作者con0415 (我不是康康)
看板NDHU-Ch101
标题[课务] 物化作业2解答
时间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
=====================正文结束!!===========================
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1F：→ r14262003 :快吐出来 不要当自私鬼>< 10/09 02:18
2F：→ MITCHLIN :大宝 快吐出来 10/09 10:33
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4F：→ con0415 :PO阿 10/09 16:52
5F：推 r14262003 :我可以问2.7的2.479KJ/MOL怎麽来的吗= =? 10/12 12:53
6F：→ r14262003 :靠 懂了= = R=8.314 10/12 12:54
7F：推 leylgoviin :学弟妹 你们的课本应该跟我们的一样 可是 不知道 10/13 11:39
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※ 编辑: con0415 来自: 114.44.201.19 (10/22 09:35)