At P = 1 atm, pure germanium melts at 1232 K and boils at 2980 K. The pressure o

Question

At P = 1 atm, pure germanium melts at 1232 K and boils at 2980 K. The pressure of the triple point is 8.4-10^-8 atm. The heat of fusion is H_m = 36.94 kJ/mol. The density of solid Ge is d_s= 5.32 g/cm^3 and the density of liquid Ge is d_L= 5.6 g/cm^3. Calculate the temperature of the triple point assuming that the enthalpy of fusion and densities are constant. Calculate the pressure required to melt germanium at room temperature 298 K assuming that the enthalpy of fusion and densities are constant. Estimate the heat of vaporization assuming germanium vapor behaves as an ideal gas and that the molar volume of liquid germanium is negligible in comparison to that of germanium gas.

Explanation / Answer

A) P1=1 atm ,T1=1232K

Hm=36.94kj/mol

P tp=8.4*10^-8 atm, T(triple pt)=Ttp=?

Using clausius clapeyron equation,

ln(p2/p1)= Hm/R(1/T1-1/T2)

ln(8.4*10^-8 atm/1atm)= 36.94kj/mol/8.314 J/Kmol(1/1232K-1/Tp)

-16.3=4443.11K=(1/1232K-1/Tp)

Tp=342.46K

B)T1=298K, P1=?

T2=1232K,P2=1 atm

ln(p2/p1)= Hm/R(1/T1-1/T2)

ln(1/P1)= 36.94kj/mol/8.314 J/Kmol(1/298K-1/1232K)

ln(1/P1)= 4443.11 K(1/298K-1/1232K)

(1/P1)=exp (11.285)

P1=1.25*10^-5 atm

C) given ds=5.32g/cm3=5.32 g/72.64 g/mol per cm3=0.0732 mol/cm3

Vm,S=molar volume solid =13.66cm3/mol=13.66 *10^-6 m3/mol

Dl=5.6g/cm3=5.6g/72.64 g/mol per cm3=0.0771 mol/cm3=12.97 cm3/mol=12.97*10^-6 m3/mol

Using clausius clapeyro equation in the form,dp/dt=Hm/Vm *1/Tm (at p=1 atm, Tm=1232K)

dp/dt=Hm/Vm *1/Tm=36.94 *1000 j/mol/(Vm,L-Vm,s)*1/1232K

[ (Vm,L-Vm,s)= 12.97*10^-6 m3/mol- 13.66 *10^-6 m3/mol=-0.69 *10^-6 m3/mol}                                         

dp/dt=36.94 *1000 j/mol/(0.69 *10^-6 m3/mol)*1/1232K=-43.45 *10^6 pa K-1=-43.45atmK-1

Also we can write, dp/dt=Hvap/Vm,vap *1/Tb

                                -43.45atmK-1==Hvap/Vm,v *1/Tb              

[Vm,vap =Vm,v-Vm.l, but molar volume of liquid =0 so Vm,vap =Vm,v]

Vm,v=RT/P(ideal gas)

Vm,v=0.0821 L atm/K mol *(2980K)/1 atm=244.66 L

                                -43.45atmK-1==Hvap/Vm,v *1/Tb

                                              -43.45atmK-1=Hvap/244.66L *1/2980K

                                                    Hvap= -43.45atmK-1 *2980K*244.66L

                                                                  =-31678821.5 L atm mol-1(8.314J

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