9.4 The First Law of Thermodynamics

The first law of thermodynamics is conservation of energy for thermal systems: the change in a system's internal energy equals the heat added plus the work done on it, or $\Delta U = Q + W$. For an ideal monatomic gas, internal energy depends only on temperature ($U = \frac{3}{2}nRT$), and PV diagrams let you track processes and read off work as the area under the curve. In AP Physics 2, this topic connects heat, work, internal energy, and process graphs.

Why This Matters for the AP Physics 2 Exam

Thermodynamics is one of the heavier-weighted units on the AP Physics 2 exam, and the first law shows up constantly because it ties temperature, heat, and work into a single energy-conservation statement. You will need to move between words, equations, and PV diagrams, which is exactly the kind of translation the exam rewards. Being able to explain why internal energy changes (or stays the same) during a process, and to back that explanation with the right equation, is core reasoning for free-response questions that ask you to make a claim and support it with physics.

Key Takeaways

• Internal energy $U$ is the sum of particle kinetic energy plus the potential energy of the configuration. An ideal gas has no internal potential energy, so its internal energy is purely kinetic.
• For an ideal monatomic gas, $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$, so internal energy depends only on absolute temperature.
• First law: $\Delta U = Q + W$, where $Q$ is heat added to the system and $W$ is work done on the system.
• Work done on a gas by external pressure is $W = -P\Delta V$. The negative sign means expansion ($\Delta V > 0$) does negative work on the gas.
• On a PV diagram, the area under the curve gives the magnitude of the work, and isotherms are lines of constant temperature.
• Memorize the four special processes: isovolumetric ($W=0$), isothermal ($\Delta U = 0$ for an ideal gas), isobaric ($W=-P\Delta V$), and adiabatic ($Q=0$).

Internal Energy of a System

Internal energy is the total energy stored inside a system at the microscopic level. It is the sum of:

• Kinetic energy of the particles moving within the system
• Potential energy from the interactions and configuration of those particles

Ideal gases are a simpler case:

• Atoms move independently with no attractive or repulsive forces between them, except during collisions.
• The internal structure of individual atoms is not considered.
• Because there are no interaction forces to store energy, an ideal gas has no internal potential energy.
• The internal energy of an ideal monatomic gas depends only on temperature:

$U=\frac{3}{2} n R T=\frac{3}{2} N k_{B} T$

Where:

• $n$ is the number of moles
• $R$ is the universal gas constant
• $T$ is the absolute temperature (in kelvin)
• $N$ is the number of atoms
• $k_B$ is the Boltzmann constant

A key idea: changes to a system's internal energy can change its internal structure and behavior without changing the motion of the system's center of mass. Heating a gas in a fixed box speeds up its particles even though the box itself is not going anywhere.

Thermodynamic Processes

Thermodynamic processes describe how a system moves from one state to another. The first law of thermodynamics restates energy conservation for these processes.

For an isolated system (no exchange of energy or matter with surroundings):

• The total energy stays constant over time.

For a closed system (exchanges energy but not matter):

• The change in internal energy equals the heat added to the system plus the work done on the system:

$\Delta U = Q + W$

• Work done on the system by a constant or average external pressure that changes its volume is:

$W = -P \Delta V$

• The negative sign matters: when a gas expands ($\Delta V > 0$), work done on the gas is negative, so the gas is doing work on its surroundings.

Reading PV Diagrams

PV diagrams (pressure vs. volume graphs) are the main tool for analyzing these processes:

• Each point represents a specific state of the system.
• A path between points represents a process.
• Isotherms are curves where temperature stays constant.
• The area under the curve equals the magnitude of the work done.

The Four Special Processes

These cases let you predict how temperature, heat, and work connect in real systems like pistons, engines, and gas-filled containers.

How to Use This on the AP Physics 2 Exam

Problem Solving

• Pin down your sign convention before plugging in. In $\Delta U = Q + W$, $Q$ is positive when heat goes into the system and $W$ is positive when work is done on the system.
• For an ideal monatomic gas, link $\Delta U$ directly to temperature change: $\Delta U = \frac{3}{2}nR\Delta T$. If $\Delta U$ is zero, the temperature did not change.
• Use $W = -P\Delta V$ only when pressure is constant or you are using an average pressure. For other paths, find work from the area under the PV curve.

