Asked by Luke
When a sample of neon with a volume of 683 mL and a pressure of 0.792 atm was heated from 19.0 to 76.0 °C, its volume became 733 mL. What was its final pressure (in atm)?
Answers
Answered by
DrBob222
(P1V1/T1) = (P2V2/T2)
Answered by
Luke
p2=.792atm x 683mL/ 292.15 x 349.15/733mL I got .882 and it says it wrong. Do you know what I done wrong?
Answered by
DrBob222
I don't think you did anything wrong.
v1 = 683 mL
p1 = 0.792 atm
t1 = 19C = 273+19 = 292K
v2 = 733 mL
p2 = ? in atm.
t2 = 76C = 273+76 = 349K
(0.792*683/292) = (p2*733/349)
If I used 292.15 and 349.15 the answer is 0.8819 which rounds to 0.882 atm.
Usually when something like this happens it's because of the number of significant figures. Technically that 683 is really 700 (rounded to 1 s.f.) and the 733 is 700 rounded to 1 s.f. but I doubt that is the problem.
p2 = (0.792*683*349)/(292*733)
p2 = 0.882 atm.
v1 = 683 mL
p1 = 0.792 atm
t1 = 19C = 273+19 = 292K
v2 = 733 mL
p2 = ? in atm.
t2 = 76C = 273+76 = 349K
(0.792*683/292) = (p2*733/349)
If I used 292.15 and 349.15 the answer is 0.8819 which rounds to 0.882 atm.
Usually when something like this happens it's because of the number of significant figures. Technically that 683 is really 700 (rounded to 1 s.f.) and the 733 is 700 rounded to 1 s.f. but I doubt that is the problem.
p2 = (0.792*683*349)/(292*733)
p2 = 0.882 atm.
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