Currently, we are studying thermo physics in our IB Physics class, and have just concluded our unit on energetics in our IB Chemistry class. The amount of overlap between the two classes is so striking that I could not resist to explore the practical explorations in more detail and of course, using mathematics.
Latent heat is energy released or absorbed by a substance, during an isothermal (constant-temperature) process, usually a first-order phase transition. Common examples include the latent heat of vaporisation $L_v$ and the latent heat of fusion $L_f$.
Latent heat is in contrast to sensible heat, which is the quantity we have come to love in both chemistry and physics, $q_{sensible}=mc\Delta T$ where $c$ is the specific heat capacity.
By the First Law of Thermodynamics, the change in internal system of a system is the heat added to the system minus the work done by the system, $\Delta U=Q-W$ or in differential form, $dU=dQ-dW$. Of course, this is the fundamental law, so it cannot be "derived".
Moreover, $dQ=TdS$ where $S$ is the entropy of the system. If you're wondering why this is so (as you should be!), then I will refer you to this amazing answer on chemistry stack exchange. As a result, the first law of thermodynamics is often written as $\bbox[#ed8e93]{dU=TdS-PdV}$
Work is the area under the pressure-volume graph, so the work done by the gas can be defined as $W=\int_{V_1}^{V_2}PdV$. Assuming an isobaric process (constant-pressure), the area is simply the area of a rectangle, $W=PdV$.
Now, the enthalpy of a thermodynamic system is defined as the sum of its internal energy and the work required to achieve its pressure and volume, so $H=U+W$, or since $W=PdV$ assuming constant pressure, $H=U+Pv$. Differentiating this yields $dH=dU+d(PV)$, so we can substitute for $dU$ from the equation in pink above, and by the product rule on $d(PV)$, $dH=TdS-PdV+PdV+VdP$. The $PdV$ terms cancel, so $\bbox[#83d6eb]{dH=TdS+VdP}$.
As we have learnt in our IB HL Chemistry class, the Gibbs free energy is defined as $\Delta G=\Delta H-T\Delta S$, so differentiating yields $dG=dH-d(TS)$. We can substitute for $dH$ from the equation in blue above, and differentiate $d(TS)$ using the product rule like before, to get $\Delta G=TdS+VdP-TdS-SdT$. Again, the $TdS$ terms cancel, so $\bbox[yellow]{\Delta G=VdP-SdT}$.
We are looking for a state of equilibrium (thermal, chemical and mechanical) where pressure, temperature and the chemical potential of the two phases are identical, so looking at the equation in yellow above, Gibbs free energy $dG=0$ since both $dP$ and $dT$ are $0$. This means that $G_{liq}=G_{vap}$, or to reduce clutter, $G_1=G_2$.
Therefore, $$V_1dP-S_1dT=V_2dP-S_2dT$$ Rearranging, $$\bbox[#be96e3]{(S_2-S_1)dT=(V_2-V_1)dP}$$
Now, going back to the definition of Gibbs free energy as $\Delta G=\Delta H-T\Delta S$, since $G_{vap}=G_{liq}$, $$H^{vap}-TS^{vap}=H^{liq}-TS^{liq}$$ Rearranging, $$H^{vap}-H^{liq}=(S^{vap}-S^{liq})T$$ $$\frac{\Delta H^{vap}}{T}=\Delta S_{vap}$$
Now we can substitute for $\Delta S^{vap}$ from the equation in purple to get $\frac{\Delta H^{vap}}{T}=\frac{(V_2-V_1)dP}{dT}$.
Finally, rearranging the above yields the Clapeyron equation which states that $\frac{dP}{dT}=\frac{\Delta H^{vap}}{T\Delta v}$. One thing to note, however, is that quantitatively, although they have different meanings, the enthalpy of vaporisation is exactly equal to the latent heat of vaporisation. Read this fantastic article by TEC Science to see why.
This means that the above equation can also be written as $\bbox[#ff745e]{\frac{dP}{dT}=\frac{L_v}{T\Delta v}}$, which is a beautiful connection between the topics we have learnt in chemistry ($\Delta H$) and the topics we have learnt in physics ($L_v$).
When a substance is between the gas and condensed phase, the specific volume of the gas phase $v_g$ greatly exceeds that of the condensed phase $v_c$, $v_g>>v_c$. Therefore, $\Delta v=v_g-v_c=v_g(1-\frac{v_c}{v_g})≈v_g(1-0)≈v_g$.
