Page 42 - BASIC CONCEPTUAL OF THERMOFLUID

CHARLES’ LAW

V 1 = V 2 = constant
T 1 T 2

FIGURE 3.9: p - v Graph for constant pressure process

EXAMPLE 8

A quantity of gas at 0.54 m and 345 C undergoes a constant pressure process that causes the
3
o
volume of the gas to decreases to 0.32 m . Calculate the temperature of the gas at the end of the
3
process.
SOLUTION : EXAMPLE 2

V = 0.54 m 3 V = 0.32 m 3
1
2
0
T1 = 345 C + 273 T 2 = ? C
0
= 618 K

V 1 = V 2
T 1 T 2
0.54 0.32
=
618 T 2

T2 = 366.222 K

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