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Context

The temperature dependence on the enthalpy of vaporization can be found by evaluating the enthalpy before and after the transition. The definition of the specific heat capacity is

$\left(\frac{dH}{dT}\right)_{P} = c_{p}$,

so

$H(T) = H_{0}+c_{p}T$,

where $H_{0}$ is the enthalpy at zero Kelvin. The enthalpy change is therefore

$\Delta H(T) = \Delta H_{0}+ \Delta c_{p}T$,

where, for example, for water $\Delta c_{p} \simeq -40~{\rm J~mol^{-1}~K^{-1}}$. This suggests that the enthalpy of vaporization decreases with increasing temperature.

Problem

My problem is when I do the same calculation for the enthalpy of ionization I get that the enthalpy increases with increasing temperature because $\Delta c_{p}$ is positive (the specific heat of plasma is higher than that of gas). Conceptually this has baffled me because this suggests that increasing the temperature makes ionization harder.

Could someone explain to me what is going on? Please try to be detailed.

Thank you.

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1 Answer 1

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Conceptually this has baffled me because this suggests that increasing the temperature makes ionization harder.

Does it suggest that? The most stable phase (at constant temperature and pressure) is the one with the lowest Gibbs free energy $G=H-TS$, with the driving force increasing with decreasing $\Delta G=\Delta H-T\Delta S$:

With increasing temperature, the driving force increases as $-\frac{d\Delta G}{dT}=-\left(\frac{d\Delta H}{dT}\right)_P+\Delta S$. Thus, we can't say anything conclusively about the sign of $-\frac{d\Delta G}{dT}$ solely from the sign of $\frac{d\Delta H}{dT}$.

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