91Ó°ÊÓ

We have been given that

$T=298k$

The${∆}_{f}G$values indicate that at room temperature and atmospheric pressure, kyanite has the lowest Gibbs free energy of the three phases. Furthermore, it also has the lowest molar volume of the three; this means that increasing the pressure causes its $G$value to increases slowest, it will never intersect the $G$values of the other phases.

We have been given

$∆G$is the difference in Gibbs energy

$T$is dependence of $∆G$

$S$is the function of heat capacity

For either phase,${\left(\raisebox{1ex}{$∂G$}\!\left/ \!\raisebox{-1ex}{$∂T$}\right.\right)}_P=-S$. Integrating this relation from $T_{1}to{T}_2$gives

$∆G\left(T_2\right)=∆G\left(T_1\right)-{∫}_{T_1}^{T_2}∆S\left(T\right)dT$

If we write this equation separately for two different phases and then subtract one equation from the other, the $G\text{'}s$become $∆G\text{'}s$and the $S$become $∆S$, so we obtain the desired result,

$∆G\left(T_2\right)=∆G\left(T_1\right)-{∫}_{T_1}^{T_2}∆S\left(T\right)dT$

We have been given

$∆S$to be independent of $T$

Our thermodynamic data is at $298k$so we'll use this value for $T_1$,$∆G$is the difference in Gibbs energy

Taking $∆S$to be independent, of $T,$we can pull it outside the integral to obtain

$∆G\left(T_2\right)=∆G\left(T_1\right)-(T_2-T_1).∆S$

If we take $T_2$to be the temperature at which the phase transition occurs (the two phases are in equilibrium), then $∆G\left(T_2\right)=0$Solving for $T_2$then gives simply

$T_2=T_1+\frac{∆G\left(T_1\right)}{∆S}$

Our thermodynamic data is at $298k$, so we'll use this value for $T_1$, For the kyanite-andalusite transition, localid="1650353248124" $∆G\left(T_1\right)=1.22kj$(for one mole of material), while $∆S=9.41j/k$.

Therefore, the temperature at which andalusite becomes more stable than kyanite should be approximately $T_2=298K+\frac{1.22kJ}{9.41J/k}=428K$

Similarly, for the kyanite-sillimanite transition,

$T_2=298k+\frac{2.89kJ}{12.30J/k}=533K$

And for the andalusite-sillimanite transition,

$T_2=298K+\frac{1.67kJ}{2.89J/k}=876K$So, at atmospheric pressure, kyanite should be stable up to $428k$, andalusite should be stable from $428k$up to $876k$and sillimanite should be stable above$876k$.

We have been given,

$∆S=cons\tan t$

To make a signifying improved calculation, However, we would really need heat capacity data over the whole temperature range.

The change in entropy as the temperature is increased is given by

$S\left(T_2\right)=S\left(T_1\right)+{∫}_{T_1}^{T_2}\frac{C_P}{T}dT$

If we write this equation for two different phases and then subtract, we obtain

$∆S\left(T_2\right)=∆S\left(T_1\right)+{∫}_{T_1}^{T_2}\frac{∆C_P}{T}dT$

Suppose, for the sake of a rough estimate that $∆C_P$is independent of temperature

Then, $∆S\left(T_2\right)=∆S\left(T_1\right)+∆C_p.\ln \left(\raisebox{1ex}{$T_2$}\!\left/ \!\raisebox{-1ex}{$T_1$}\right.\right)$

For $kyanite→andalusite$, $∆C_P=1.01J/k$(at room temperature). Taking $T_1=298K$and $T_2=428k$, I find that the final term in this equation is $0.37J/k$,Compared to $∆S\left(T_1\right)=9.41J/k$

So, over this range, $∆S$is reasonably independent of temperature. However for $andalusite→sillimanite$, $∆C_P=1.80J/k$and we should take $T_2=876K$with these number I get a correction term of $1.94J/k$, compared to $∆S\left(T_1\right)=289J/k$, With $∆S$varying by nearly $70\%$over the temperature range of interest I conclude that the calculated temperature of the transition could be off by as much as $100-200K$. To make a significantly improved calculation however we would really need heat capacity data over the whole temperature range.