A quantity of supercooled liquid tin is adiabatically contained at 495 K. Calculate the fraction of the tin which spontaneously freezes, given
\Delta H_{m,(Sn)}=7070 J at T_m=505 K
c_{p,Sn(l)} =34.7-9.2\times 10^{-2}T J/K
c_{p,Sn(s)} =18.5+26\times 10^{-3}T J/K
The equilibrium state of the adiabatically contained system is that in which the solid, which has formed spontaneously, and the remaining liquid coexist at 505 K. Thus, the fraction of the liquid which freezes is that which releases just enough heat to increase the temperature of the system from 495 to 505 K.
Consider 1 mole of tin and let the molar fraction which freezes be x . In Figure 6.16, the process is represented by a change of state from a to c , and, as the process is adiabatic, the enthalpy of the system remains constant; that is,
\Delta H=H_c-H_a=0
Either of two paths can be considered.
Path 1 : a → b → c , during which the temperature of the 1 mole of liquid is increased from 495 to 505 K and then x moles freeze. In this case,
\Delta H_{(a\rightarrow b)} = -\Delta H_{(b\rightarrow c)}\Delta H_{(a\rightarrow b)} =\int_{495}^{505}{c_{p,Sn(l)}dT=34.7\times (505-495)-\frac{9.2}{2} \times 10^{-3}(505^2-495^2) }
=301 J
\Delta H_{(b\rightarrow c)} = -7070x J
and thus,
x=\frac{301}{7070}=0.0426
That is, 4.26 mol% of the tin freezes.
Path 2 : a → d → c ; that is, the fraction x freezes at 495 K, and then, the temperature of the solid and the remaining liquid is increased from 495 to 505 K. In this case,
\Delta H_{(a\rightarrow d)}=-\Delta H_{(d\rightarrow c)}
\Delta H_{(a\rightarrow d)}= the head of freezing moles of tin at 495 K
=-x\Delta H_m(495 K)
But,
\Delta H_m(495 K)=\Delta H_m(505 K)+\int_{505}^{495}{\Delta c_{p(s\rightarrow l)}dT }=7070+16.2(495-505)-\frac{35.2}{2}\times 10^{-3}(495^2-505^2)
=7084 J
Thus,
\Delta H_{(a\rightarrow d)}=-7084x J\Delta H_{(d\rightarrow c)}=-x\int_{495}^{505}{c_{p(s)}dT+(1-x) } \int_{495}^{505}{c_{p(l)}dT }
=x\left[18.5(505-495)+\frac{26}{2}\times 10^{-3} (505^2-495^2) \right]
+(1-x)\left[34.7(505-495)-\frac{9.2}{2}\times 10^{-3}(505^2-495^2) \right]
=301+14x
Thus,
-7084x=-14x-301which gives
x=\frac{301}{7070}=0.0426
The actual path the process follows is intermediate between paths 1 and 2; that is, the process of freezing and increase in temperature occur simultaneously.
The entropy produced by the spontaneous freezing is
\Delta S_{(a\rightarrow b)} + \Delta S_{(b\rightarrow c)}=34.7\times \ln \left\lgroup\frac{505}{495} \right\rgroup -\left[9.2\times 10^{-3}(505-495) \right]-\left\lgroup0.0426\times \frac{7070}{505} \right\rgroup
=0.602-0.596=0.006 J/K.mole
