6 Basic Thermodynamic Processes

Processes and the energy balance

A thermodynamic process changes a system from one equilibrium state to another. It can be represented on a pressure–volume diagram; for a quasi-equilibrium process, the area under the path gives the work done by the system.

Use the sign convention that heat entering the system is positive and work done by the system is positive. For a closed system with negligible changes in kinetic and potential energy, the first law is

The relations in this guide focus on a fixed amount of ideal gas. Let $m$ be its mass, $R$ its specific gas constant, and $c_v$ and $c_p$ its specific heats. When specific heats are constant, the internal-energy change is $\Delta U=mc_v\Delta T$; for an ideal gas, $c_p-c_v=R$.

Constant-pressure process

In an isobaric process, pressure remains fixed:

The boundary work is

For an ideal gas, $P\Delta V=mR\Delta T$. With constant specific heats, $\Delta U=mc_v\Delta T$, so the first law gives $Q=mc_p\Delta T$. During heating and expansion, some energy raises the gas’s internal energy and some is transferred as work. A weighted piston can approximate constant pressure as the gas expands.

Constant-volume process

In an isochoric process, volume remains fixed:

Because the boundary does not move, there is no boundary work, and the heat transfer equals the internal-energy change:

For an ideal gas with constant specific heat, $Q=mc_v\Delta T$. For example, heating gas in a rigid, sealed tank raises its internal energy and temperature without boundary work.

Isothermal process

In an isothermal process, temperature remains fixed:

For an ideal gas, $PV=mRT$, so $PV=\text{constant}$. Because ideal-gas internal energy depends only on temperature, $\Delta U=0$, and the first law gives $Q=W$.

For a quasi-equilibrium expansion or compression, the work is

During expansion, the gas does positive work and must receive an equal amount of heat to remain at the same temperature. This idealization requires enough time and heat transfer to maintain a uniform temperature.

Adiabatic process

An adiabatic process has no heat transfer, so $Q=0$. The first law then becomes

Consequently, an expanding gas that does work cools, while a gas compressed by work done on it warms. Adiabatic does not by itself mean reversible or constant entropy.

For a reversible (or quasi-static) adiabatic process of an ideal gas with constant specific heats, the process is isentropic and obeys

These pressure–volume and temperature–volume relations do not generally apply to an irreversible adiabatic process.

Polytropic process

A polytropic process follows the model relation

where $n$ is the polytropic exponent. For a fixed mass of ideal gas, boundary work in a quasi-equilibrium process is

The special case $n=1$ is isothermal for an ideal gas. Constant pressure corresponds to $n=0$, while a reversible adiabatic ideal-gas process with constant specific heats corresponds to $n=\gamma$. For any polytropic path, heat transfer can be found from $Q=\Delta U+W$. The relation is a useful model for processes such as gas compression and expansion; the exponent describes the path, not a universal property of the gas.

Choosing and applying a process model

Identify the process from the quantity held fixed or its defining relation: pressure for an isobaric process, volume for an isochoric process, temperature for an isothermal process, no heat transfer for an adiabatic process, and $PV^n=\text{constant}$ for a polytropic process.

Then calculate boundary work and apply $\Delta U=Q-W$. Use special ideal-gas relations only when their stated assumptions apply.

• specific heat at constant volume

• specific heat at constant pressure