Page 70 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

From equation 2.10 the gas constant for any gas can be found when the molecular weight

is known, e.g. for oxygen of molecular weight 32, the gas constant is

R 8314 4 .
R = o = = 259 8 . J/kg K
M 32

Example 2.13

3
2
0.046 m of gas are contained in a sealed cylinder at a pressure of 300 kN/m and a

2
o
temperature of 45 C. The gas is compressed until the pressure reaches 1.27 MN/m

o
and the temperature is 83 C. If the gas is assumed to be a perfect gas, determine:

a) the mass of gas (kg)

b) the final temperature of gas (K)

Given:

R = 0.29 kJ/kg K

Solution to Example 2.13

From the question
3
V1 = 0.046 m
P1 = 300 kN/m 2

T1 = 45 + 273 K = 318 K

2
2
3
P2 = 1.27 MN/m = 1.27 x 10 kN/m
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K

From equation 2.14

PV = mRT
P V 300 x 0.046
m = 1 1 = = 0.1496 kg
RT 1 0.29 x 318

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