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๐Ÿ”ฅThermodynamics I Unit 2 Review

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2.1 Phase changes and phase diagrams

2.1 Phase changes and phase diagrams

Written by the Fiveable Content Team โ€ข Last updated August 2025
Written by the Fiveable Content Team โ€ข Last updated August 2025
๐Ÿ”ฅThermodynamics I
Unit & Topic Study Guides

Phases of Matter and Their Characteristics

Solid, Liquid, and Gas Phases

Matter exists in three main phases: solid, liquid, and gas. Each phase has distinct properties tied to how its particles are arranged and how they move.

  • Solid: Particles are tightly packed in a fixed arrangement, vibrating in place. This gives solids a definite shape and definite volume (think of an ice cube).
  • Liquid: Particles are still close together but can slide past one another. Liquids have a definite volume but take the shape of their container (water in a glass).
  • Gas: Particles are widely separated and move randomly. Gases have no fixed shape or volume and expand to fill whatever container they're in (steam filling a pot).

In thermodynamics, you'll mostly work with pure substances transitioning between these three phases, so getting comfortable with these distinctions matters.

Plasma and Other Phases

Plasma is sometimes called a fourth state of matter. It consists of ionized particles at very high energy levels (stars, lightning bolts). Other exotic phases exist too, like Bose-Einstein condensates and supercritical fluids. For this course, though, the focus is on solid, liquid, and gas phases and the transitions between them. Supercritical fluids come up later in this unit because they're relevant to engineering applications.

Interpretation of Phase Diagrams

Key Features and Boundaries

A phase diagram is a plot of pressure (y-axis) versus temperature (x-axis) that shows which phase a pure substance will be in at any given combination of PP and TT. It also shows where phase transitions occur.

There are three important features to identify on any phase diagram:

  • Phase boundaries (also called coexistence curves): These are the lines on the diagram where two phases coexist in equilibrium. There are three of them:
    • Sublimation curve: separates solid and gas regions
    • Melting curve (fusion curve): separates solid and liquid regions
    • Vaporization curve: separates liquid and gas regions
  • Triple point: The single unique point where all three phase boundaries meet. At this specific TT and PP, solid, liquid, and gas all coexist in equilibrium. For water, the triple point is at 0.01ยฐC and 611.73 Pa.
  • Critical point: The endpoint of the vaporization curve. Above this temperature and pressure, there's no distinct boundary between liquid and gas. For water, the critical point is at 374ยฐC and 22.06 MPa. For carbon dioxide, it's at 31.1ยฐC and 7.38 MPa.

One thing that trips students up: the critical point doesn't mean the substance stops existing. It means the liquid and gas phases become indistinguishable. The substance enters what's called the supercritical fluid region.

Solid, Liquid, and Gas Phases, Major Features of a Phase Diagram | Introduction to Chemistry

Reading a Phase Diagram

To figure out what phase a substance is in, find the point on the diagram corresponding to your TT and PP values:

  1. Locate the temperature on the x-axis and the pressure on the y-axis.
  2. Find where those values intersect on the diagram.
  3. If the point falls within a region (not on a boundary line), the substance is entirely in that single phase.
  4. If the point falls on a boundary line, two phases coexist in equilibrium.
  5. If the point is exactly at the triple point, all three phases coexist.

To trace a phase change, imagine moving across the diagram by changing TT at constant PP (horizontal line) or changing PP at constant TT (vertical line). Every time your path crosses a boundary line, a phase transition occurs.

Supercritical Fluids

At temperatures and pressures above the critical point, a substance becomes a supercritical fluid. It has properties of both liquids and gases: liquid-like density but gas-like viscosity and diffusivity. Supercritical CO2CO_2 is widely used in industrial decaffeination of coffee because it can dissolve compounds like a liquid solvent but penetrate materials like a gas.

Phase Transitions and Energy Changes

Types of Phase Transitions

A phase transition happens when a substance changes from one phase to another due to a change in temperature, pressure, or both. There are six transitions to know, grouped into three pairs:

Transitions that absorb energy (endothermic):

  • Melting (solid โ†’ liquid): Occurs at the melting point when thermal energy overcomes the rigid structure of the solid. For water at 1 atm, this happens at 0ยฐC.
  • Vaporization (liquid โ†’ gas): Can happen at the surface (evaporation) or throughout the bulk (boiling). Boiling occurs when the substance's vapor pressure equals the external pressure. For water at 1 atm, the boiling point is 100ยฐC.
  • Sublimation (solid โ†’ gas): The solid converts directly to gas, skipping the liquid phase entirely. Dry ice (solid CO2CO_2) does this at atmospheric pressure because 1 atm is below CO2CO_2's triple point pressure.

