Isothermal process

Squeeze the plunger of a bicycle pump slowly, with the nozzle open to the air, and something quietly strange happens: the gas trapped inside gets compressed, its pressure climbs, yet the gas never really heats up — because the heat generated leaks out through the metal walls into the room as fast as it is made. Do the opposite — let a gas expand slowly against a piston while sitting in a giant water bath — and its temperature refuses to drop, because the bath keeps feeding heat in from outside. Either way, one variable in the ideal gas equation is being held on a leash while the other two are set free to change. That leashed variable is temperature, and the process is called isothermal — literally "same heat," though the more precise word is "same temperature." The isothermal process is the gentlest and most forgiving change of state a gas can undergo, and for exactly that reason it is the workhorse of the CBSE Thermodynamics chapter: it is where Boyle's law, learned in earlier classes as a rule of thumb, reappears as a direct consequence of the ideal gas equation and the first law of thermodynamics. It is also the process where the mathematics is friendliest — no unfamiliar exponent $\gamma$ , just a natural logarithm — which is precisely why board papers lean on it so heavily for numerical questions. This article rebuilds the isothermal process from the ground up: why $PV$ stays constant, why the internal energy refuses to budge, why every joule of heat put in comes back out as work, and how to carry that logic cleanly through both a straightforward board-style numerical and a sharper, multi-step competitive problem — while flagging the handful of sign and log-vs-linear mistakes that quietly cost the most marks.

Key takeaways

An isothermal process is any change of state of a gas in which the temperature $T$ is held constant throughout, usually by keeping the gas in contact with a large heat reservoir and changing its volume slowly (quasi-statically).
Because $T$ and the number of moles $n$ are both fixed, the ideal gas equation $PV = nRT$ collapses to $PV = \text{constant}$ — this is exactly Boyle's law, and on a $P$ – $V$ diagram it traces a rectangular hyperbola called an isotherm.
The internal energy of an ideal gas depends only on temperature, so an unchanged $T$ means $\Delta U = 0$ for the whole process — no matter how large the volume change is.
Applying the first law $\Delta Q = \Delta U + \Delta W$ with $\Delta U = 0$ gives $\Delta Q = \Delta W$ : every joule of heat absorbed becomes work done by the gas, and every joule of work done on the gas leaves as heat.
The work done in an isothermal change from $V_1$ to $V_2$ is not $P\Delta V$ but a logarithm: $W = nRT\ln(V_2/V_1)$ , obtained by integrating $P\,dV$ with $P = nRT/V$ .
Expansion ( $V_2 > V_1$ ) gives positive $W$ and heat absorbed; compression ( $V_2 < V_1$ ) gives negative $W$ and heat released — the magnitudes are always equal because $\Delta U = 0$ .
Key idea: holding temperature constant forces $PV = \text{constant}$ and $\Delta U = 0$ , so the first law reduces to $\Delta Q = \Delta W = nRT\ln(V_2/V_1)$ — heat in equals work out, exactly, with no leftover for internal energy.

The core insight

In an isothermal process $T$ is fixed, so $PV = \text{constant}$ (Boyle's law) and $\Delta U = 0$ , which forces $\Delta Q = \Delta W = nRT\ln(V_2/V_1)$ — all heat exchanged is converted directly into work, with none left to change the gas's internal energy.

Where it fits in the syllabus

The isothermal process is named explicitly in the CBSE Class 11 Physics Thermodynamics unit, in the same breath as adiabatic, reversible, irreversible, and cyclic processes, with the relation $PV = \text{constant}$ given directly as prescribed content. You are expected to derive this relation from the ideal gas equation of state, connect it through the first law $\Delta Q = \Delta U + \Delta W$ to show $\Delta Q = \Delta W$ , and compute the work done during isothermal expansion or compression using the integral of $P\,dV$ . The chapter deliberately keeps heat engines, Carnot efficiency, and quantitative entropy out of the Class 11 prescription — this article stays inside that boundary and treats the isothermal process purely as a state change governed by the first law, not as a step inside an engine cycle.

The physical picture

Start from the equation every state of an ideal gas must obey, the ideal gas equation of state for $n$ moles:

PV = nRT

where: $P$ is pressure (pascal), $V$ is volume (cubic metre), $n$ is the number of moles (fixed for a sealed sample), $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant, and $T$ is the absolute temperature in kelvin. This single equation links all four quantities; an isothermal process is simply the special case where you freeze one of them, $T$ , and watch how the other two trade off. If $T$ and $n$ are both held fixed, the right-hand side $nRT$ is a single fixed number, so the equation says the product on the left must also stay fixed:

PV = \text{constant}, \qquad P_1V_1 = P_2V_2

where: the subscripts 1 and 2 label the initial and final states of the same gas sample. This is exactly Boyle's law, rediscovered here not as an empirical rule but as a direct consequence of the equation of state when temperature is pinned down. Physically it says pressure and volume move in lockstep but in opposite directions — squeeze the volume to half and the pressure exactly doubles, provided the temperature genuinely does not change. On a pressure–volume graph this relationship traces a curve, not a straight line: $P = (nRT)/V$ is an inverse proportionality, so the plotted curve is a rectangular hyperbola, called an isotherm. A different, fixed temperature gives a different isotherm — hotter gas sits on a hyperbola farther from the origin, because for the same volume a hotter gas must have higher pressure.

