Page 117 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Feedback To Activity 3C

3.7 Data: p1 = 1.2 bar; T1= 25 + 273 = 298 K

T2 = 600 + 273 = 873 K; p2 = ?

We can idealize this process at constant volume heating of a perfect gas.

Applying the general property relation between states 1 and 2

p V p V
1 1 = 2 2
T 1 T 2

in this case V2 = V1

p p
Hence, 1 = 2
T 1 T 2

T
or p = p 2
2
1
T 2
873
= 1.2 x
298

= 3.52 bar
Since the aerosol cannot withstand pressures above 3 bar, it will clearly explode

in the fire.

3
3.8 Data: m = 0.5 kg; p = 2 bar; V2 = 0.0658 m ;
T1 = 130 + 273 =403 K

Using the characteristic gas equation at state 2
p V
T2 = 2 2
mR

2 x 10 5 x . 0 0658
=
. 0 05 x . 0 287 x 10 3
= 917 K

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