Isothermal Process on a PV Diagram

In an isothermal process, the gas changes state (volume changes, pressure changes) but its temperature stays exactly constant — because the gas is in thermal contact with something large enough to keep it from warming up or cooling down. That "something large" is called a heat reservoir or thermal bath, and the whole assumption is that it's so big that a little heat flowing in or out doesn't change its temperature noticeably.

Isothermal expansion is one of the most important processes in thermodynamics. It shows up in the Carnot cycle (problem 179), in real-world refrigeration and heat pumps, and anywhere you want a gas to do work while staying at a fixed temperature.

The key equation

For an ideal gas, $PV = nRT$. If $T$ is constant, then $P V = \text{const}$, which is Boyle's Law. On a PV diagram (pressure on the vertical axis, volume on the horizontal), constant $PV$ traces out a hyperbola:

$P = \dfrac{nRT}{V}$

A higher temperature gives a hyperbola that sits farther from the origin. Lower temperature, closer to the origin. Each temperature has its own unique isotherm — a fingerprint curve on the PV diagram.

Work done during an isothermal process

Here's where the geometry gets interesting. The work done by the gas on its surroundings during any process is the area under the PV curve from the starting volume to the ending volume:

$W = \int_{V_{1}}^{V_{2}} P \, dV$

For an isothermal process, $P = nRT/V$, so:

$W_{\text{iso}} = \int_{V_{1}}^{V_{2}} \dfrac{nRT}{V} \, dV = nRT \ln\!\left(\dfrac{V_{2}}{V_{1}}\right)$

That natural logarithm in the answer is the signature of an isothermal process. If $V_{2} > V_{1}$ (expansion), $W$ is positive — the gas is doing work on the environment. If $V_{2} < V_{1}$ (compression), $W$ is negative — the environment is doing work on the gas.

Heat transfer during an isothermal process

Now here's a subtle point that often confuses students. For an isothermal process:

• Internal energy doesn't change. $\Delta U = 0$, because internal energy of an ideal gas depends only on $T$, and $T$ is held constant.
• The first law says $\Delta U = Q - W$. So $0 = Q - W$, which means $Q = W$.

Interpretation: during isothermal expansion, the gas is doing positive work, so $Q$ must be positive too. Heat flows INTO the gas from the reservoir at exactly the rate needed to keep the temperature from dropping (because expansion normally cools a gas).

During isothermal compression, the gas has negative work done on it. $Q$ is also negative, meaning heat flows OUT of the gas into the reservoir — compressing normally heats a gas, and the reservoir absorbs that excess heat.

So in an isothermal process, heat and work are equal. All the heat that flows in becomes useful work done on the environment; none of it sticks around to raise the temperature.

Worked example: 300 K, expanding from 20 L to 40 L, with 1 mol

Let's compute everything for the default setup.

Step 1: Find the initial pressure.

$P_{1} = \dfrac{nRT}{V_{1}} = \dfrac{(1)(8.314)(300)}{0.020} = 124{,}710\;\text{Pa} \approx 1.231\;\text{atm}$

Step 2: Find the final pressure.

Since $PV$ is constant (isothermal):

$P_{2} = P_{1} \cdot \dfrac{V_{1}}{V_{2}} = 124{,}710 \cdot \dfrac{0.020}{0.040} = 62{,}355\;\text{Pa} \approx 0.615\;\text{atm}$

Pressure halves when volume doubles — Boyle's Law exactly.

Step 3: Compute the work done.

$W = nRT \ln\!\left(\dfrac{V_{2}}{V_{1}}\right) = (1)(8.314)(300)\,\ln\!\left(\dfrac{40}{20}\right)$

$= 2494.2 \times \ln 2$

$\approx 2494.2 \times 0.6931$

$\approx 1728.6\;\text{J}$

So the gas does about 1729 joules of work on its surroundings during this expansion. That's enough to lift about 176 kg by 1 meter — a meaningful amount of work for a single mole of gas.

Step 4: Find the heat flow.

Since $\Delta U = 0$ and $W = Q$:

$Q = W \approx 1729\;\text{J}$

1729 J of heat flows from the reservoir into the gas during the expansion. The gas never warms up because every joule of heat that enters is instantly converted into work pushing the piston outward.

