The first law of thermodynamics

1. Overview

The first law of thermodynamics is a fundamental statement of the principle of conservation of energy specifically adapted for thermal systems. It dictates that energy cannot be created or destroyed, only transferred between a system and its surroundings in the form of heating ($q$) or work ($W$). This law allows us to track the internal energy ($U$) of a gas as it undergoes various physical processes. In the context of Cambridge 9702 Physics, the first law provides the mathematical link between the microscopic behavior of gas molecules (kinetic and potential energy) and the macroscopic variables we can measure, such as pressure, volume, and temperature.

Key Definitions

To secure full marks in descriptive questions, definitions must include specific technical keywords:

• Internal Energy ($U$): The sum of the random distribution of kinetic and potential energies associated with the molecules of a system.
  • Note: For an ideal gas, the potential energy is zero because there are no intermolecular forces; therefore, internal energy is solely a function of the total kinetic energy (and thus temperature).
• Work Done ($W$): The energy transferred to or from a system when a force moves through a distance. In thermodynamics, this is typically the result of a gas changing its volume against an external pressure.
• Heating ($q$): The transfer of energy between a system and its surroundings due to a temperature difference. Energy always flows from a region of higher temperature to a region of lower temperature.
• System: A specific mass of substance (usually an ideal gas) enclosed by a boundary (like a cylinder with a piston).
• Surroundings: Everything outside the system boundary that can exchange energy with the system.
• Thermal Equilibrium: A state where two objects in thermal contact have the same temperature, resulting in no net transfer of thermal energy ($q = 0$).

Content

3.1 Work Done by a Gas at Constant Pressure

When a gas expands, it must push against the external pressure of its surroundings. This requires the gas to exert a force over a distance, thereby transferring energy.

Derivation of $W = p\Delta V$:

1. Imagine a gas at pressure $p$ trapped inside a cylinder by a movable piston of cross-sectional area $A$.
2. The gas exerts an outward force $F$ on the piston: $$F = p \times A$$
3. If the gas expands at a constant pressure, the piston moves outward by a small distance $\Delta x$.
4. The work done by the gas is defined as: $$\text{Work} = \text{Force} \times \text{distance moved in direction of force}$$ $$W_{by} = (p \times A) \times \Delta x$$
5. The product of the area $A$ and the displacement $\Delta x$ is the change in volume, $\Delta V$: $$\Delta V = A \times \Delta x$$
6. Therefore, the work done by the gas is: $$W_{by} = p\Delta V$$

The "Work Done ON" Convention: The Cambridge 9702 syllabus specifically uses $W$ to represent the work done ON the system. This is a critical distinction:

• Expansion: The gas does work on the surroundings. Energy leaves the gas. Therefore, the work done on the gas is negative. $$W = -p\Delta V$$
• Compression: The surroundings do work on the gas (e.g., a piston is pushed in). Energy enters the gas. Therefore, the work done on the gas is positive. $$W = +p\Delta V$$

3.2 The First Law of Thermodynamics Equation

The law is expressed as: $$\mathbf{\Delta U = q + W}$$

Where:

• $\Delta U$: The increase in internal energy of the system.
• $q$: The energy transferred to the system by heating.
• $W$: The work done on the system.

3.3 Sign Conventions (The "Golden Rule" for Exams)

Success in Paper 2 and Paper 4 depends entirely on applying the correct signs. Always use the following table as a reference:

Symbol	Meaning of Positive (+) Sign	Meaning of Negative (-) Sign
$\Delta U$	Internal energy increases (Temp rises)	Internal energy decreases (Temp falls)
$q$	Energy is supplied to the system	Energy is removed from the system
$W$	Work is done on the system (Compression)	Work is done by the system (Expansion)

3.4 Application to Specific Thermodynamic Processes

The first law takes different forms depending on the constraints of the process:

1. Isochoric (Constant Volume):

   • If volume is constant, $\Delta V = 0$.
   • Therefore, $W = p\Delta V = 0$.
   • The law becomes: $\Delta U = q$.
   • Physical meaning: All energy added via heating goes directly into increasing the internal energy (and temperature) of the gas.

