A balloon is filled with helium. Its volume is 3.40 L at 36 degrees C. What will its volume be at 93 ∘C, assuming no pressure change?
回答 (2)
2016-03-01 7:17 am
V₁ = 3.40 L, T₁ = (273 + 36) K = 309 K
V₂ = ? L, T₂ = (273 + 93) K = 366 K
For a fixed amount of gas at constant pressure :
V₁/T₁ = V₂/T₂
3.40/309 = V₂/366
Final volume = 366 × (3.40/309) = 4.03 L
V₂ = ? L, T₂ = (273 + 93) K = 366 K
For a fixed amount of gas at constant pressure :
V₁/T₁ = V₂/T₂
3.40/309 = V₂/366
Final volume = 366 × (3.40/309) = 4.03 L
2016-03-01 9:29 am
P1V1/T1 = P2V2/T2 , here P1 = P2 as there is no change in pressure , V1 = 3.40 lL , V2 to be found , T1 = 36C + 273 = 309 , T2 = 273 + 93C = 366 .
from the above equation of universal gas law , we get V1/T1= V2/T2 when P1 = P2 , so Final volume V2 at 93C = V1/T1 X ( T2 ) = 3.4X 366/309 = 4.027184466L
from the above equation of universal gas law , we get V1/T1= V2/T2 when P1 = P2 , so Final volume V2 at 93C = V1/T1 X ( T2 ) = 3.4X 366/309 = 4.027184466L
收錄日期: 2021-04-18 14:39:29
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https://hk.answers.yahoo.com/question/index?qid=20160229210949AAUAF8h