Geometric intuition: the area under the curve

Turn on Show work area in the controls. You'll see the region between the hyperbola and the V-axis, from $V_{1}$ to $V_{2}$, filled with a shaded color. That shaded area is literally the work done — its numerical area (in pressure × volume units, which have units of energy) equals $W$.

Here's a beautiful property: for any two isothermal processes that start and end at the same volumes but at different temperatures, the higher-temperature one does more work. That's because the higher-$T$ isotherm sits higher on the PV diagram, so the area under it is larger. Visually, the hyperbola is "taller" everywhere, so the area underneath is bigger.

Toggle Show reference isotherms to see three hyperbolas drawn at $T = 200\;\text{K}$, $300\;\text{K}$, and $400\;\text{K}$ on the same axes. Each one has the same shape (a hyperbola), but they're scaled up as temperature increases.

Comparison to other processes

On the same PV diagram, we can compare isothermal to:

• Isobaric (constant pressure): a horizontal line. Work is $W = P \Delta V$, which is just a rectangle.
• Isochoric (constant volume): a vertical line. Work is zero — no volume change, no area under the curve.
• Adiabatic (no heat exchange, $Q = 0$): a steeper curve than isothermal (we'll explore this in problem 178). For the same starting state and same expansion, the adiabatic curve drops faster because the gas cools as it expands (no heat flowing in to sustain the temperature), so pressure drops faster too.

Real-world isothermal processes

True isothermal processes are idealizations — they require infinitely slow ("quasi-static") changes so that the gas always has time to equilibrate with the reservoir. Real processes are only approximately isothermal. But the approximation is useful:

• A slowly inflating balloon (over many seconds in a room at constant temperature) is roughly isothermal. The balloon walls let heat flow in/out fast enough that the air inside stays at room temperature.
• The "cold" stroke of a refrigerator compressor is designed to approach isothermal — the compressor dumps heat to the environment as it compresses the refrigerant, keeping the working fluid near ambient temperature.
• The Carnot engine's power stroke and exhaust stroke (you'll see this in problem 179) are both isothermal by design — they touch hot and cold reservoirs respectively.

Common mistakes

• Thinking "isothermal" means "insulated." It's the opposite. Isothermal means the gas is in thermal contact with a reservoir so it can exchange heat. An insulated process (where $Q = 0$) is called adiabatic, and it's different.
• Forgetting the natural logarithm. The isothermal work formula has $\ln(V_{2}/V_{1})$, not $V_{2}/V_{1}$ or $V_{2} - V_{1}$. The log is there because the pressure drops as you expand — and it drops proportionally, so integrating gives a log.
• Getting the sign of $W$ wrong. Physicists (and this problem) use the convention that $W$ is work done BY the gas. If the gas expands, $W > 0$; if compressed, $W < 0$. Chemists sometimes flip this convention — be careful which textbook you're reading.
• Expecting $\Delta U \ne 0$. For an ideal gas, internal energy depends only on temperature. Isothermal means $\Delta T = 0$, which means $\Delta U = 0$ period, end of story.
• Computing work as $P \Delta V$ for an isothermal process. That formula only works for constant pressure (isobaric). In an isothermal process, $P$ is changing continuously, so you have to integrate.

Try it

• Drag $V_{2}$ to watch the process slide along the isotherm. The shaded work area updates live.
• Drag temperature $T$ and watch the whole hyperbola shift up or down. Higher $T$ means the whole curve is higher, and the area under it (= work) grows.
• Try compression instead of expansion by setting $V_{2} < V_{1}$. The work becomes negative — the environment is doing work on the gas. Notice the HUD flips to show $Q < 0$ as well.
• Enable "Show reference isotherms" to see 3 isotherms at different temperatures superimposed. Your chosen process will sit on one of them.
• Enable "Animate particles in piston" to see a side-by-side view of the PV diagram plus an actual particle-filled box showing the gas expanding while particles keep the same speed (because $T$ is constant).
• Compare isothermal to adiabatic: the next problem (178) will put them side by side on the same PV axes. Preview: adiabatic drops faster.