# Clausius Clapeyron Equation

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## Clausius Clapeyron Equation Thermodynamics

The Clausius Clapeyron equation is a way of describing a discontinuous phase transformation between two phases of matter of a single constituent, named after Rudolf Clausius and Benoît Paul Émile Clapeyron. A straight line does not exist between a liquid's temperature and its vapour pressure.

The vapour pressure in the water, for example, rises at a much faster rate than the system's temperature. The Clausius equation will describe this action. The temperature of a system determines the balance between a liquid and its vapour; an increase in temperature induces a subsequent rise in the vapour pressure of the liquid.

The clausius clapeyron equation thermodynamics is as follow,

ln($\frac{P_{2}}{P_{1}}$) = $\frac{\Delta H_{vap}}{R}$ ($\frac{1}{T_{1}}$ - $\frac{1}{T_{2}}$)

To determine the ranges of hydrate stability, the Clausius Clapeyron equation can be applied to a hydrating system and used to estimate the equilibrium water behaviour for a hydrate pair occurring in equilibrium at various temperatures.

For hydrate systems, the Clausius Clapeyron equation was obtained by assuming that the higher and lower hydrates occur in equilibrium as three pure phases of water vapour and that the average volume transition involved in this equilibrium process is approximated by the volume of the emitted water vapour which behaves ideally.

A thermogravimetric analyzer with an attached water vapour distribution system was used to dynamically assess the equilibrium water vapour pressure for the nedocromil sodium monohydrate and trihydrate systems at various temperatures.

The enthalpy of dehydration obtained by applying the Clausius Clapeyron equation to experimentally defined equilibrium water vapour pressures agrees very well with the enthalpy of dehydration resulting from differential scanning calorimetry (13.7 +/- 0.6 kcal/mol of water loss, n = 5), meaning that the Clausius Clapeyron equation can be applied to organic hydrate structures.

### The Clausius Clapeyron Equation Derivation

The Clausius Clapeyron equation predicts the rate at which vapour pressure increases per unit increase in temperature for a substance's vapour pressure (P) and temperature (T).

$\frac{\text{d ln p}}{dT}$ = $\frac{\Delta H_{vap}}{RT^{2}}$

The molar enthalpy of vaporisation of the liquid, the ideal gas constant, and the temperature of the system determines the rate at which the natural logarithm of the vapour pressure of a liquid varies with temperature, according to this equation.

If H$_{vap}$ is assumed to be independent of the system's temperature, the Clausius Clapeyron equation can be written in the integrated form below, where C is a constant.

Where Hvap is the liquid's enthalpy of vaporisation, R is the gas constant, and A is a constant whose value is determined by the substance's chemical identity.

In this equation, the temperature (T) must be in kelvin. Since the relationship between vapour pressure and temperature is not linear, the equation is often rewritten in the logarithmic form to provide the following linear equation:

ln(P) = - $\frac{\Delta H_{vap}}{RT}$ + ln A

If the enthalpy of vaporisation and vapour pressure at a given temperature is defined for every liquid, the Clausius Clapeyron equation can be used to calculate the vapour pressure at a different temperature.

The linear equation can be formulated in a two-point format to accomplish this. If the vapour pressure at temperature T$_{1}$ is P$_{1}$ and the vapour pressure at temperature T$_{2}$ is P$_{2}$ , the corresponding linear equations are:

ln(P$_{1}$) =  - $\frac{\Delta H_{vap}}{RT_{1}}$ + ln A

And

ln(P$_{2}$) =  - $\frac{\Delta H_{vap}}{RT_{2}}$ + ln A

Since the constant, A, is the same in both equations, they can be rearranged to separate ln A and then made equal:

ln(P$_{1}$) =  - $\frac{\Delta H_{vap}}{RT_{1}}$ + ln A = ln(P$_{2}$) =  - $\frac{\Delta H_{vap}}{RT_{2}}$ + ln A

ln(P$_{1}$) =  - $\frac{\Delta H_{vap}}{RT_{1}}$ = ln(P$_{2}$) = - $\frac{\Delta H_{vap}}{RT_{2}}$

which can be combined into:

ln ($\frac{P_{2}}{P_{1}}$) = $\frac{\Delta H_{vap}}{R}$ ($\frac{1}{T_{1}}$ - $\frac{1}{T_{2}}$)

Hence it is the Clausius clapeyron equation.

