Question

The expressions for the vapour pressure of solid (\( p_1 \)) and vapour pressure of liquid (\( p_2 \)) phases of a pure substance, respectively, are \[ \ln p_1 = \frac{-2000}{T} + 5 \quad {and} \quad \ln p_2 = \frac{-4000}{T} + 10 \] The triple point temperature of this substance is __________ K (in integer).

Hint:

At the triple point, the solid and liquid phases of a substance coexist in equilibrium with the vapor. Set \( \ln p_1 = \ln p_2 \) and solve for \( T \) to find the triple point temperature.

Answer 1

Correct Answer: 400

At the triple point, the vapour pressures of the solid and liquid phases are equal: \[ \ln p_1 = \ln p_2 \] Equating the two given expressions: \[ \frac{-2000}{T} + 5 = \frac{-4000}{T} + 10 \] \[ \Rightarrow \left(\frac{-2000}{T} + 5\right) - \left(\frac{-4000}{T} + 10\right) = 0 \Rightarrow \frac{2000}{T} - 5 = 0 \Rightarrow \frac{2000}{T} = 5 \Rightarrow T = \frac{2000}{5} = \boxed{400 { K}} \]