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Why does the triple point temperature have very similar values to the freezing point, in most substances?

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This question is ambiguous. Originally, I thought the question was why different material have similar triple point temperatures and similar freezing temperatures. After the last edit by @ChickenP, it sounds like it is a comparison between triple point and freezing point for a given material. Please specify what is your question. – fffred May 26 '13 at 21:40
Why are the triple point temperature and normal freezing point very close in temperature for most substance? this means for most given substance, why are they close? – el psy Congroo May 26 '13 at 21:58
@JamesNgaiChunTat do you mean the freezing point at $1\: \mathrm{atm}$ of pressure compared to the triple point temperature? If that is your question then you need to first provide the data that shows it's actually true. What does "close" mean and what materials have this property? Without clarification and data the question can't be answered. – Brandon Enright May 27 '13 at 1:51
It is, of course, true that the 1 atm freezing point for pure water is without 0.01 Kelvin of the triple point but you have not provided any evidence that this is a common situation. – dmckee Mar 13 '15 at 1:00

From the Clausius Clapeyron equation,

$$ \frac{dP}{dT}_{i \rightarrow j} = \frac{\ell_{i \rightarrow j}}{T(v^{j}-v^{i})} $$

Where $ \ell_{i \rightarrow j} $ is the latent heat of going from phase $i$ to phase $j$, $\, v$ is the specific volume, and all other variables have their common meaning.

Taking for example water, this equation gives the change in temperature as the pressure is increased from the low value of the triple point ($P_{tp}=610 \, Pa$) to atmospheric pressure, where the normal melting point of water is defined ($P=1.01 \times 10^{5} \, Pa$).

Notice that the specific volume is the reciprocal of the density. The density for liquid water is $1000 \, kg\,m^{-3}$, and the density of ice is $916 \, kg\,m^{-3}$. The latent heat is $3.34 \times 10^{5} J\,kg^{-1}$.

Now, assuming the change of temperature is small, which is the case,

$$ \Delta T = \frac{T(v^{j}-v^{i})}{\ell_{i \rightarrow j}} \Delta P $$

$$ = \frac{273 (1.00 \times 10^{-3} - 1.09 \times 10^{-3})(1.01\times10^{5})}{3.34\times 10^{5}} = -0.0074\, K$$

It helps to give you an idea of why the change in temperature is so small. (The previous example was borrowed from Ashley Carter's Classical and statistical thermodyanmics).

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