Modern Engineering Thermodynamics

154 CHAPTER 5: First Law Closed System Applications
5.6 IDEAL GASES
Ideal gas equations are usually quite familiar to engineering students. You see them in chemistry courses, fluid
mechanics courses, and of course thermodynamics courses. They are perhaps the most used equations of state
ever devised. The next example illustrates the use of the ideal gas equations in conjunction with solving a basic
thermodynamics problem.
EXAMPLE 5.5
A new radiation heat transfer sensor consists of a small, closed, rigid, insulated 0.0400 m 3 box containing a 0.0100 m 3 rubber
balloon. Initially, the box is evacuated but the balloon contains argon (an ideal gas) at 20.0°C and 0.0100 MPa. When
the balloon receives 0.100 kJ of radiation energy through an uninsulated window in the box, it bursts. The resulting pressure
change is sensed by a transducer and an alarm is sounded. Determine the pressure and temperature inside the box after the
balloon bursts.
Solution
First, draw a sketch of the system (Figure 5.5).
The unknowns here are the final pressure and temperature inside the box, p 2 and T 2 . The system is closed and consists of the
argon in the balloon. Initially, it is at T 1 = 20:0°C and p 1 = 0:0100 MPa. Then, it undergoes a process in which
1Q 2 = 0:100 kJ of heat is added until the balloon bursts. After it bursts, the argon occupies the volume of the entire box, but
its mass has not changed.
Window
Rigid insulated box
with a volume of
0.0400 m 3
Argon filled
balloon initially
with a volume
of 0.0100 m 3 at
20.0°C and
10.0 kPa
FIGURE 5.5
Example 5.5.
The basic conservation of energy is
2
3
1Q 2 − 1 W 2 = m4ðu 2 − u 1 Þ + V2 − 2 V2 1
+ g ðZ 2 − Z 1 Þ5
2g c g c
and since we are given no indication that the box is moving either before or after the heat transfer process, we set the
changes in kinetic and potential energy equal to zero. Three of the relevant auxiliary equations for an ideal gas are (a) the
equation of state, pV = mRT, (b) the internal energy equation, Eq. (3.38), u 2 − u 1 = c v ðT 2 − T 1 Þ, and (c) the work mode
equation for moving boundary work, Eq. (4.26), 1 W 2 =
Z 2
1
argon
pdV. In this problem, the argon certainly changes volume as the
balloon bursts, but the argon expands into a vacuum (the box was initially evacuated), so it expands against zero resistance.
Therefore, the moving boundary work is zero. Putting these results into the basic energy balance equation gives
from which we can solve for T 2 as
1Q 2 − 0 = mc v ðT 2 − T 1 Þ + 0 + 0
T 2 = T 1 + 1 Q 2
mc v
where m is the mass of argon present and 1 Q 2 = 0:100 kJ. Since this is a closed system, the mass can be determined from the
ideal gas equation of state:
m = p 1V 1
RT 1
= p 2V 2
RT 2