First Law of Thermodynamics: Heat, Work, Internal Energy & Processes

What is Thermodynamics?

Thermodynamics is the branch of physics that studies heat, work, and energy transfer between systems. It helps us understand how engines work, why refrigerators cool, and how energy is conserved in nature.

Key Terms

1. System: The part of the universe we study (e.g., gas in a cylinder, water in a container).
2. Surroundings: Everything outside the system that can exchange energy with it.
3. Boundary: The imaginary line separating the system from its surroundings.

classDiagram
    class System {
        +Gas in a cylinder
        +Water in a container
    }
    class Surroundings {
        +Everything outside
        +Can exchange energy
    }
    System --> Surroundings : "Boundary"

First Law of Thermodynamics

The First Law states:

"Energy cannot be created or destroyed, only transferred or converted." Mathematically: ΔU = Q – W

• ΔU = Change in internal energy (Joules)
• Q = Heat added to the system (Joules)
• W = Work done by the system (Joules)

Sign Conventions

Example:

• If a gas expands, it does work on the surroundings (W is positive).
• If heat is added to the gas, Q is positive.

Internal Energy (U)

• Definition: The total energy (kinetic + potential) of all molecules in a system.
• Factors affecting U:
  • Temperature (higher T → higher U)
  • Volume (for gases, higher V → higher U)
  • Phase (solid → liquid → gas → higher U)

Example:

• When you heat water, its internal energy increases (ΔU > 0).
• When a gas expands, it may lose internal energy (ΔU < 0).

Types of Thermodynamic Processes

Different processes occur when heat and work are exchanged. Let’s compare them:

pie
    title Thermodynamic Processes
    "Isochoric" : 30
    "Isobaric" : 25
    "Isothermal" : 20
    "Adiabatic" : 25

Work Done in Thermodynamic Processes

Work done by a gas is given by: W = ∫P dV (for small changes, W ≈ PΔV)

Example 1: Isochoric Process (ΔV = 0)

• A gas is heated in a rigid container (volume cannot change).
• Work done (W) = 0 (since ΔV = 0).
• ΔU = Q (all heat goes into increasing internal energy).

Calculation: If Q = 500 J is added, ΔU = 500 J (since W = 0).

Example 2: Isobaric Process (ΔP = 0)

• A gas expands at constant pressure (e.g., piston moving slowly).
• Work done (W) = PΔV.

Calculation:

• Initial pressure, P = 1 atm = 101,325 Pa
• Volume changes from V₁ = 2 m³ → V₂ = 4 m³
• W = P(V₂ – V₁) = 101,325 × (4 – 2) = 202,650 J
• If Q = 500,000 J is added, ΔU = Q – W = 500,000 – 202,650 = 297,350 J

Example 3: Isothermal Expansion (ΔU = 0)

• A gas expands slowly while keeping temperature constant (e.g., ideal gas in a cylinder with a heat reservoir).
• ΔU = 0 (since temperature is constant).
• Q = W (all heat added is converted to work).

Calculation: If W = 300 J, then Q = 300 J (since ΔU = 0).

Example 4: Adiabatic Process (Q = 0)

• A gas expands rapidly (no time for heat exchange, e.g., sudden release of a piston).
• Q = 0, so ΔU = –W (internal energy decreases as work is done).

Calculation: If W = 200 J (work done by gas), then ΔU = –200 J (internal energy decreases).

P-V Diagrams (Pressure-Volume Graphs)

A P-V diagram shows how pressure and volume change in a process.

graph LR
    A["P-V Diagram"] --> B["Area under curve = Work done (W)"]
    A --> C["Steeper curve = Adiabatic"]
    A --> D["Horizontal line = Isobaric"]
    A --> E["Vertical line = Isochoric"]
    A --> F["Hyperbola = Isothermal"]

Example:

• Isobaric process: Horizontal line (constant pressure).
• Isochoric process: Vertical line (constant volume).
• Adiabatic process: Steeper than isothermal (since no heat exchange).

Applications of the First Law

1. Heat Engines (e.g., Car engines, Steam engines)

   • Convert heat into mechanical work.
   • Efficiency (η) = W_out / Q_in (always < 100%).

2. Refrigerators & Air Conditioners

   • Transfer heat from cold to hot regions (requires work input).

3. Human Body

   • Food provides energy (Q), body does work (W), and some energy is stored as internal energy (U).

Common Mistakes to Avoid

❌ Mixing signs of Q and W:

• Q > 0 when heat is added to the system.
• W > 0 when work is done by the system.

❌ Assuming ΔU = 0 in all processes:

• Only isothermal processes have ΔU = 0 (for ideal gases).

❌ Ignoring units:

• Always use Joules (J) for Q, W, and ΔU.

Exam Tip: How to Score Full Marks

✅ Understand the First Law equation (ΔU = Q – W) and its sign conventions. ✅ Practice P-V diagrams – label axes, processes, and calculate work from the area. ✅ Memorize the four processes (isochoric, isobaric, isothermal, adiabatic) and their key features. ✅ Solve numerical problems step-by-step:

1. Identify the process (isochoric/isobaric/etc.).
2. Write down given values (Q, W, P, V, ΔU).
3. Apply ΔU = Q – W correctly. ✅ For NEB exams, expect:

• Short questions on definitions (First Law, internal energy).
• Numerical problems (calculate Q, W, or ΔU).
• P-V diagram questions (identify processes, calculate work).

NEB Board-Style Questions (Practice)

Short Questions (2 marks each)

1. State the First Law of Thermodynamics in words and symbols.
2. What is the difference between an isothermal and an adiabatic process?
3. Define internal energy. How does it change in an isochoric process?
4. Why is work done in an isochoric process always zero?

Numerical Problems (5-7 marks)

5. A gas undergoes an isobaric process where 500 J of heat is added, and the gas does 200 J of work. Calculate the change in internal energy (ΔU).

   • Solution: ΔU = Q – W = 500 J – 200 J = 300 J

6. In an adiabatic process, a gas does 150 J of work. What is the change in internal energy?

   • Solution: Q = 0 (adiabatic), so ΔU = –W = –150 J.

7. A monatomic ideal gas is heated at constant volume. If 300 J of heat is added, what is the change in internal energy? (For monatomic gases, ΔU = (3/2)nRΔT, but here Q = ΔU since W = 0.)

   • Solution: ΔU = Q = 300 J

8. A gas expands isothermally from 2 m³ to 4 m³ at a constant pressure of 100 kPa. Calculate the work done by the gas.

   • Solution: W = PΔV = 100,000 Pa × (4 – 2) m³ = 200,000 J

Conceptual Questions (3-5 marks)

9. Explain why a refrigerator cannot have 100% efficiency.

   • Answer: According to the First Law, energy must be conserved. A refrigerator must do work (W) to transfer heat (Q_cold) from the cold reservoir to the hot reservoir. Since Q_hot = Q_cold + W, some energy is always lost as work, making 100% efficiency impossible.

10. Draw a P-V diagram for an isobaric expansion followed by an isochoric cooling.

    • Answer:

      graph LR
          A["Start: P1, V1"] -->|"Isobaric Expansion"| B["P1, V2"]
          B -->|"Isochoric Cooling"| C["P2, V2"]

      (Horizontal line for isobaric, vertical downward for isochoric cooling.)

Keep practicing P-V diagrams and numerical problems—they appear frequently in NEB exams! 🚀