AP Physics 2: First Law of Thermodynamics

The First Law of Thermodynamics is the cornerstone of thermal physics, translating the universal principle of energy conservation into the language of heat, work, and internal energy. Mastering it allows you to predict how engines operate, why refrigerators cool, and how the energy of the universe is perpetually rearranged. For the AP Physics 2 exam and future engineering challenges, proficiency with this law is non-negotiable.

The Components of the Equation: $\Delta U=Q-W$

To apply the First Law, you must first understand its three variables. The law is expressed as $\Delta U=Q-W$, where each term has a precise meaning.

Internal energy ($U$) is the total microscopic energy of a system's molecules. This includes their random translational, rotational, and vibrational kinetic energies, plus any potential energy from intermolecular forces. For an ideal gas, where molecules don't interact, the internal energy depends only on temperature. A change in internal energy, $\Delta U$, signifies a change in the system's temperature or phase. Heat ($Q$) is energy transferred between a system and its surroundings due to a temperature difference. Heat added to a system is positive; heat released by a system is negative. Work ($W$) is energy transferred by a force acting through a distance. In thermodynamics, we often focus on pressure-volume work done by or on a gas. When a gas expands (increases its volume), it does work on its surroundings; we define this as positive work. When a gas is compressed (volume decreases), work is done on the gas, which is negative work.

The First Law Decoded: Sign Conventions and Interpretation

The equation $\Delta U=Q-W$ is a precise accounting statement. Think of the system's internal energy ($U$) as a bank account. Heat ($Q$) is a deposit (positive Q) or withdrawal (negative Q). Work ($W$) is a payment the system makes (positive W) or income it receives (negative W). The change in your account balance ($\Delta U$) equals deposits minus payments.

This sign convention is critical:

• $Q>0$: Heat flows into the system.
• $Q<0$: Heat flows out of the system.
• $W>0$: The system does work on the surroundings (expands).
• $W<0$: The surroundings do work on the system (compresses).
• $\Delta U>0$: The system's internal energy (and typically temperature) increases.
• $\Delta U<0$: The system's internal energy decreases.

The law states that energy cannot be created or destroyed; it can only change forms or be transferred. The system's stored energy ($U$) changes only due to energy crossings its boundary as heat ($Q$) or work ($W$).

Applying the Law to Standard Thermodynamic Processes

The power of the First Law becomes clear when applied to specific processes an ideal gas undergoes. The work done, $W$, is often calculated as the area under the curve on a Pressure-Volume (PV) diagram.

Isobaric Process (Constant Pressure): Here, work is straightforward: $W=P\Delta V$. If the gas expands at constant pressure ($\Delta V>0$), it does positive work. To find $\Delta U$, you need the temperature change. For a monatomic ideal gas, $\Delta U=\frac{3}{2}nR\Delta T$. Then, the First Law directly gives the heat transferred: $Q=\Delta U+W$. Example: A gas in a cylinder with a movable piston is heated, expanding at 2 atm from 1.0 m³ to 1.5 m³. Work done by* the gas is $W=P\Delta V=(2.0\times 10^5\text{ Pa})(0.5{\text{ m}}^3)=1.0\times 10^5\text{ J}$.*

Isochoric Process (Constant Volume): If volume is constant, $\Delta V=0$, so $W=0$. The First Law simplifies to $\Delta U=Q$. All heat added goes directly to increasing internal energy (temperature). *Example: Heating a gas in a rigid, sealed container. No work is done, so $Q_V=\Delta U$.*

Isothermal Process (Constant Temperature): For an ideal gas, constant temperature means $\Delta U=0$. The First Law becomes $0=Q-W$, or $Q=W$. Any heat added to the gas is completely converted into work done by the gas during expansion. Example: A gas expands slowly in thermal contact with a large reservoir, maintaining temperature. The work done equals the heat absorbed.

