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Thermal Physics
Clausius-Clapeyron Equation
The Clausius-Clapeyron equation describes the phase equilibrium curve for a pure substance undergoing a first-order phase transition. It relates the rate of change of vapour pressure with temperature to the latent heat and specific volume change.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Qualitative understanding of phase transitions, effect of pressure on boiling and melting points, and simple identification of latent heat terms.
PGT extension
Rigorous thermodynamic derivation using Maxwell's relations, integration for liquid-vapour and solid-liquid transitions, and volume approximations for gases.
Prerequisites
Concept of Temperature
First and Second Laws of Thermodynamics
Maxwell's Thermodynamic Relations
Thermodynamic Foundation of Phase Equilibrium
Along a coexistence curve separating two phases (e.g., liquid and vapour), the chemical potentials of both phases are equal: mu_1(T, P) = mu_2(T, P).
A first-order phase transition is characterized by discontinuities in first-order derivatives of the Gibbs free energy, specifically volume and entropy.
The slope of the coexistence curve dP/dT represents the rate of change of equilibrium pressure with temperature for the two phases.
Physical Interpretation of the Slope (dP/dT)
For a vaporization process, the latent heat L is positive and the specific volume of the gas v_g is much greater than that of the liquid v_l, making dP/dT > 0.
For melting (fusion) in most substances, volume increases upon melting, so dP/dT > 0; however, for anomalous substances like water, ice contracts upon melting (v_l < v_s), resulting in a negative slope (dP/dT < 0).
An increase in external pressure lowers the melting point of ice due to the negative slope of the solid-liquid coexistence curve.
Clapeyron vs. Clausius-Clapeyron Form
The exact Clapeyron equation accounts for the specific volume change delta $v = v_2 - v_1 without approximations$.
The Clausius-Clapeyron equation introduces approximations valid for vaporization/sublimation: neglecting the specific volume of the condensed phase and treating the vapour phase as an ideal gas.
The integrated form of the Clausius-Clapeyron equation allows calculation of vapour pressure at an arbitrary temperature given the vapour pressure at a reference temperature and the latent heat.
Limitations and Scope of Application
The standard integrated form assumes the latent heat L is independent of temperature over the range of integration.
The ideal gas approximation fails near the critical point where the density of liquid and vapour become identical and delta v approaches zero.
It applies strictly to pure single-component systems in thermodynamic equilibrium.
Derivations
Derivation of the Clapeyron Equation from Maxwell Relations
Conditions: Pure substance in two-phase equilibrium, constant temperature and pressure.
Start with the Maxwell relation derived from Gibbs free energy: ( dP / dT )_$v = ( dS / dV )_T = L / (T \cdot \Delta v)$, where L is the latent heat (L = T \Delta S).
Substitute the change in entropy during the phase transition as Delta $S = L / T$, where L is the latent heat absorbed per mole or per unit mass.
Express the change in volume as Delta $v = v_2 - v_1$, where v_2 and v_1 are the specific volumes of the two phases respectively.
Result: \frac{dP}{dT} = \frac{L}{T(v_2 - v_1)}
Derivation of the Clausius-Clapeyron Equation for Liquid-Vapour Equilibrium
Conditions: Vapour behaves as an ideal gas, and specific volume of liquid is negligible compared to vapour (v_g >> v_l).
Take the exact Clapeyron equation: dP/d$T = L / (T (v_g - v_l))$.
Apply approximations: v_g - v_l \approx v_g and use the ideal gas law for one mole of vapour: $v_g = RT / P$.
Substitute v_g into the equation to get dP/d$T = L P / (R T^2)$, and separate the variables to obtain dP/$P = (L / R T^2) dT$.
Result: \frac{d( \ln P )}{dT} = \frac{L}{R T^2}
Formula sheet
Formula / law
Equation
Variables and units
Conditions
General Clapeyron Equation
$$\frac{dP}{dT} = \frac{L}{T \Delta v}$$
P = pressure, T = absolute temperature, L = latent heat, Delta v = change in specific volume Pa/K or atm/K
Draw: Pressure (P) on y-axis versus Temperature (T) on x-axis, showing sublimation, fusion, and vaporization curves meeting at the triple point.
