AP Physics 2: Thermodynamic Processes

Understanding thermodynamic processes is the key to analyzing how engines, refrigerators, and our atmosphere function. These processes describe the specific pathways a gas takes as its pressure, volume, and temperature change, governed by fundamental conservation laws. Mastering how to characterize them, calculate energy transfers, and apply the first law of thermodynamics is essential for both AP Physics 2 success and grasping real-world engineering principles.

Systems, States, and the First Law

Before analyzing specific processes, we must define the playing field. A thermodynamic system is the specific portion of matter we choose to study, such as a gas inside a piston. Everything outside is the surroundings. The condition of the system is described by state variables: pressure ($P$), volume ($V$), and temperature ($T$). These are related by the ideal gas law, $PV=nRT$, where $n$ is the number of moles and $R$ is the universal gas constant.

The first law of thermodynamics is the statement of energy conservation for thermal systems: $\Delta U=Q-W$. Here, $\Delta U$ is the change in the system's internal energy (which, for an ideal gas, depends only on temperature), $Q$ is the heat added to the system, and $W$ is the work done by the system on its surroundings. The sign convention is critical: positive $Q$ means heat flows in, and positive $W$ means the system expands and does work on the outside world.

Characterizing Processes: The "Constant" Variable

A thermodynamic process is defined by which state variable—or combination thereof—is held constant. We analyze four fundamental types: isothermal, isobaric, isochoric, and adiabatic.

The Isothermal Process: Constant Temperature

In an isothermal process, the temperature $T$ of the gas remains constant ($\Delta T=0$). Since internal energy for an ideal gas is a function of temperature only, $\Delta U=0$. The process must occur slowly, allowing heat exchange with a thermal reservoir to maintain constant $T$.

• Gas Law Constraint: From $PV=nRT$, if $T$ and $n$ are constant, then $PV=\text{constant}$. This means pressure and volume are inversely proportional: $P\propto 1/V$.
• Work Done: The work done by the gas is not simply $P\Delta V$ because pressure changes. It must be calculated using calculus, resulting in $W=nRT\ln \left(\frac{V_f}{V_i}\right)$. If the gas expands ($V_f>V_i$), work $W$ is positive.
• First Law Application: With $\Delta U=0$, the first law simplifies to $0=Q-W$, or $Q=W$. All heat added to the system is converted directly into work done by the system.

Example: A cylinder with a movable piston contains 2.0 moles of ideal gas at 300 K. It expands isothermally until its volume doubles. The work done is $W=(2.0\text{ mol})(8.31\text{ J/molcdotpK})(300\text{ K})\ln (2)\approx 3460\text{ J}$. Therefore, $Q=+3460\text{ J}$ of heat was transferred into the gas.

The Isobaric Process: Constant Pressure

In an isobaric process, pressure $P$ is held constant. This is common in processes open to the atmosphere or with a freely moving piston under a constant weight.

• Gas Law Constraint: With $P$ constant, $V\propto T$. If volume increases, temperature must increase proportionally.
• Work Done: Calculating work is straightforward: $W=P\Delta V$. Since $P$ is constant, it can be factored out.
• First Law Application: All terms are active. Heat transfer $Q$ changes both the internal energy and allows the system to do work: $Q=\Delta U+W$. For an ideal monatomic gas, $\Delta U=\frac{3}{2}nR\Delta T$.

Example: A gas at 200 kPa expands from 1.0 m³ to 3.0 m³ at constant pressure. The work done is $W=(200,000\text{ Pa})(2.0{\text{ m}}^3)=400,000\text{ J}$. This work, plus the increase in internal energy, must be supplied by the heat $Q$ added.

The Isochoric Process: Constant Volume

An isochoric (or isovolumetric) process occurs at constant volume ($\Delta V=0$). This happens in a sealed, rigid container.

