Question

An element transforms from \( \alpha \) to \( \beta \) at 773 K and 1 atm pressure with 912 J mol\(^{-1}\) as enthalpy of transformation. The molar volumes of \( \alpha \) and \( \beta \) phases are 7.377 cm³ and 7.317 cm³, respectively. Assume that the difference in molar volumes of \( \alpha \) and \( \beta \) is independent of pressure. The pressure (in atm) required for \( \alpha \) to \( \beta \) transformation to occur at 723 K is \(\underline{\hspace{2cm}}\) (round off to nearest integer).

Hint:

Use the equation \( \Delta H = V \Delta P \) to calculate the pressure required for phase transformation.

Answer 1

Correct Answer: 9600

The pressure required for the phase transformation can be calculated using the following equation:
\[ \Delta H = V \Delta P \] Where: - \( \Delta H = 912 \, \text{J/mol} \), - \( \Delta V = 7.377 - 7.317 = 0.06 \, \text{cm}^3/\text{mol} \). First, we convert \( \Delta V \) to \( \text{m}^3/\text{mol} \):
\[ \Delta V = 0.06 \times 10^{-6} \, \text{m}^3/\text{mol} \] Now, using \( \Delta H = V \Delta P \), the pressure required is:
\[ \Delta P = \frac{\Delta H}{\Delta V} = \frac{912}{0.06 \times 10^{-6}} = 9.6 \times 10^3 \, \text{atm}. \] Thus, the pressure is approximately \( 9600 \, \text{atm} \).