Interpret: The slopes (dP/dT) of each curve correspond directly to the values given by the Clapeyron equation for the respective phase boundaries.
Exam use: Directly tested in TGT/PGT for identifying why the fusion curve of water has a negative slope while most substances have positive slopes.
ln P versus 1/T Linear Plot
Draw: Natural logarithm of vapour pressure (ln P) on y-axis plotted against inverse temperature (1/T) on x-axis.
Interpret: A straight line whose slope is equal to -L_v / R. This allows experimental determination of the latent heat of vaporization.
Exam use: Tested numerically and conceptually to find latent heat from graphical data.
Volume Discontinuity across First-Order Transition
Draw: Specific volume v versus temperature T at constant pressure, showing a sharp jump (delta v) at the phase transition temperature.
Interpret: Highlights that first-order transitions feature a discontinuous change in volume and entropy.
Exam use: Conceptual distinction between first-order and second-order phase transitions.
Solved numericals
TGT core
If the specific volume of water changes from 1.00 x 10^-3 m^3/kg to 1.671 m^3/kg when it vaporizes at 100 °C under 1 atm pressure, and the latent heat of vaporization is 2.26 x 10^6 J/kg, find the rate of change of vapour pressure with temperature (dP/dT).
Identify given values: T = 373.15 K, $L = 2.26 x 10^6 J/kg$, $v_g = 1.671 m^3/kg$, $v_l = 0.001 m^3/kg$.
Apply Clapeyron equation: dP/d$T = L / (T * \delta v) = (2.26 x 10^6) / (373.15 * 1.670)$.
Compute the numerical value: dP/d$T = 3626 Pa/K$.
Answer: 3626 Pa/K
PGT extension
The vapour pressure of a liquid doubles when its temperature is raised from 300 K to 320 K. Calculate the molar latent heat of vaporization of the liquid. (Assume $R = 8.314 J/mol K)$
Use the integrated Clausius-Clapeyron equation: ln(P_2 / $P_1) = (L_v / R) * [(T_2 - T_1) / (T_1 T_2)]$.
Calculate final value: $L_v = 27725 J/mol = 27.73 kJ/mol$.
Answer: 27.73 kJ/mol
PGT extension
At 0 °C, the latent heat of fusion of ice is 3.34 x 10^5 J/kg. The specific volume of ice is 1.09 x 10^-3 m^3/kg and of water is 1.00 x 10^-3 m^3/kg. Find the change in melting point of ice under an applied pressure increase of 10 atmospheres (1.013 x 10^6 Pa).
Use Clapeyron equation for fusion: dP/d$T = L_f / [T * (v_l - v_s)]$.
Substitute values: $L_f = 3.34 x 10^5 J/kg$, T = 273.15 K, $v_l = 1.00 x 10^-3$, $v_s = 1.09 x 10^-3$.
Calculate Delta $v = v_l - v_s = 1.00 x 10^-3 - 1.09 x 10^-3 = -0.09 x 10^-3 m^3/kg$.
Calculate dP/d$T = (3.34 x 10^5) / [273.15 * (-0.09 x 10^-3)] = -1.359 x 10^7 Pa/K$.
Calculate change in temperature: delta $T = (1 / (dP/dT)) * \delta P = (1 / (-1.359 x 10^7)) * (1.013 x 10^6) = -0.0745 K$.
Answer: -0.0745 K
Assumptions and limitations
The vapour phase is assumed to behave as an ideal gas, which fails at high pressures and near the critical point.
The specific volume of the condensed phase (liquid or solid) is neglected compared to the vapour phase, which becomes invalid near the critical temperature.
The latent heat L is treated as constant over the small temperature interval of integration, though it varies slightly with temperature in reality.
Common misconceptions
Common error: An increase in pressure always raises the melting point of a substance.
Correct: An increase in pressure lowers the melting point of substances like water and bismuth that contract upon melting.
