Properties of Pure Substances

Why it matters

Understanding the properties of pure substances is essential for analyzing and designing thermodynamic systems such as engines, refrigerators, and power plants. These properties help engineers predict how substances will behave under different conditions, which is crucial for efficiency and safety.

Key ideas

• Pure Substance: A substance with a uniform and invariable chemical composition. It can exist in more than one phase but its chemical composition remains the same.
• Phases of a Pure Substance: Solid, liquid, and gas. Each phase has distinct properties and behaviors.
• Phase Change: Transition between different phases, such as melting, boiling, or sublimation.
• Saturation: A state where a substance is about to undergo a phase change. For example, saturated liquid and saturated vapor coexist at the saturation pressure and temperature.
• Critical Point: The end point of the liquid–vapor coexistence curve, beyond which distinct liquid and gas phases do not exist.
• Triple Point: The unique set of conditions where all three phases coexist in equilibrium.

Formulas

• v = V / m
  • v: Specific volume (m³/kg)
  • V: Volume (m³)
  • m: Mass (kg)
• u = U / m
  • u: Specific internal energy (kJ/kg)
  • U: Internal energy (kJ)
  • m: Mass (kg)
• h = u + Pv
  • h: Specific enthalpy (kJ/kg)
  • u: Specific internal energy (kJ/kg)
  • P: Pressure (kPa)
  • v: Specific volume (m³/kg)

Phase identification and quality

For a simple compressible pure substance, use two independent intensive properties to fix the state. In the liquid–vapor two-phase region, saturation pressure and temperature are dependent; specify quality x = mass_vapor/mass_total or another independent property. For y = v, u, h or s, y = y_f + x(y_g-y_f). Quality is defined only in the saturated mixture region.

Worked example

For a property-conversion exercise, a state is specified by P = 2 MPa, v = 0.127 m³/kg and u = 2600 kJ/kg. Calculate h. These are supplied exercise values, not a claimed steam-table lookup at a specified temperature.

P = 2000 kPa and 1 kPa·m³ = 1 kJ. Pv = 2000(0.127) = 254 kJ/kg. h = u + Pv = 2600 + 254 = 2854 kJ/kg.

The conversion 2000 kPa = 2 kJ/m³ would be wrong: 2000 kPa = 2000 kJ/m³. For a real state specified by pressure and temperature, identify the phase and obtain mutually consistent properties from an appropriate table or equation of state.

Common mistakes

• Confusing specific volume with total volume.
• Using incorrect units, especially for pressure and volume.
• Not using updated steam tables for accurate data.

For GATE ME

Questions often involve calculating properties using steam tables or Mollier charts. Practice identifying phase states and using property tables effectively.

Quick check

1. What is a pure substance?
2. Define the critical point.
3. How do you calculate specific enthalpy?

Answers: 1. A substance with uniform chemical composition. 2. The end point of a phase equilibrium curve. 3. h = u + Pv.