From chemistry, we know that the mole is given by $n=\frac mM$. Also, the specific gas constant is defined as the ratio of the universal gas constant and molar mass, $R_{specific}=\frac RM$.
Therefore, substituting into the ideal gas law $pV=nRT$, $$PV=\frac{m}{M}RT$$ $$PV=mR_{specific}T$$ $$\frac Vm=v=\frac {R_{specific}T}{P}$$
We can then substitute this value into our Clapeyron equation to yield $$\frac{dP}{dT}=\frac{L_v}{T\Delta v}=\frac{L_v}{Tv_g}=\frac{PL}{T^2R}$$ This is called the Clausius-Clapeyron equation, which is valid only for low temperature and pressure.
Integrating both sides, assuming $L$ is constant, $$\int_{P_1}^{P_2}\frac{dP}{dT}=\frac LR \int_{T_1}^{T_2}\frac{dT}{T^2}$$ $$\left.\ln P\right|_{P_1}^{P_2}=-\frac LR \left.\frac 1T\right|_{T_1}^{T_2}$$ $$\ln P_2-\ln P_1=\ln \frac{P_2}{P_1}=-\frac LR(\frac {1}{T_2}-\frac{1}{T_1})$$
P.S. make sure to check out this article by MIT, which provides a cool derivation of the Clapeyron equation from the Carnot cycle.
The Clausius-Clapeyron relation can be interpreted as the slope of the tangent curves to the coexistence curve which separates two phases on a pressure-temperature (P-T) diagram.
The solid green line represents the behaviour of most substances, and the dotted green line represents the anomalous behaviour of water due to the extensive hydrogen bonding (another topic from IB SL Chemistry!)
Identifying the variables,
When you simply substitute in the values, you get that $T_2$, the boiling point of water at the top of Mt Everest, is equal to $344.12K≈71^°C$.
This is a straightforward substitution problem, $$\frac{dP}{dT}=\frac{L_f}{T\Delta v}$$ $$\frac{dP}{dT}=\frac{3.34*10^5J/kg}{273K*-9.05*10^{-5}m^3/kg}$$ $$\frac{\Delta P}{\Delta T}=-13.5MPa/K$$
Since the atmospheric water vapour dictates many meteorologic phenomena (notably precipitation), the Clausius-Clapeyron equation is a valuable tool for meteorologists.
$$\frac{de_s}{dT}=\frac{L_v(T)e_s}{R_vT^2}$$where $e_s$ is saturation vapour pressure and $R_v$ is the gas constant of water vapour.
However, when we derived the Clausius-Clapeyron equation from the Clapeyron equation, we made an assumption that the latent heat of vaporisation $L_v$ is constant. In meteorology, this assumption is significant, and so usually empirical formulae like the August-Roche-Magnus formula below are used. $$e_s(T)=6.1094e^{\frac{17.625T}{T+243.04}}$$ Note that here, $e_s$ is in $hPa$ and $T$ is in $°C$.
Many other formulae exist for the vapour pressure, including the Antoine equation, derived from the Clausius-Clapeyron equation, $\log_{10}{p}=A-\frac{B}{C+T}$. Others include the Tetens equation (identical structure as the August-Roche-Magnus equation but different coefficients), the Goff-Gratch equation, the Arden Buck equations, etc.
Since equations in meteorology can never be perfect, it is crucial that we have many equations to model the atmosphere for different conditions (it is said that meteorology is a chaotic system, and if you're wondering what that is, make sure to read @StevenOh's latest post!). For example, while the Goff-Gratch equation is an incredibly reliable, the Arden Buck equation is more accurate for temperatures in the range of -80 to 50°C. Theoretical equations like the Clausius-Clapeyron equation may be useful, but empirical formulae can often be more accurate. Still, the Clausius-Clapeyron equation serves the foundation for many of of the empirical formulae used in meteorology today.
The research paper here offers an excellent explanation of how the Clausius-Clapeyron equation can be used for the "different exotic clouds present in other planets" in our solar system. Definitely worth checking out.
It is incredibly fascinating to see how chemistry and physics are related in this single equation, all just derived from the First Law of Thermodynamics. I hope to be able to continue sharing such connections between my IB subjects.
Also, now that we are talking about thermodynamics, I just have to show this video by UC Santa Cruz Physics.
When I first saw it, I didn't know that a substance could coexist in all three states of matter (called the triple point, see the phase diagram above), so I was incredibly amazed. What about you? Comment your reaction down below!
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html http://home.iitk.ac.in/~suller/lectures/lec28.htm