Transitions that release energy (exothermic):

  • Freezing (liquid โ†’ solid): The reverse of melting.
  • Condensation (gas โ†’ liquid): The reverse of vaporization.
  • Deposition (gas โ†’ solid): The reverse of sublimation. Frost forming on a cold surface is an example.
Solid, Liquid, and Gas Phases, Phase Diagrams ยท Chemistry

Enthalpy Changes and Latent Heats

During a phase transition, the temperature of the substance stays constant even though energy is being added or removed. This is a critical concept. The energy goes into breaking (or forming) intermolecular bonds rather than changing the kinetic energy of the particles. This energy is called latent heat.

Two specific latent heats you'll use in calculations:

  • Heat of fusion (hsfh_{sf}): Energy required to melt a unit mass of substance at its melting point. For water: hsf=333.55ย kJ/kgh_{sf} = 333.55 \text{ kJ/kg} (or 6.01 kJ/mol).
  • Heat of vaporization (hfgh_{fg}): Energy required to vaporize a unit mass of substance at its boiling point. For water: hfg=2257ย kJ/kgh_{fg} = 2257 \text{ kJ/kg} (or 40.65 kJ/mol).

Notice that hfgh_{fg} is much larger than hsfh_{sf} for water. That's because vaporization requires completely separating molecules from each other, while melting only loosens the rigid structure. This pattern holds for most substances.

The energy for a phase change is calculated as:

Q=mโ‹…hQ = m \cdot h

where QQ is the heat transferred, mm is the mass, and hh is the specific latent heat (fusion or vaporization). For sublimation, the heat of sublimation is approximately the sum of the heats of fusion and vaporization: hsgโ‰ˆhsf+hfgh_{sg} \approx h_{sf} + h_{fg}.

Vapor Pressure Calculations with the Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressure of a substance to temperature along a phase boundary (typically the vaporization curve). It's derived from thermodynamic principles and is useful when you need to estimate vapor pressure at a temperature where you don't have data.

The integrated form most commonly used in calculations is:

lnโก(P2P1)=ฮ”HvapR(1T1โˆ’1T2)\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

where:

  • P1P_1 and P2P_2 are the vapor pressures at temperatures T1T_1 and T2T_2
  • ฮ”Hvap\Delta H_{vap} is the molar heat of vaporization (in J/mol)
  • RR is the universal gas constant (8.314 J/molยทK)
  • T1T_1 and T2T_2 are in Kelvin (a common mistake is using ยฐC)

This equation assumes two things: (1) ฮ”Hvap\Delta H_{vap} is roughly constant over the temperature range, and (2) the vapor behaves as an ideal gas. Both assumptions work well over moderate temperature ranges but break down near the critical point.

How to Use It: Step-by-Step

Finding vapor pressure at a new temperature:

  1. Identify your known values: P1P_1, T1T_1, T2T_2, and ฮ”Hvap\Delta H_{vap}.
  2. Convert all temperatures to Kelvin (TK=TC+273.15T_K = T_C + 273.15).
  3. Plug values into the equation and solve for lnโก(P2/P1)\ln(P_2/P_1).
  4. Exponentiate both sides to isolate P2P_2: P2=P1โ‹…e[ฮ”HvapR(1T1โˆ’1T2)]P_2 = P_1 \cdot e^{\left[\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]}

Finding ฮ”Hvap\Delta H_{vap} from experimental data:

  1. Collect vapor pressure measurements at several temperatures.
  2. Plot lnโก(P)\ln(P) on the y-axis versus 1/T1/T on the x-axis.
  3. The data should form a roughly straight line. The slope of that line equals โˆ’ฮ”Hvap/R-\Delta H_{vap}/R.
  4. Solve for ฮ”Hvap\Delta H_{vap}: ฮ”Hvap=โˆ’slopeร—R\Delta H_{vap} = -\text{slope} \times R

This graphical method is especially useful in lab settings where you have multiple data points and want to determine ฮ”Hvap\Delta H_{vap} experimentally rather than relying on a single pair of measurements.