A sample of Argon has a volume of 3.2 L and the pressure is 0.9 atm. If the final temperature is 34 c, the final volume is 1 L, and the final pressure is 6 atm, what was the initial temperature of the argon in Kelvin?
Combined gas law:
P1V1/T1=P2V2/T2 ;
P1=0.9 atm
V1=3.2 L
T1 -- ?
P2=6 atm
V2=1 L
T2=34 oC= 307 K
T1=P1V1T2/P2V2=0.9*3.2*307/(6*1)=147.36 K
Answer: the initial temperature of the argon was 147.36 K
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