Topic: Thermodynamics
Vapor pressure in mm of mercury of solid and liquid ammonia is given by $\ln P = 23.1 - (3749/T)$ and $\ln P = 19.6 - (3049/T)$, find out the triple point.
- A. 700 K
- B. 231 K
- C. 200 K
- D. 180 K
Answer: C. 200 K
Explanation: At the triple point, the vapor pressures of the solid and liquid phases are equal. Equating the two expressions: $23 Jan 3749/T = 19 Jun 3049/T$. Rearranging gives $23.1 - 19.6 = 3749/T - 3049/T$, so $3.5 = 700/T$. Solving for $T$ yields $T = 700 / 3.5 = 200$ K.