A gas at 59.0 degrees C and atmospheric pressure fills a cylinder. The gas is transferred to a new cylinder with three times the volume, after which the pressure is half the original pressure.
What is the new temperature of the gas in C?
P1V1 / T1 = P2V2 / T2
T2 = (P2)(V2)(T1) / (P1)(V1)
T2 = (1/2P1)(3V)(T1) / (P1)(V1)
T2 = (1/2)(3)(T1)
T2 = 1.5(T1)
Convert 59C to K.
59C + 273.15 = 332.15K
T2 = 1.5(332.15K)
T2 = 498.225K
Convert back to C
K - 273.15 = C
T2 = 225.075C
Reply:It should be the same as before the transfer, correct?
Reply:________________________________________...
Initial pressure =p1=P
initial volume=v1=V
initial temperature=T1=273+59=332 K
final pressure =p2 =P/2
final volume=v2=3V
final temperature=T2= ?
Applying equation of state,p1v1/T1 =p2v2/T2
T2=p2v2T1/p1v1
T2=P*3V332/2PV
T2= 498 K
t =498 -273= 225 C
The new temperature of the gas is 225 C
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garden sheds
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