Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in Problem 5.28; use this information to sketch the phase diagram of calcium carbonate.

The slope of the phase boundary at T= 298K and P=3.597kBar is as follows:

$\frac{dP}{dT}=\frac{S_c-S_a}{V_c-V_a}\dots \dots \left(1\right)$

Where,

P is pressure

T is the temperature

V is the volume

S_c is the entropy of calcite

S_a is the entropy of aragonite

V_cis the volume of calcite

V_a is the volume of aragonite

From the book, we know that

$S_c=92.9\mathrm{J}/\mathrm{K},S_a=88.7\mathrm{J}/\mathrm{K},V_a=34.15{\mathrm{cm}}^3,V_c=36.93{\mathrm{cm}}^3$

Substituting these values in equation 1

role="math" localid="1646902780078" $$\begin{gathered} \frac{dP}{dT}=\frac{92.9\mathrm{J}/\mathrm{K}-88.7\mathrm{J}/\mathrm{K}}{36.93{\mathrm{cm}}^3{\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}^3-34.15{\mathrm{cm}}^3{\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}^3} \\ =1.51\times {10}^6\mathrm{Pa}/\mathrm{K}\left(\frac{{10}^{-5}\mathrm{bar}/\mathrm{K}}{1\mathrm{Pa}/\mathrm{K}}\right) \\ =15.1\mathrm{bar}/\mathrm{K} \end{gathered}$$

Imagine the phase boundary is a straight line at T = 298K. The value of pressure at temperature T= 298K is P=3.6x 10³ bar, which is well-known.

As a result, at critical temperature, the pressure differential can be represented as;

$P\left(T\right)-P(298\mathrm{K})=\frac{dP}{dT}(T-298\mathrm{K})$

P(T) is the pressure at any given temperature T

Therefore, substituting the values

role="math" localid="1646903188657" $$\begin{gathered} P\left(T\right)=P\left(298\right)+\frac{dP}{dT}(T-298) \\ P\left(T\right)=3.597\times {10}^3\mathrm{bar}+15.1\mathrm{bar}/\mathrm{K}(T-298) \end{gathered}$$

So that the relationship between pressure and temperature is not linear, there must be a point when the change in pressure is zero, such that

$$\begin{gathered} 0=3.6\times {10}^3\text{bar}+15.1\text{bar}/\mathrm{K}(T-298) \\ (T-298\mathrm{K})=-\frac{3.6\times {10}^3\mathrm{bar}}{15.1\mathrm{bar}/\mathrm{K}} \\ T=298\mathrm{K}-\frac{3.6\times {10}^3\mathrm{bar}}{15.1\mathrm{bar}/\mathrm{K}} \\ T=59.6K \end{gathered}$$

The graph will not be linear at temperature T= 59.6K

To build the graph, first get the different pressure values at various temperatures, then use the pressure and temperature values to draw the pressure vs temperature graph for the calcium carbonate phase diagram.

The calcium carbonate phase diagram would be linear up to temperature 58.2 SK, but as the temperature decreased, the pressure would not fall linearly, resulting in a non linear graph.