Re: [問題] Cp=?R at Diatomic
※ 引述《LuisGodofTW (盧彥勳是台灣之神)》之銘言:
: that we know
: Cv = 3/2R at Monatomic
: Cv = 5/2R at Diatomic
: C = 3R at Elemental solid
: and
: Cp = 5/2R Monatomic
: but how much Cp at Diatomic ?
: and at Elemental solid?
: anyone know it?
: Thanks for your giving!!
由熱力學第一定律
ΔU=ΔQ-W
內能變化=熱量變化-氣體對外作功
(1)Control Volume情況下 W=PΔV=0
單原子氣體自由度為3(x,y,z軸共5平移自由度)
所以ΔU=3/2 * nRΔT
雙原子氣體自由度為5(x,y,z軸再加其中兩軸自轉共5平移自由度)
所以ΔU=5/2 * nRΔT
而ΔQ=nCΔT
這時的Cv(單原子)=ΔQ/ΔT=3/2*R
Cv(多原子)=ΔQ/ΔT=5/2*R
(2)Control pressure情況下 W=PΔV=nRΔT
這時的Cp(單原子)=ΔQ/ΔT=3/2*R+R=5/2*R
Cp(多原子)=ΔQ/ΔT=5/2*R+R=7/2*R
(3)固體的情況下
U=Ek+Us 因為震盪運動 內能等於動能+彈性位能
有x,y,z三個方向
所以ΔQ=ΔU(因W=0)=3nRT
C=ΔQ/(nRΔT)=3R
固體分子之間排列緊密 所以體積不大會變!!
於是Cv=Cp=3R
這次我念機械系熱力學的心得...有錯請麻煩改正@@
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