Because for these substances v_l < v_s, making the Clapeyron slope negative (dP/dT < 0).
Common error: The Clausius-Clapeyron equation is exact for all temperature ranges.
Correct: It involves major approximations such as ideal gas behaviour for the vapour and neglecting liquid volume.
The exact relation is the Clapeyron equation without any approximations.
Common error: Latent heat is independent of temperature during phase change integration.
Correct: Latent heat is a function of temperature; assuming it constant is only valid for narrow temperature intervals.
Heat capacities of the two phases differ, leading to a temperature dependence of latent heat via Kirchhoff's relation.
Quick revision
The Clapeyron equation governs all first-order phase transitions via dP/d$T = L / (T \delta v)$.
General Clapeyron Equation: \frac{dP}{dT} = \frac{L}{T \Delta v}
Along a phase coexistence curve, the chemical potentials of both phases are equal: $\mu_1(T, P) = \mu_2(T, P)$.
The exact Clapeyron equation is given by $\frac{dP}{dT} = \frac{L}{T \Delta v}$.
For vaporization, $\frac{dP}{dT} > 0$ because latent heat $L$ is positive and $v_g \gg v_l$.
For anomalous substances like water, ice contracts upon melting ($v_l < v_s$), leading to a negative slope ($\frac{dP}{dT} < 0$).
The integrated Clausius-Clapeyron equation assumes constant latent heat and that the vapour behaves as an ideal gas.
Glossary
Chemical Potential ($\mu$)Gibbs free energy per mole; equal for two phases in equilibrium along a coexistence curve.
Latent Heat ($L$)Heat absorbed or released during a first-order phase transition at constant temperature and pressure.
Coexistence CurveA curve on the $P-T$ phase diagram separating two phases in thermodynamic equilibrium.
First-Order Phase TransitionA phase transition characterized by discontinuities in first-order derivatives of Gibbs free energy, such as volume and entropy.
Specific Volume ($v$)Volume per unit mass or per mole of a substance.
Clapeyron vs. Clausius-Clapeyron Equation
Feature
Clapeyron Equation
Clausius-Clapeyron Equation
Approximations
None; exact for all first-order transitions
Neglects condensed phase volume and treats vapour as ideal gas
Applicability
All phase transitions (solid-liquid, liquid-gas, solid-gas)
Vaporization and sublimation transitions
Mathematical Form
$\frac{dP}{dT} = \frac{L}{T \Delta v}$
$\frac{dP}{dT} \approx \frac{P L_v}{R T^2}$
Normal vs. Anomalous Melting (Water)
Property
Normal Substances
Water (Ice)
Volume Change on Melting ($\Delta v$)
Positive ($v_l > v_s$)
Negative ($v_l < v_s$)
Slope of Melting Curve ($\frac{dP}{dT}$)
Positive ($> 0$)
Negative ($< 0$)
Effect of Increased Pressure
Raises melting point
Lowers melting point
Practice Quiz
Choose an option to reveal the explanation.
1. Along a coexistence curve separating two phases, which thermodynamic quantity must be equal for both phases?
Explanation: Along a coexistence curve separating two phases, the chemical potentials of both phases are equal: $\mu_1(T, P) = \mu_2(T, P)$.
2. What is the exact form of the Clapeyron equation for a first-order phase transition?
Explanation: The exact Clapeyron equation is given by $\frac{dP}{dT} = \frac{L}{T \Delta v}$, where $L$ is the latent heat and $\Delta v$ is the change in specific volume.
3. For a vaporization process, why is the slope of the coexistence curve $\frac{dP}{dT}$ positive?
Explanation: For vaporization, the latent heat $L$ is positive and the specific volume of the gas $v_g$ is much greater than that of the liquid $v_l$, making $\frac{dP}{dT} > 0$.
4. Why does an increase in external pressure lower the melting point of ice?
Explanation: For anomalous substances like water, ice contracts upon melting ($v_l < v_s$), resulting in a negative slope $\frac{dP}{dT} < 0$, which means increased pressure lowers the melting point.
