In the vicinity of the triple point, the equation of liquid-vapour boundary in the 𝑷 − 𝑻 phase diagram for ammonia is 𝐥𝐧 𝑷 = 𝟐𝟒. 𝟑𝟖 − 𝟑𝟎𝟔𝟑/𝑻, where P is pressure (in Pa) and 𝑻 is temperature (in K). Similarly, the solid-vapour boundary is given by 𝐥𝐧 𝑷 = 𝟐𝟕. 𝟗𝟐 − 𝟑𝟕𝟓𝟒/𝑻. The temperature at the triple point is ___________ K (round off to one decimal place).

The correct answer is 195.2 (or 195.19 K, rounding off to one decimal place as 195.2 K).

Understanding the Triple Point:
A triple point on a phase diagram is the unique temperature and pressure at which three phases of a substance (solid, liquid, and gas/vapour) coexist in thermodynamic equilibrium. Since all three phases coexist at this point, the temperature $T$ and pressure $P$ must be the same along the phase boundaries. Therefore, the liquid-vapour boundary and the solid-vapour boundary intersect at the triple point.

To find the temperature $T$ at the triple point, we equate the pressure $P$ (or $\ln P$) from both boundary equations.

Step-by-Step Derivation:

The equation for the liquid-vapour boundary is:
$\ln P=24.38-\frac{3063}{T}$

The equation for the solid-vapour boundary is:
$\ln P=27.92-\frac{3754}{T}$

Equating the two expressions for $\ln P$ at the triple point:
$24.38-\frac{3063}{T}=27.92-\frac{3754}{T}$

Rearranging the terms to group the temperature-dependent terms on one side:
$\frac{3754}{T}-\frac{3063}{T}=27.92-24.38$

Simplify both sides:
$\frac{3754-3063}{T}=3.54$

$\frac{691}{T}=3.54$

Solving for $T$:
$T=\frac{691}{3.54}$

$T\approx 195.1977\text{K}$

Rounding off to one decimal place yields 195.2 K (or keeping the high-precision value of 195.19 K as provided).