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理想气体 压强公式 的量纲分析 - PowerPoint PPT Presentation


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理想气体 压强公式 的量纲分析. 设气体的压强为 p , 分子数密度为 n ; 分子的质量为 m , 平均速率为 v , 由于气体压强是一种动力学效应,所以 p 应与 n 、 m 、 v 有关,. 它们之间的关系可以表示为 p  n  m  v  式中的  、  、  均为待定常数。. dim n = L -3 ( 曾用[ n ] 表示 dim n ), dim m = M, dim v = LT -1 , dim p = (dim F )/L 2 = L -1 MT -2.

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理想气体压强公式的量纲分析


P n m v p n m v
设气体的压强为 p,分子数密度为n;分子的质量为m,平均速率为v,由于气体压强是一种动力学效应,所以p应与 n、m、v有关,


P n m v
它们之间的关系可以表示为p  nmv式中的、、均为待定常数。


Dim n l 3 n dim n dim m m dim v lt 1 dim p dim f l 2 l 1 mt 2
dimn = L-3(曾用[n]表示dimn),dimm = M,dimv = LT-1,dimp = (dimF)/L2 = L-1MT-2.


Dim p dim n m v dim n dim m dim v l 3 m lt 1 l 3 m t
dimp = dim(nmv) =(dimn)(dimm)(dimv) = (L-3)M(LT-1) = L -3MT-.


1 2 1 3 1 p nmv 2
 = 1, = 2, = (1+)/3 = 1.故由量纲分析得: p  nmv2.


Dim n l 3 dim m m dim v lt 1 dim p dim f l 2 l 1 mt 2
[另一解法] dimn = L-3,dimm = M,dimv = LT-1, dimp = (dimF)/L2 = L-1MT-2


Dim p l 1 mt 2 l 2 3 mt 2 l 3 m l 2 t 2 dim n dim m dim v 2 dim nmv 2 p nmv 2
dimp = L-1MT-2= L2-3MT-2 = (L-3)(M)(L2T-2) =(dimn)(dimm)[dim(v2)] = dim(nmv2).故由量纲分析得:p  nmv2.


A p a nmv 2 a 8
A为一无量纲量,则有压强公式:p = Anmv2.可以证明在此压强公式中的A = /8.


经验表明:由量纲分析所得结果中的无量纲量A的数量级往往为1.


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