作者adogacat (ˊ口ˋ)
看板NDHU-Ch101
標題[課務] 物化作業2解答
時間Sat Oct 17 01:43:22 2009
Problem 1.11
The molar volume is obtained from
a)
密度= M/Vm =molar mass/molar volume
or Vm=M/密度=18.02g mol^(-1)/133.2gL^(-1)=0.1353Lmol^(-1)
b)
Z=pVm/RT[1.20]
=(327.6atm)*(0.1353 L mol^-1)/(0.08206L atm K^(-1) mol^(-1))*(776.4K)
=0.6957
c)The van der Waals equation is
p=(RT/Vm-b)-a/(Vm^2)[1.25b]
Substituting this expression for p into Z [1.20]gives
Z=Vm/(Vm-b)-a/VmRT={0.1353Lmol^(-1)}/{(0.1353Lmol^(-1))-(0.0305Lmol^(-1)}
-(5.464L^2 atm mol^-2)/(0.1353Lmol^-1)*(0.08206L atm K^-1 mol^-1)*776.4k
=1.291-0.633=0.658
1.15
This expansion has already been given in the solutions to Exercise 1.24(a)
and Problem 1.17;the result is
p=(RT/Vm) (1+[b-a/RT]1/Vm + b^2/Vm^2 + ......)
Compare this expansion with p=(RT/Vm)(1+B/Vm+C/Vm^2 + .....) [1.22]
and hence find
B=b-a/RT and
C=b^2
Since C=1200 cm^6mol^-2 , b=C^(1/2)=34.6 cm^3mol^-1
a=RT(b-B)=(8.206*10^-2)*(273 L atm mol^-1)*(34.6+21.7)cm^3mol^-1
=(22.40L atm mol^-1)*(56.3*10^-3 L mol^-1)=1.26 L^2 atm mol^-2
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1F：→ r14262003 :說好的物化一解答呢 怎麼沒有PO 10/20 00:58
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