KAHIBARO
Table of Contents
Constant temperature change
An isothermal process is a thermodynamic process that happens at constant temperature. The word "isothermal" literally means "same temperature." If a gas changes its volume or pressure while its temperature stays fixed, the process is isothermal.
For an ideal gas, temperature is directly related to the average kinetic energy of the molecules. So in an isothermal process, the average molecular kinetic energy does not change. This means the internal energy of an ideal gas does not change either, because ideal gas internal energy depends only on temperature.
For an ideal gas in an isothermal process,
$$\Delta T = 0 \quad \Rightarrow \quad \Delta U = 0$$
This makes isothermal processes especially important, because the first law of thermodynamics becomes simpler.
Relation between pressure and volume
For an ideal gas, the ideal gas law is
$$PV = nRT$$
If the temperature $T$ is constant, and the amount of gas $n$ is also constant, then $nRT$ is constant. Therefore,
$$PV = \text{constant}$$
This means that pressure and volume change in opposite ways. If the volume increases, the pressure decreases. If the volume decreases, the pressure increases.
This relation is often called Boyle's law for a fixed amount of gas at constant temperature:
$$P_1 V_1 = P_2 V_2$$
In an isothermal process for an ideal gas,
$$PV = \text{constant}$$
or equivalently
$$P_1V_1 = P_2V_2$$
The pressure-volume graph
On a $P$ versus $V$ graph, an isothermal process is a curved line, not a straight line. Since
$$P = \frac{nRT}{V}$$
pressure is inversely proportional to volume. So the graph is a hyperbola.
Moving to the right along the curve means expansion. Moving to the left means compression.
Work in an isothermal process
When a gas expands or is compressed, work is done. For a general process, the work done by the gas is
$$W = \int_{V_1}^{V_2} P \, dV$$
For an isothermal ideal gas, we use
$$P = \frac{nRT}{V}$$
so
$$W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV$$
Since $n$, $R$, and $T$ are constant,
$$W = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV$$
Therefore,
$$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$
This is the work done by the gas.
If the gas expands, then $V_2 > V_1$, so the logarithm is positive and $W > 0$.
If the gas is compressed, then $V_2 < V_1$, so the logarithm is negative and $W < 0$.
Work done by an ideal gas during an isothermal process:
$$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$
Using $P_1V_1 = P_2V_2 = nRT$, the same formula can also be written as
$$W = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)$$
Heat transfer in an isothermal process
From the first law of thermodynamics,
$$\Delta U = Q - W$$
For an ideal gas in an isothermal process, $\Delta U = 0$, so
$$0 = Q - W$$
and therefore
$$Q = W$$
This means the heat added to the gas is exactly equal to the work done by the gas.
If the gas expands isothermally, it does work on the surroundings. To keep the temperature constant, heat must enter the gas.
If the gas is compressed isothermally, work is done on the gas. To keep the temperature constant, heat must leave the gas.
For an ideal gas in an isothermal process,
$$Q = W$$
because
$$\Delta U = 0$$
Physical picture
Imagine a gas inside a cylinder with a movable piston, with the cylinder in contact with a large thermal reservoir. The reservoir can supply or absorb heat so that the gas temperature remains constant.
If the gas expands slowly, its pressure drops as its volume increases. The gas does work on the piston. Without heat entering, the gas would cool. But because it is in thermal contact with the reservoir, heat flows into the gas and keeps the temperature constant.
If the gas is compressed slowly, the surroundings do work on the gas. Without heat leaving, the gas would warm up. But heat flows out to the reservoir, and the temperature stays constant.
Sign of heat and work
It is useful to keep track of signs carefully.
| Process | Volume change | Work done by gas $W$ | Heat $Q$ |
|---|---|---|---|
| Isothermal expansion | increases | positive | positive |
| Isothermal compression | decreases | negative | negative |
This table uses the common convention that $W$ is the work done by the gas, so the first law is
$$\Delta U = Q - W$$
Example using Boyle's law
Suppose a gas goes from volume $V_1 = 2.0 \, \text{m}^3$ and pressure $P_1 = 300 \, \text{kPa}$ to a new volume $V_2 = 5.0 \, \text{m}^3$ in an isothermal process. The final pressure is found from
$$P_1V_1 = P_2V_2$$
So,
$$P_2 = \frac{P_1V_1}{V_2}$$
$$P_2 = \frac{(300 \times 10^3)(2.0)}{5.0} = 120 \times 10^3 \, \text{Pa}$$
Thus,
$$P_2 = 120 \, \text{kPa}$$
The pressure decreases because the volume increased.
Example of work and heat
Suppose $n = 1.0 \, \text{mol}$ of an ideal gas expands isothermally at temperature $T = 300 \, \text{K}$ from $V_1 = 0.010 \, \text{m}^3$ to $V_2 = 0.020 \, \text{m}^3$.
The work done by the gas is
$$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$
$$W = (1.0)(8.31)(300)\ln\left(\frac{0.020}{0.010}\right)$$
$$W = 2493 \ln(2)$$
$$W \approx 2493 \times 0.693$$
$$W \approx 1728 \, \text{J}$$
So the gas does about $1.73 \times 10^3 \, \text{J}$ of work.
Since the process is isothermal for an ideal gas,
$$Q = W \approx 1728 \, \text{J}$$
The gas absorbs the same amount of heat.
Important limitations
The simple results in this chapter apply directly to an ideal gas. Real gases can behave differently, especially at high pressure or low temperature.
Also, an isothermal process is easiest to achieve when the change happens slowly enough for heat transfer to keep the temperature constant throughout the process. In practice, very rapid expansions or compressions are usually not perfectly isothermal.
Key results for an ideal gas in an isothermal process:
$$PV = \text{constant}$$
$$\Delta U = 0$$
$$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$
$$Q = W$$
Distinguishing isothermal processes
An isothermal process is recognized by constant temperature, not by constant pressure or constant volume. In this process, pressure and volume both change, but in such a way that their product stays constant for an ideal gas.
This makes isothermal processes a special and very useful model for studying how heat, work, and state variables are connected in thermodynamics.