Page 68 - DJJ20063- Thermodynamics 1
P. 68
DJJ20063- Thermodynamics 1
Example 2.12
3
o
A quantity of gas at 0.54 m and 345 C undergoes a constant pressure process that
3
causes the volume of the gas to decreases to 0.32 m . Calculate the temperature of
the gas at the end of the process.
Solution to Example 2.12
From the question
3
V1 = 0.54 m
T1 = 345 + 273 K = 618 K
3
V2 = 0.32 m
V 1 = V 2
T 1 T 2
V
T 2 = T 1 x 2
V 1
0.32 m
3
= (618 K ) 3
0.54 m
= 366 K
2.4.3 Explain the law of ideal gas.
Universal Gases Law
PV = constant = R (2.11)
T
P V P V
i.e. 1 1 = 2 2 (2.12)
T 1 T 2
R- gas constant (Nm/kg K or J/kg K). Each perfect gas has a different gas constant.
59 | P a g e