2. Isobaric (Constant Pressure):

   • Pressure remains constant while volume changes.
   • Work $W = -p\Delta V$ (for expansion).
   • The law remains: $\Delta U = q + W$.
   • Physical meaning: Energy added by heating is split between increasing the internal energy and doing work on the surroundings.

3. Isothermal (Constant Temperature):

   • For an ideal gas, internal energy depends only on temperature. If $\Delta T = 0$, then $\Delta U = 0$.
   • The law becomes: $0 = q + W$, or $q = -W$.
   • Physical meaning: If a gas expands isothermally, it does work ($W$ is negative). To keep the temperature constant, an equal amount of energy must be supplied by heating ($q$ is positive).

4. Adiabatic (No Heat Exchange):

   • The system is perfectly insulated, or the process happens so quickly that heat has no time to flow. Thus, $q = 0$.
   • The law becomes: $\Delta U = W$.
   • Physical meaning: If you compress a gas adiabatically ($W$ is positive), the internal energy must increase ($\Delta U$ is positive), causing the temperature to rise. This is why a bicycle pump gets hot during use.

3.5 Internal Energy and the Ideal Gas

In Topic 15 (Kinetic Theory), we learn that for an ideal gas:

1. There are no intermolecular forces (except during collisions).
2. Therefore, there is no potential energy ($E_p = 0$).
3. Internal energy is the sum of the kinetic energies of the molecules.
4. Since average kinetic energy is proportional to thermodynamic temperature ($E_k \propto T$), then $U \propto T$.

Exam Link: If a question states that the temperature of an ideal gas is constant, you must immediately conclude that $\Delta U = 0$.

---

4. Worked Examples

Worked Example 1 — Basic Application of the First Law

A fixed mass of an ideal gas is held in a container. $650\text{ J}$ of thermal energy is removed from the gas. At the same time, the gas is compressed, and $420\text{ J}$ of work is done on the gas. Calculate the change in internal energy and state whether the temperature rises or falls.

Step 1: Identify the variables with correct signs.

• Energy removed by heating: $q = -650\text{ J}$
• Work done on the gas: $W = +420\text{ J}$

Step 2: State the First Law of Thermodynamics. $$\Delta U = q + W$$

Step 3: Substitute and calculate. $$\Delta U = (-650) + (+420)$$ $$\Delta U = -230\text{ J}$$

Step 4: Interpret the result. Since $\Delta U$ is negative, the internal energy of the gas decreases. For an ideal gas, internal energy is proportional to temperature. Therefore, the temperature falls.

Answer: $\Delta U = -230\text{ J}$; Temperature falls.

---

Worked Example 2 — Work Done during Expansion

A gas expands from a volume of $2.0 \times 10^{-3}\text{ m}^3$ to $5.5 \times 10^{-3}\text{ m}^3$ at a constant pressure of $1.4 \times 10^5\text{ Pa}$. During this expansion, $900\text{ J}$ of thermal energy is supplied to the gas. Calculate the change in internal energy.

Step 1: Calculate the work done on the gas ($W$).

• $\Delta V = V_{final} - V_{initial} = (5.5 \times 10^{-3}) - (2.0 \times 10^{-3}) = 3.5 \times 10^{-3}\text{ m}^3$
• Since it is an expansion, work is done by the gas. Work done on the gas is: $$W = -p\Delta V$$ $$W = -(1.4 \times 10^5\text{ Pa}) \times (3.5 \times 10^{-3}\text{ m}^3)$$ $$W = -490\text{ J}$$

Step 2: Identify the heating ($q$).

• Energy supplied to the gas: $q = +900\text{ J}$

Step 3: Use the First Law. $$\Delta U = q + W$$ $$\Delta U = 900 + (-490)$$ $$\Delta U = +410\text{ J}$$

Answer: The internal energy increases by $410\text{ J}$.

---

Worked Example 3 — Unit Conversions and p-V Graphs

A cylinder contains a gas that is compressed from $800\text{ cm}^3$ to $200\text{ cm}^3$ at a constant pressure of $2.5 \times 10^5\text{ Pa}$. The internal energy of the gas increases by $120\text{ J}$. Calculate the thermal energy transfer $q$.