Free Response

• When a question asks you to make a claim and justify it without equations first, then derive the math, lean on conservation of energy in words: heat in and work done on the gas both add to internal energy.
• Watch for the identity hidden in each special process. "The gas is heated at constant volume" should immediately tell you $W=0$ and $\Delta U = Q$.
• Connect your final equation back to your verbal claim. For example, if you argued the temperature rises, show that your derived $\Delta U > 0$ confirms it.

Common Trap

• The area under a PV curve gives the magnitude of work, but you still have to decide the sign from whether the gas expanded or compressed.

Practice Problem 1: Internal Energy Change

Solution

Apply the first law to find the change in internal energy:

$\Delta U = Q + W$

Here $Q$ is heat absorbed (positive) and $W$ is work done on the gas (negative when the gas expands and does work).

Given:

• $Q = 2500 \text{ J}$ (heat absorbed)
• $W = -1500 \text{ J}$ (gas does work, so work on the gas is negative)
• $n = 2.0 \text{ mol}$

$\Delta U = 2500 \text{ J} + (-1500 \text{ J}) = 1000 \text{ J}$

For a monatomic ideal gas:

$U = \frac{3}{2}nRT \quad\Rightarrow\quad \Delta U = \frac{3}{2}nR(T_f - T_i)$

Solving for the final temperature:

$T_f = T_i + \frac{2\Delta U}{3nR}$

$T_f = 300 \text{ K} + \frac{2 \times 1000 \text{ J}}{3 \times 2.0 \text{ mol} \times 8.31 \text{ J/(mol}\cdot\text{K})}$

$T_f = 300 \text{ K} + \frac{2000 \text{ J}}{49.86 \text{ J/K}}$

$T_f = 300 \text{ K} + 40.1 \text{ K} = 340.1 \text{ K}$

Practice Problem 2: Work in a Thermodynamic Process

Solution

For expansion against a constant external pressure, the work done by the gas is:

$W_{by} = P_{ext}(V_f - V_i)$

This is the work done by the gas, which is the negative of the work done on the gas in the first law.

Given:

• Initial volume, $V_i = 2.0 \text{ L} = 2.0 \times 10^{-3} \text{ m}^3$
• Final volume, $V_f = 5.0 \text{ L} = 5.0 \times 10^{-3} \text{ m}^3$
• External pressure, $P_{ext} = 1.0 \times 10^5 \text{ Pa}$

$W_{by} = (1.0 \times 10^5 \text{ Pa})[(5.0 \times 10^{-3} \text{ m}^3) - (2.0 \times 10^{-3} \text{ m}^3)]$

$W_{by} = (1.0 \times 10^5 \text{ Pa})(3.0 \times 10^{-3} \text{ m}^3) = 300 \text{ J}$

The gas does 300 J of work during the expansion. In terms of the first law, the work done on the gas is $W = -300 \text{ J}$.

Common Misconceptions

• "Internal energy includes the motion of the whole container." Internal energy is microscopic. It tracks particle kinetic and potential energy, not the kinetic energy of the system's center of mass.
• "Ideal gases have potential energy too." An ideal gas has no internal potential energy because its particles only interact during collisions. Its internal energy is entirely kinetic and depends only on temperature.
• "Positive work always means the gas expanded." In $\Delta U = Q + W$, positive $W$ means work is done on the gas, which happens during compression. Expansion gives negative $W$.
• "Heat and temperature are the same thing." Heat $Q$ is energy transferred because of a temperature difference. Temperature reflects average kinetic energy. You can add heat with no temperature change, like in an isothermal process where that energy leaves as work.
• "In an isothermal process no energy moves." Constant temperature means $\Delta U = 0$ for an ideal gas, but heat and work are both nonzero and exactly cancel: $Q = -W$.
• "Adiabatic means the temperature stays constant." Adiabatic means $Q = 0$, no heat transfer. The temperature usually does change because all the internal energy change comes from work: $\Delta U = W$.

Related AP Physics 2 Guides

• 9.1 Kinetic Theory of Temperature and Pressure
• 9.2 The Ideal Gas Law
• 9.3 Thermal Energy Transfer and Equilibrium
• 9.6 Entropy and the Second Law of Thermodynamics
• 9.5 Specific Heat and Thermal Conductivity