• Gas Law Constraint: With $V$ constant, $P\propto T$. An increase in temperature leads to a direct increase in pressure.
• Work Done: Since the volume doesn't change, $W=0$. No area is enclosed under the process curve on a PV diagram.
• First Law Application: The law simplifies to $\Delta U=Q$. Any heat added or removed goes entirely into changing the internal energy (and thus temperature) of the gas.

Example: Heating air in a sealed, rigid bottle. As you add heat ($Q>0$), the temperature and pressure rise dramatically ($\Delta U>0$), but the gas does no work.

The Adiabatic Process: No Heat Transfer

An adiabatic process is defined by zero heat transfer: $Q=0$. The system is perfectly thermally insulated from its surroundings. This is an idealization approximating very fast processes (like a compression stroke in an engine) where there isn't time for significant heat exchange.

• Gas Law Constraint: The relationship is more complex. For an ideal gas, $P{V}^{\gamma }=\text{constant}$, where $\gamma$ (gamma) is the ratio of specific heats, $\gamma =C_P/C_V$. For a monatomic ideal gas, $\gamma =5/3$. This steeper curve on a PV diagram means that during an adiabatic compression, temperature increases more than in an isothermal compression.
• Work Done: Work is done at the expense of internal energy. The calculation requires integration using the adiabatic condition.
• First Law Application: With $Q=0$, we have $\Delta U=-W$. If work is done on the gas (a compression, $W<0$), then $\Delta U$ is positive and the gas heats up. If the gas does work on the surroundings (an expansion, $W>0$), it cools down ($\Delta U<0$).

Example: Compressing air rapidly in a bicycle pump. You do work on the gas ($W$ on the gas is positive, so $W$ by the gas is negative). Since the process is nearly adiabatic, $Q\approx 0$, so $\Delta U=-W>0$. The internal energy and temperature of the air increase, which you feel as the pump getting hot.

Common Pitfalls

1. Confusing Adiabatic and Isothermal Curves on a PV Diagram: Both show curved lines, but the adiabatic curve is steeper ($P{V}^{\gamma }=constant$) than the isothermal one ($PV=constant$). Remember: for a given starting point, an adiabatic expansion leads to a lower pressure and temperature than an isothermal expansion to the same final volume.
2. Misapplying the Work Formula: Using $W=P\Delta V$ only works for an isobaric (constant pressure) process. For non-constant pressure processes, work is the area under the curve on a PV diagram, which often requires integration or geometric calculation.
3. Incorrect First Law Sign Interpretation: The equation $\Delta U=Q-W$ is non-negotiable. A common mistake is to confuse the sign of $W$. If a gas is compressed, the surroundings do work on the gas. This means the gas does negative work on the surroundings. So, for a compression, $W$ (by the gas) is negative, making $-W$ positive and adding to $\Delta U$.
4. Forgetting What "Constant" Defines the Process: It’s easy to assume "constant" refers to all variables. Remember the definitions: isothermal (constant $T$), isobaric (constant $P$), isochoric (constant $V$), adiabatic (constant $Q=0$). The other variables change according to the gas law and first law constraints.

Summary

• A thermodynamic process is defined by which macroscopic variable (pressure, volume, temperature, or heat) is held constant, leading to specific constraints like $PV=\text{constant}$ (isothermal) or $P{V}^{\gamma }=\text{constant}$ (adiabatic).
• The work done by an ideal gas, $W$, is the area under the process curve on a Pressure-Volume (PV) diagram, directly calculable as $P\Delta V$ only for constant-pressure processes.
• The first law of thermodynamics, $\Delta U=Q-W$, governs all energy accounting. Internal energy change $\Delta U$ for an ideal gas depends solely on temperature change.
• In an isothermal process ($\Delta T=0$), $\Delta U=0$, so $Q=W$. In an adiabatic process ($Q=0$), $\Delta U=-W$. In an isobaric process, all terms are active, and in an isochoric process ($W=0$), $\Delta U=Q$.
• Successfully analyzing any process requires a stepwise approach: 1) Identify the constant parameter, 2) Apply the corresponding ideal gas law relationship, 3) Calculate work done, 4) Use the first law to find heat transfer or internal energy change.