Question

Write down the Clapeyron's equation.
How does it explain the effect of
pressure on the melting point of
solids? 5 marks
​

Answer 1

Explanation:

The equilibrium between water and water vapor depends upon the temperature of the system. If the temperature increases the saturation pressure of the water vapor increases. The rate of increase in vapor pressure per unit increase in temperature is given by the Clausius-Clapeyron equation. Let p be the saturation vapor pressure and T the temperature. The Clausius-Clapeyron equation for the equilibrium between liquid and vapor is then

dp/dT = L/(T(Vv-Vl))

where L is the latent heat of evaporation, and Vv and Vl are the specific volumes at temperature T of the vapor and liquid phases, respectively.

More generally the Clausius-Clapeyron equation pertains to the relationship between the pressure and temperature for conditions of equilibrium between two phases. The two phases could be vapor and solid for sublimation or solid and liquid for melting.

The material below is an examination of the application of the Clausius-Clapeyron equation to meteorology where the most relevant systems are water and water vapor and salt water and water vapor. But the equation also applies to ice and water vapor and ice and water. Finally there is a derivation of the Clausius-Clapeyron equation from thermodynamic principles.

Note that Vv is much greater than Vl so that to a good approximation

dp/dT = L/(TVv)

Furthermore the ideal gas equation applies to the vapor; i.e.,

pVv = RT

and hence

Vv = RT/p

where R is the universal gas constant.

Thus

dp/dT = L/(RT²/p)

or, equivalently

(1/p)(dp/dT) = L/(RT²)

In differential form this is

dp/p = (L/R)(dT/T²)

or, equivalently

d(ln(p)) = (L/R)d(-1/T)

If L is independent of temperature then the solution to the differential equation is

ln(p) = c0 - (L/RT)

or, equivalently

p = c1exp(-L/RT)

where c1 is a constant.

The shape of this function is given below:

The empirical curves have the opposite curvature but clearly the equation dp/dT = L/(RT²/p) = Lp/T² indicates that as temperature increases the slope (dp/dT) decreases. Of course, the empirical curves take into account that the latent heat L can change with temperature, but the change in L with T is relative small. There is a fundamental discrepancy between the Clausius-Clapeyron Equation and the empirical relationships and the published derivations of the equation blithely show an empirical curve in which dp/dT increases with T and an equation in which dp/dT is inversely proportional to T. However this discrepancy will be left unresolved here and the implications of the Clausius-Clapeyron equation will will be further explored.

For some purposes the density of the vapor is of more interest than the pressure. By the ideal gas equation, the molecular density D is given by

D(T) = 1/Vv = p/RT

which for the above

pressure-temperature

relationship is

D(T) = c1exp(-L/RT)/RT

The shape of this latter relationship for some arbitrary values of the parameters is shown below:

It is a surprise that there would be a case such that as the temperature goes up the density of decreases. This needs to be checked for generality.

The ideal gas equation indicates that density would go down as temperature increases if pressure remained constant. Even if pressure increases with temperature the density will decrease if the increase in pressure is not enough to offset the direct effect of the temperature increase on density. In differential terms the ideal gas equation is

dD/D = dp/p - dT/T

The Clausius Clapeyron equation, when the specific volume of the liquid is assumed to be zero, gives

dp/p = (LD/p)(dT/T) = (L/RT)(dT/T)

so

dD/D = [L/RT − 1](dT/T)

Thus the effect of a temperature increase on molecular density depends upon the magnitude of the latent heat of vaporization compared to RT.

The latent heat of vaporization for water is 2.257×106 J/kg. The gas constant for water vapor in SI units is 461.5 J/(kg K) so at T=300 K, RT=1.3845×105 J/kg. Thus L/RT=16.3 and hence dD/D=15.3(dT/T) so if (dT/T)=1/300 then dD/D = 15.3/300=.051.

Answer 2

Answer:

The shape of the Clapeyron equation most customarily used is

dP/dT = ΔS/ΔV

Explanation

This equation states that the slope (rise/run) of a univariant equilibrium plotted on a P-T diagram is the same as the entropy change (ΔS) of the response divided through the extent change (ΔV) of the response.

Using the Clapeyron equation, most effective the slope of the equilibrium is decided, now no longer the real function of the response in stress-temperature space. The function should be decided through in addition thermodynamic calculations or experiments.

The Clapeyron equation has many capacity uses. Some examples are:

• We can decide the slope of a metamorphic response from thermodynamic statistics to decide if it may be a capacity geothermometer or geobarometer. A response with a shallow dP/dT slope is greater touchy to stress adjustments and will be a geobarometer. A response with a steep (almost vertical) slope is touchy to temperature and will be a geothermometer.
• If we've got experimental outcomes on a few responses at 1 temperature (or stress), we will calculate the slope and extrapolate to different situations in place of doing greater time-ingesting experiments.
• The Clapeyron equation enables us to decide thermodynamic values for reactions or stages. When blended with extent statistics, we will use the slope of an experimentally-decided response to calculate the Δ S of the response and to calculate the entropy of formation (ΔSf) of a selected phase. Often the volumes of stages are very well-known, however, the entropy statistics might also additionally have massive uncertainties.
• If we carry out a Schreinemakers evaluation of an invariant point, we will use the Clapeyron equation to put reactions efficiently approximately the point.

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