Asked by Thomas
Pure helium is closed into cylinder with a movable piston. The initial volume, pressure, and temperature of the gas are 15 dm3, 2·105 Pa, and 300 K, respectively. If volume is decreased to 12 dm3 and the pressure increase to 3.5·105 Pa find the final temperature of the gas.
Answers
Answered by
Damon
P V/T = n R
n and R are the same for both
P1 V1/T1 = P2 V2/T2
P1 V1 T2 = P2 V2 T1
T2 = T1 (P2/P1)(V2/V1)
luckily using degrees Kelvin so
T2 = 300 (3.5/2)(12/15)
n and R are the same for both
P1 V1/T1 = P2 V2/T2
P1 V1 T2 = P2 V2 T1
T2 = T1 (P2/P1)(V2/V1)
luckily using degrees Kelvin so
T2 = 300 (3.5/2)(12/15)
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