4 Basic Thermodynamic Processes

States and process conditions

A thermodynamic process takes a system from one equilibrium state to another. For a fixed amount of ideal gas, pressure, volume, and absolute temperature are related by

where $n$ is the amount of gas in moles and $R$ is the ideal-gas constant. Each process name identifies a property that is held constant, except for an adiabatic process, which is defined by the absence of heat transfer.

Work, heat, and internal energy

Use the sign convention that $W$ is work done by the gas and $Q$ is heat added to it. The first law of thermodynamics is

For a quasi-static process, the gas pressure is well defined throughout, and the work is

Expansion gives positive work, while compression gives negative work. On a pressure–volume ($p$-$V$) diagram, work is the signed area under the process curve.

For an ideal gas, internal energy depends only on temperature. If the gas has constant heat capacity at constant volume, its internal-energy change is

The heat capacity at constant pressure is $C_P$. For an ideal gas, $C_P-C_V=R$, and the heat capacity ratio is defined as $\gamma=C_P/C_V$.

Isobaric process

An isobaric process holds pressure $p$ constant. For a fixed amount of ideal gas, the ideal-gas law then gives $V/T=\text{constant}$. Work and heat are

Heating gas while a frictionless piston moves under a fixed load can approximate an isobaric expansion. The gas expands and does work as its temperature and internal energy rise.

Isochoric process

An isochoric process holds volume constant. Since the volume does not change, there is no pressure–volume work:

Heating gas in a rigid, sealed container is an example. Its temperature and pressure rise while its volume stays fixed.

Isothermal process

An isothermal process holds temperature constant. For a fixed amount of ideal gas, $pV=nRT=\text{constant}$, so pressure decreases as volume increases. Because ideal-gas internal energy depends only on temperature, $\Delta U=0$.

For a reversible isothermal process, work and heat are

During expansion, heat must enter the gas to replace the energy it transfers as work. The logarithmic work formula applies to a reversible (or quasi-static) path, not to every process whose temperature happens to remain constant.

Adiabatic process

An adiabatic process has no heat transfer, so $Q=0$. This can occur when a system is well insulated or changes quickly enough that little heat is exchanged.

For a reversible adiabatic process of an ideal gas,

Its work and internal-energy change are

Work during adiabatic expansion comes at the expense of the gas’s internal energy, so the gas cools. During adiabatic compression, work is done on the gas and it warms. The relations involving $\gamma$ require a reversible adiabatic ideal-gas process; adiabatic alone means only that heat transfer is zero.

Compare and solve process problems

The processes differ in the condition they impose and in how heat and work relate to the state change:

• Isobaric: Pressure $p$ is constant; $W=p\Delta V$ and $Q=nC_P\Delta T$.

• Isochoric: Volume $V$ is constant; $W=0$ and $Q=nC_V\Delta T$.

• Isothermal, ideal gas: Temperature $T$ is constant; for a reversible path, $W=nRT\ln\!\left(\frac{V_2}{V_1}\right)$ and $Q=W$.

• Adiabatic: Heat transfer is zero; work depends on the path, with the reversible ideal-gas case given by the relations above.

To solve a process problem, identify what is held constant, use $pV=nRT$ to relate the initial and final states, calculate work using the appropriate path, and apply $\Delta U=Q-W$. Keep temperatures in kelvins and use consistent units.

In general, work depends on the path between states, whereas an ideal gas’s internal-energy change depends only on its temperature change. The first law connects heat, work, and internal energy for each process.