5. What approximations are introduced to derive the Clausius-Clapeyron equation from the exact Clapeyron equation for vaporization?
Explanation: The Clausius-Clapeyron equation introduces approximations by neglecting the specific volume of the condensed phase ($v_l$) and treating the vapour phase as an ideal gas.
6. What is the differential form of the Clausius-Clapeyron equation for liquid-vapour transition?
Explanation: Using the ideal gas law $P v_g = R T$ for the vapour phase, the differential form becomes $\frac{dP}{dT} \approx \frac{P L_v}{R T^2}$.
7. Which of the following represents the integrated form of the Clausius-Clapeyron equation?
8. A first-order phase transition is characterized by discontinuities in which derivatives of the Gibbs free energy?
Explanation: A first-order phase transition is characterized by discontinuities in first-order derivatives of the Gibbs free energy, specifically volume and entropy.
9. Under what condition does the ideal gas approximation in the Clausius-Clapeyron equation fail?
Explanation: The ideal gas approximation fails near the critical point because the density of liquid and vapour become identical, making $\Delta v$ approach zero.
10. What does the latent heat expression $L = T \cdot \frac{dP}{dT} \cdot \Delta v$ represent?
Explanation: Rearranging the exact Clapeyron equation $\frac{dP}{dT} = \frac{L}{T \Delta v}$ yields the latent heat expression $L = T \cdot \frac{dP}{dT} \cdot \Delta v$.
11. For sublimation (solid-vapour transition), what is the differential relation given by the Clausius-Clapeyron equation?
Explanation: For sublimation, assuming ideal vapour behavior, the equation is $\frac{d(\ln P)}{dT} = \frac{L_s}{R T^2}$, where $L_s$ is the molar latent heat of sublimation.
12. Which assumption is made when deriving the standard integrated form of the Clausius-Clapeyron equation?
Explanation: The standard integrated form assumes the latent heat $L$ is constant and independent of temperature over the integration range.
13. What is the slope $\frac{dP}{dT}$ for fusion (melting) in terms of specific volumes $v_l$ and $v_s$?
Explanation: The Clapeyron equation for fusion is $\frac{dP}{dT} = \frac{L_f}{T (v_l - v_s)}$, where $L_f$ is the latent heat of fusion.
14. Which of the following systems is the Clausius-Clapeyron equation strictly applicable to?
Explanation: The Clausius-Clapeyron equation applies strictly to pure single-component systems in thermodynamic equilibrium.
15. If the vapour pressure of a liquid at temperature $T_1$ is $P_1$, which formula estimates the vapour pressure $P(T)$ at temperature $T$ assuming constant latent heat $L_v$?
Explanation: Integrating the differential form with constant $L_v$ gives the exponential vapour pressure formula $P(T) = P_0 \exp\left[ -\frac{L_v}{R} \left(\frac{1}{T} - \frac{1}{T_0}\right) \right]$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The Clapeyron equation gives the slope of the coexistence curve in a $P-T$ diagram. For the liquid-vapour transition, if $L_v = 40 \text{ kJ/mol}$ and boiling point $T = 373 \text{ K}$, what primarily determines the magnitude of $\frac{dP}{dT}$ under standard assumptions?
Explanation: Using the Clausius-Clapeyron form $\frac{dP}{dT} \approx \frac{P L_v}{R T^2}$, the magnitude is determined by the vapour pressure $P$, latent heat $L_v$, and temperature $T$, resting on the ideal gas assumption for the vapour.
2. Why does ice float on water and why does applying pressure decrease the melting point of ice?
Explanation: Water expands upon freezing (ice is less dense than water, so $v_l < v_s$). By the Clapeyron equation, this gives $\frac{dP}{dT} < 0$, meaning increased pressure lowers the melting point.
3. In the derivation of the Clausius-Clapeyron equation from the exact Clapeyron equation $\frac{dP}{dT} = \frac{L}{T(v_g - v_l)}$, which approximation is applied to the specific volumes?
