A volume of 8 liters of gas has a pressure of 50 kPa and a temp of 27oC.
a) If the pressure were increased to 300 kPa without changing he temperature
what would be the resulting volume? b) If the temperature was raised to 627o C while maintaining
the original pressure, what would be the final volume? c) What would be the resulting volume
if the temperature was raised to 627oC and the pressure increased to 300 kPa simultaneously?
Hint 1
If the temperature is constant, then pressure is inversely proportional to
the volume ( Boyle's Law) that is P a 1/V or PV = constant
b) If the pressure is kept constant, then volume is directly proportional
to temperature (Charle's Law) that is V a T or V/T = constant
Answer 1
V1 = 8 liters
P1 = 50 kPa
T1 = 27o C = 300 K
If the temperature is constant, then pressure is inversely proportional to
the volume ( Boyle's Law) that is P a 1/V or PV = constant
P1V1 = P2V2
P2 = 300 kPa
50 x 103 x 8 = 300 x 103 x V2
V2 = (50 x 103 x 8)/(300 x 103) = 1.33 liters
b) If the pressure is kept constant, then volume is directly proportional to temperature (Charle's Law) that is V a T or V/T = constant
V1 / T1 = V2 / T2 T2 = 627o C = 900K
8 / 300 = V2 / 900
V2 = (900 x 8)/300 = 24 liters
c) P1V1/T1 = P2V2/T2
Now T2 = 627oC = 900 K and P2 = 300 x 103 Pa
(50 x 103 x 8)/300 = (300 x 103 x V2)/900
V2 = (50 103 x 8 x 900)/(300 x 300 x 103) = 4 liters