### Potential Uses of Clausius Clapeyron Equation

The Clausius clapeyron equation has many potential uses. Some examples are:

• We can use thermodynamic data to calculate the slope of a metamorphic reaction to see whether it could be used as a geothermometer or geobarometer. A geobarometer could be a reaction with a shallow dP/dT slope because it is more sensitive to pressure changes. A reaction with a steep (nearly vertical) slope is vulnerable to temperature and maybe a geothermometer.

• We can quantify the slope and extrapolate to other conditions instead of conducting more time-consuming tests if we have experimental effects on a reaction at one temperature (or pressure).

• The Clausius equation is used to calculate the thermodynamic values of reactions and phases. The slope of an experimentally defined reaction can be used to measure the S of the reaction and the entropy of formation (Sf) of a given process when combined with volume results. The amounts of phases are often well-known, but the entropy data can be subject to significant uncertainty.

• We can use the Clausius equation to accurately place reactions around an invariant point if we do a Schreinemakers analysis.

### Clausius Clapeyron Equation Derivation in Thermodynamics

The temperature of the system affects the balance between water and water vapour.

The saturation pressure of water vapour rises as the temperature rises. The Clausius Clapeyron equation calculates the rate of increase in vapour pressure per unit increase in temperature. Let T be the temperature and p be the saturation vapour pressure. The Clausius Clapeyron equation for liquid-vapour equilibrium is then used.

$\frac{dp}{dT}$ = $\frac{L}{(T(V_{v} - V_{l}))}$

where L is the latent heat of evaporation and V$_{v}$ and $_{l}$ are the vapour and liquid phases' real volumes at temperature T, respectively.

The Clausius Clapeyron equation, in a larger perspective, describes the relationship between pressure and temperature in two-phase equilibrium.

For sublimation, the two phases could be vapour and solid, or solid and liquid for melting.

Finally, the Clausius Clapeyron equation is derived using thermodynamic principles.

Note that V$_{v}$ is very much greater than V$_{l}$ so that is a good approximation.

$\frac{dp}{dT}$ = $\frac{L}{TV_{v}}$

Furthermore, the ideal gas equation is applied to the vapour, i.e,

V$_{v}$ = $\frac{RT}{p}$

where R is the universal gas constant.

Thus,

($\frac{1}{p}$) ($\frac{dp}{dT}$) = $\frac{L}{RT^{2}}$

In differential form this is,

d(ln(p)) = $\frac{L}{R}$ d$\frac{-1}{T}$

If L is independent of temperature then the solution of the above differential equation is,

P = c$_{1}$ exp (- $\frac{L}{RT}$)

where c$_{1}$ is a constant.

Q.1) Who Discovered the Clausius Clapeyron Equation ?

Answer: The Clausius-Clapeyron equation is named after Rudolf Clausius, a 19th-century German physicist, and Émile Clapeyron, a 19th-century French engineer.

This relationship suggests that an improvement in the ability of air to retain water vapour is a result of the rising temperature of the amount of air, assuming there is a liquid water surface in equilibrium with the atmosphere.

Q.2) What is the Unit of Clausius Clapeyron?

1. The natural log term on the left-hand side (LHS) is unitless. It makes no difference what pressure units you use; the only requirement is that P₁ and P₂ must be represented in the same pressure unit.

2. The unit on the temperature term will be K⁻¹. The unit on the R is J mol⁻¹ K⁻¹.

3. The K⁻¹ in the temperature will cancel with the K⁻¹ associated with R.

4. Hence it means that the unit on ΔH must be J/mol.