Explanation: The specific volume of the liquid phase ($v_l$) is much smaller than that of the gas phase ($v_g$) and is neglected, while $v_g$ is approximated using the ideal gas law $v_g = \frac{RT}{P}$.
4. If the latent heat of vaporization $L_v$ is assumed constant, the integrated form $\ln\left(\frac{P_2}{P_1}\right) = -\frac{L_v}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ implies that a plot of $\ln P$ versus $\frac{1}{T}$ is:
Explanation: Rearranging $\ln P = -\frac{L_v}{R}\left(\frac{1}{T}\right) + \text{constant}$ shows that $\ln P$ versus $\frac{1}{T}$ yields a straight line with slope $-\frac{L_v}{R}$.
5. What happens to the slope of the liquid-vapour coexistence curve $\frac{dP}{dT}$ as the system approaches the critical point?
Explanation: In the exact Clapeyron equation $\frac{dP}{dT} = \frac{L}{T \Delta v}$, near the critical point the difference in specific volume $\Delta v = v_g - v_l$ approaches zero, causing the ideal gas approximation to fail and the behaviour of the slope to become problematic.
6. Which of the following statements is correct regarding first-order phase transitions?
Explanation: A first-order phase transition is defined by discontinuities in the first-order derivatives of Gibbs free energy, which are volume ($V = \frac{\partial G}{\partial P}$) and entropy ($S = -\frac{\partial G}{\partial T}$).
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. The chemical potentials of liquid and vapour phases along the coexistence curve satisfy which relation?
Explanation: Along a coexistence curve separating two phases, the chemical potentials of both phases are equal: $\mu_1(T, P) = \mu_2(T, P)$.
2. What is the physical reason behind the negative slope of the solid-liquid coexistence curve for water?
Explanation: For water, ice contracts upon melting ($v_l < v_s$), making the denominator in the Clapeyron equation negative, resulting in a negative slope $\frac{dP}{dT} < 0$.
3. Which differential equation correctly describes the Clausius-Clapeyron relation for sublimation?
Explanation: For sublimation (solid-vapour), the Clausius-Clapeyron equation is expressed as $\frac{d(\ln P)}{dT} = \frac{L_s}{R T^2}$.
4. If the latent heat $L_v$ is treated as constant, what is the integrated form of the Clausius-Clapeyron equation used to find $P_2$ at $T_2$?
Explanation: Using $\ln\left(\frac{P_2}{P_1}\right) = -\frac{L_v}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{L_v}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$, this is the correct integrated form.
5. Why does the Clausius-Clapeyron equation fail near the critical point?
Explanation: Near the critical point, the specific volume difference $\Delta v$ approaches zero and the ideal gas approximation breaks down, rendering the Clausius-Clapeyron equation invalid.
6. What does the term $\Delta v$ represent in the general Clapeyron equation $\frac{dP}{dT} = \frac{L}{T \Delta v}$?
Explanation: $\Delta v = v_2 - v_1$ represents the change in specific or molar volume between phase 2 and phase 1.
7. Which of the following equations represents the mass-based (specific) Clausius-Clapeyron form?
Explanation: The mass-based form uses specific latent heat per unit mass ($l_v$) and specific volumes: $\frac{dP}{dT} = \frac{l_v}{T (v_g - v_l)}$.
8. What is the primary assumption regarding the vapour phase in the derivation of the Clausius-Clapeyron equation?
Explanation: The Clausius-Clapeyron equation treats the vapour phase as an ideal gas to simplify $v_g$ as $\frac{RT}{P}$.
9. If a substance has a positive change in volume upon melting ($v_l > v_s$), how does an increase in pressure affect its melting point?
Explanation: For normal substances where $v_l > v_s$, the slope $\frac{dP}{dT}$ is positive, meaning an increase in pressure raises the melting point.
10. Which thermodynamic derivative is discontinuous across a first-order phase transition?
Explanation: First-order phase transitions are defined by discontinuities in the first-order derivatives of the Gibbs free energy, which include volume ($V$) and entropy ($S$).