Clapeyron Equation (Example)

The Clapeyron equation can be used (for example) to determine how much the melting point of ice changes when the pressure is increased.

[@user-ec6mo4sj4z]

That's so informative and it solves my current problem, truly appreciate it.

[@minguyen6743]

i cannot thank you enough!!!

[@PhysicalChemistry]

On the contrary... you already have

[@wadhax5191]

Thank you soo much you explained things in a very clear and organized way, I have physical chemistry exam tomorrow I hope I make it 😅

[@PhysicalChemistry]

You're welcome. I'm glad to help. I hope the exam went smoothly

[@bigqx]

For T in the denominator, will this always be the temperature of the phase change? On one-hand it makes sense because the original equation involves the latent heat but say I wanted to compute the slope of the coexistence curve at a particular temperature, say 20 degrees C. Would I replace T with 293 K instead of having 273 K as you have in your example? It just seems strange that a curve like that would only have one value for the slope, which because it's non-linear curve, the slope should be the tangent line to the curve at any point. So for example, the slope of the coexistence curve at the liquid-gas boundary, at lower temperatures should have a smaller slope while near the critical point, the slope should be larger. Do we account for this by simply plugging the temperature at which we would like to compute the slope?

[@PhysicalChemistry]

You're right on all counts. The T in the denominator is the temperature at which you are interested in the phase transition. If you want the slope dP/dT at 20 °C, you would use Tᵩ = 293 K. You're also right that the P(T) coexistence curve is nonlinear. Some of that nonlinearity comes from the Tᵩ in the denominator. But some of it also comes from the fact that the ΔHᵩ and ΔVᵩ also depend on temperature. The main reason that the dP/dT slope gets steeper as you approach the critical point (as you correctly observe) is that the densities of the liquid and gas are becoming more similar as the fluid nears criticality, so ΔVᵩ is becoming small. Great questions!

[@bigqx]

@@PhysicalChemistry Thank you so much! This really is a great channel. Very clear derivations and examples.

[@PhysicalChemistry]

@@bigqx Thanks, I appreciate that

[@abimbolaadisa5401]

Merci

[@PhysicalChemistry]

De rien

[@user-or1ed9on2w]

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