3 and that of

vesselYis 3.0 m^3. Determine the final

pressure at 27°C when the valve is opened

andthegasesareallowedtomix.AssumeR

for the gas to be 300 J/(kg K).

For vesselX:

pX= 750 × 103 Pa,TX=( 27 + 273 )K=300 K,

VX= 2 .0m^3 andR=300 J/(kg K)

From the characteristic gas equation,

pXVX=mXRTX.

Hence ( 750 × 103 )( 2. 0 )=mX( 300 )( 300 )

from which, mass of gas in vesselX,

mX=

( 750 × 103 )( 2. 0 )

( 300 )( 300 )

= 16 .67 kg

For vesselY:

pY= 1. 2 × 106 Pa,TY=( 27 + 273 )K=300 K,

VY= 3 .0m^3 andR=300 J/(kg K)

From the characteristic gas equation,

pYVY=mYRTY.

Hence ( 1. 2 × 106 )( 3. 0 )=mY( 300 )( 300 )

from which, mass of gas in vesselY,

mY=

( 1. 2 × 106 )( 3. 0 )

( 300 )( 300 )

=40 kg

When the valve is opened, mass of mixture,

m=mX+mY

= 16. 67 + 40 = 56 .67 kg.

Total volume,V=VX+VY= 2. 0 + 3. 0 = 5 .0m^3 ,

R=300 J/(kg K),T=300 K.

From the characteristic gas equation,

pV=mRT

p( 5. 0 )=( 56. 67 )( 300 )( 300 )

from which,final pressure,

p=

( 56. 67 )( 300 )( 300 )

5. 0

= 1 .02 MPa