作者con0415 (我不是康康)
看板NDHU-Ch101
標題[課務] 物化作業2解答
時間Thu Oct 8 20:00:28 2009
2.3(a)
a. [ΔU=ΔH=0]
w = -nRT*ln(Vf/Vi)
= (-1.00 mol)*(8.314 J/(K*mol))*(273 K)*ln(44.8L/22.4L)
= -1.57*10^3 J = -1.57 kJ
q = ΔU-w = 0+1.57 kJ = 1.57kJ
b. [ΔU=ΔH=0]
w = -PexΔV ΔV = (44.8-22.4)L =22.4L
Pex = Pf = nRT/Vf = (1.00 mol*0.08206 L*atm/(K*mol) * 273K)/44.8L
= 0.500atm
w = (- 0.500atm)*(1.013*10^5 Pa/ 1 atm)*22.4 L*(1 m^3/10^3 L)
= 1.13*10^3 Pa m^3) = -1.13kJ
q = ΔU - w = 0 + 1.13kJ = 1.13kJ
c. [ΔU=ΔH=0]
w = 0 (free expansion) q = ΔU-w = 0-0 = 0
2.7(a)
the chemical reaction that occur is
Mg(s) + 2HCl(aq) -> H2(g) +MgCl2(aq)
M(Mg)=24.31 g per mol
Work is done against by the expansion of the hydrogen gas
produced in the reaction
w = PexΔV [2.10]
Vi = 0 Vf = nRT/Pf Pf = Pex
w = -Pex(Vf-Vi) = (-Pex)*nRT/Pex = -nRT
n = 15(g)/24.31(g/mol) = 0.617 mol RT = 2.479 kJ/mol
w = -1.5 kJ
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◆ From: 114.44.200.161
1F：→ r14262003 :快吐出來 不要當自私鬼>< 10/09 02:18
2F：→ MITCHLIN :大寶 快吐出來 10/09 10:33
3F：→ adogacat :在我這了~有人拿給我了~所以我還要PO嗎~愣 10/09 10:52
4F：→ con0415 :PO阿 10/09 16:52
5F：推 r14262003 :我可以問2.7的2.479KJ/MOL怎麼來的嗎= =? 10/12 12:53
6F：→ r14262003 :靠 懂了= = R=8.314 10/12 12:54
7F：推 leylgoviin :學弟妹 你們的課本應該跟我們的一樣 可是 不知道 10/13 11:39
8F：→ leylgoviin :你們有沒有注意到單位不太一樣 請大家抄的時候注意唷 10/13 11:40
※ 編輯: con0415 來自: 114.44.201.19 (10/22 09:35)