Adiabatic Process (No Heat Transfer): Here, $Q=0$. The law becomes $\Delta U=-W$. If the gas expands adiabatically ($W>0$), it does work by using its own internal energy, so $\Delta U$ and temperature decrease. In compression ($W<0$), work done on the gas increases its internal energy and temperature. Example: Rapid compression in a diesel engine cylinder, where there isn't time for significant heat exchange.

Solving Multi-Step Problems: Thermodynamic Cycles

Many real devices, like heat engines and refrigerators, operate in cycles—sequences of processes that return the system to its initial state. Since internal energy $U$ is a state function (it depends only on the current state, not the path taken), $\Delta U_{cycle}=0$ for any complete cycle.

This is where multi-step problem-solving shines. For a cycle, the First Law implies: $$\Delta U_{total}=0=Q_{net}-W_{net}$$ Therefore, $$Q_{net}=W_{net}$$

The net heat added over the cycle equals the net work done by the system. If $Q_{net}>0$, the system converts some net heat into net work (an engine). If $W_{net}<0$, net work is done on the system to eject heat from a cold reservoir (a refrigerator).

Step-by-Step Approach:

1. Identify each process in the sequence (e.g., A→B is isobaric, B→C is adiabatic).
2. Analyze each process individually. Use the appropriate rules (e.g., $W=0$ for isochoric, $\Delta U=0$ for isothermal) with the First Law to find $Q$, $W$, and $\Delta U$ for that step.
3. Track quantities carefully. Keep a table. Remember signs are based on the system's perspective.
4. Combine for the whole path. For any overall change from state i to state f, $\Delta U_{i\to f}$, $Q_{i\to f}$, and $W_{i\to f}$ are the sums of the changes for each intervening step.
5. For a cycle, verify that $\Delta U_{total}=0$ and $Q_{net}=W_{net}$.

Common Pitfalls

1. Incorrect Work Sign for Compression/Expansion: The most frequent error is misassigning the sign of $W$. Remember the system-centric definition: $W=P\Delta V$ for constant pressure. If the gas expands ($\Delta V>0$), it does work on the world, so $W>0$. If it is compressed ($\Delta V<0$), work is done on it, so $W<0$. Confusing this will derail every calculation.
2. Treating Heat and Work as State Functions: Heat ($Q$) and work ($W$) are path functions. Their values depend on how the system gets from initial to final state. You cannot look up "the heat" of a state; you can only calculate heat transfer for a specific process. Only $U$ and $\Delta U$ are state functions.
3. Assuming $\Delta U=0$ Means No Heat or Work: A zero change in internal energy ($\Delta U=0$) does not imply $Q=0$ and $W=0$. It means $Q=W$. In an isothermal expansion, for example, heat is added ($Q>0$) and an equal amount of work is done ($W>0$), with no net change in stored energy.
4. Forgetting the Ideal Gas Assumptions: The simplifications $\Delta U=\frac{f}{2}nR\Delta T$ and $U\propto T$ apply only to ideal gases. The First Law itself ($\Delta U=Q-W$) is universal, but calculating $\Delta U$ for real substances requires different methods.

Summary

• The First Law of Thermodynamics, $\Delta U=Q-W$, is the law of energy conservation for thermal systems, where $\Delta U$ is change in internal energy, $Q$ is heat added, and $W$ is work done by the system.
• Work done by an expanding gas is positive and is often calculated as the area under a PV curve. The sign convention is system-centric: positive for energy leaving as work, negative for energy entering as work.
• Internal energy for an ideal gas is a function of temperature only. In isochoric processes ($W=0$), $Q$ equals $\Delta U$. In isothermal processes ($\Delta U=0$), $Q$ equals $W$.
• For any complete thermodynamic cycle, the net change in internal energy is zero ($\Delta U_{cycle}=0$), which means the net heat transfer equals the net work output: $Q_{net}=W_{net}$.
• Solving multi-step problems requires analyzing each process sequentially, tracking signs meticulously, and summing contributions to find total $Q$, $W$, and $\Delta U$ for the overall path.