Step 1: Convert volume to SI units ($m^3$).

• $V_1 = 800 \times 10^{-6}\text{ m}^3$
• $V_2 = 200 \times 10^{-6}\text{ m}^3$
• $\Delta V = V_2 - V_1 = (200 - 800) \times 10^{-6} = -600 \times 10^{-6}\text{ m}^3$

Step 2: Calculate work done on the gas.

• $W = -p\Delta V$ (Note: the formula $W = -p\Delta V$ automatically handles the sign if $\Delta V$ is negative). $$W = -(2.5 \times 10^5) \times (-600 \times 10^{-6})$$ $$W = +150\text{ J}$$
• (Logic check: Compression means work is done ON the gas, so $W$ should be positive. $150\text{ J}$ is positive. Correct.)

Step 3: Use the First Law to find $q$.

• $\Delta U = +120\text{ J}$ (given as an increase) $$\Delta U = q + W$$ $$120 = q + 150$$ $$q = 120 - 150$$ $$q = -30\text{ J}$$

Answer: $30\text{ J}$ of thermal energy is removed from the gas ($q = -30\text{ J}$).

---

Key Equations

Equation	Description	Status
$\Delta U = q + W$	The First Law of Thermodynamics.	Memorise
$W = p\Delta V$	Work done at constant pressure.	Memorise
$U \propto T$	Internal energy is proportional to Kelvin temperature (Ideal Gas only).	Understand
$1\text{ cm}^3 = 10^{-6}\text{ m}^3$	Essential unit conversion for volume.	Memorise

Note: The First Law is often not provided on the 9702 Data Sheet. You must be able to state it and define the symbols.

---

Common Mistakes to Avoid

• ❌ The $cm^3$ to $m^3$ Trap: Many students multiply by $10^{-2}$ or $10^{-3}$.
  • ✅ Right: $1\text{ cm} = 10^{-2}\text{ m}$, so $1\text{ cm}^3 = (10^{-2}\text{ m})^3 = 10^{-6}\text{ m}^3$.
• ❌ Sign Confusion in $\Delta U = q + W$: Using the engineering convention $\Delta U = q - W$.
  • ✅ Right: In A-Level Physics (9702), $W$ is work done ON the system. Always use the plus sign and adjust the value of $W$ (negative for expansion, positive for compression).
• ❌ Confusing Temperature ($T$) with Heating ($q$): Thinking that if $q=0$, the temperature cannot change.
  • ✅ Right: In an adiabatic compression, $q=0$ but $W$ is positive, so $\Delta U$ increases and the temperature rises.
• ❌ Ignoring "Constant Pressure": Using $W = p\Delta V$ when the pressure is changing.
  • ✅ Right: If pressure changes, work done is the area under the p-V graph. You cannot simply multiply the final pressure by the change in volume.
• ❌ Forgetting Kelvin: Using Celsius in any proportionality involving temperature.
  • ✅ Right: Internal energy is proportional to the thermodynamic (Kelvin) temperature. While $\Delta T$ is the same in both scales, $U$ itself is not.

---

Exam Tips

1. The "First Mark" Strategy: In any calculation involving the first law, write down $\Delta U = q + W$ as your first line. Even if you mess up the signs later, you often secure a mark for the correct formula.
2. Keyword Check for Definitions: When defining internal energy, you must mention "random distribution," "sum," and "kinetic and potential energies." Missing any of these usually results in zero marks.
3. The "Rigid Container" Hint: If a question mentions a "rigid container" or "fixed volume," immediately write $W = 0$.
4. The "Vacuum" Scenario: If a gas expands into a vacuum (free expansion), no external pressure is being pushed against. Therefore, $p = 0$ and $W = 0$.
5. Graph Interpretation:
   • If the process on a $p-V$ graph moves to the right, the gas is expanding ($W$ is negative).
   • If the process moves to the left, the gas is being compressed ($W$ is positive).
   • The area under the line represents the magnitude of the work done.
6. State the Sign: When asked to "Describe the energy changes," don't just give the number. Explicitly state: "Work is done by the gas," or "Energy is supplied to the gas by heating." This shows the examiner you understand the direction of energy flow.