Figure

A pressure volume graph showing three isothermal curves as hyperbolas for three different fixed temperatures, with the hotter isotherm lying farther from the origin and an arrow marking rightward motion along one curve as expansion

Now bring in energy bookkeeping. For an ideal gas, internal energy $U$ is purely kinetic energy of random molecular motion, and it depends only on temperature — not on pressure or volume separately. Since an isothermal process starts and ends at the same $T$ :

\Delta U = 0

where: this holds for the entire process, however large the volume change, as long as the gas is ideal and $T$ truly returns to (or never leaves) its starting value. Substituting into the first law of thermodynamics,

\Delta Q = \Delta U + \Delta W \quad\Rightarrow\quad \Delta Q = \Delta W

where: $\Delta Q$ is heat absorbed by the gas and $\Delta W$ is work done by the gas on its surroundings. The physical reading is stark: in an isothermal process, heat and work become interchangeable, joule for joule — none of the incoming energy is ever diverted into raising the gas's internal store, because that store is temperature-locked.

Derivation AWork done by a gas expanding or compressing isothermally

The goal of this derivation is to obtain a usable formula for the work $W$ done by an ideal gas as it changes volume from $V_1$ to $V_2$ while its temperature is held fixed at $T$ , starting from nothing more than the mechanical definition of work done by a gas against a piston and the ideal gas equation of state — the result is the single most-used formula in this entire topic, and unlike the constant-pressure case it cannot be found by simple multiplication because the pressure itself is changing throughout the process.

Step 1. Write the general definition of work done by a gas during a small volume change. For an infinitesimal expansion $dV$ against a piston at pressure $P$ , the work done by the gas is $dW = P\,dV$ . To get the total work over a finite change from $V_1$ to $V_2$ , this must be summed — that is, integrated — because $P$ itself changes as $V$ changes: $W = \displaystyle\int_{V_1}^{V_2} P\,dV$ .

Step 2. Substitute the isothermal pressure. Since the process is isothermal, at every instant the gas still obeys $PV = nRT$ with $T$ fixed, so $P = nRT/V$ . Putting this into the integral gives $W = \displaystyle\int_{V_1}^{V_2} \frac{nRT}{V}\,dV$ . Because $n$ , $R$ , and $T$ are all constants during the process, they can be pulled outside the integral sign, leaving only the variable part: $W = nRT\displaystyle\int_{V_1}^{V_2} \frac{dV}{V}$ .

Step 3. Evaluate the integral. The standard result $\int dV/V = \ln V$ gives $W = nRT\,\big[\ln V\big]_{V_1}^{V_2} = nRT\big(\ln V_2 - \ln V_1\big)$ . Using the logarithm subtraction rule $\ln a - \ln b = \ln(a/b)$ , this collapses to the compact final form:

W = nRT\ln\left(\frac{V_2}{V_1}\right)

Step 4. Check the signs make physical sense. If the gas expands, $V_2 > V_1$ , so $V_2/V_1 > 1$ and $\ln(V_2/V_1) > 0$ , giving positive $W$ — the gas does positive work on its surroundings, exactly as expected for expansion. If the gas is compressed, $V_2 < V_1$ , so $V_2/V_1 < 1$ and $\ln(V_2/V_1) < 0$ , giving negative $W$ — work is done on the gas rather than by it. Because $\Delta U = 0$ throughout, the first law immediately gives the matching heat: $\Delta Q = W = nRT\ln(V_2/V_1)$ , positive (heat absorbed) for expansion and negative (heat released) for compression, always equal in magnitude to the work.

Worked examples

Board level - isothermal expansion of a fixed gas sample

2 moles of an ideal gas at 300 K expand isothermally and reversibly from a volume of $0.01\ \text{m}^3$ to $0.03\ \text{m}^3$ . Find the work done by the gas and the heat absorbed. (Take $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ .)

Step 1. Identify the quantities: $n = 2\ \text{mol}$ , $T = 300\ \text{K}$ , $V_1 = 0.01\ \text{m}^3$ , $V_2 = 0.03\ \text{m}^3$ . Since the process is isothermal, the work formula to use is $W = nRT\ln(V_2/V_1)$ , not $P\Delta V$ .

Step 2. Compute the volume ratio and its logarithm: $V_2/V_1 = 0.03/0.01 = 3$ , and $\ln 3 = 1.0986$ .

Step 3. Substitute into the work formula: $W = 2 \times 8.31 \times 300 \times \ln(3) = 4986 \times 1.0986$ . This gives $W \approx 5478\ \text{J}$ .

Step 4. Find the heat absorbed. Since the process is isothermal, $\Delta U = 0$ , so by the first law $\Delta Q = \Delta W$ . The gas therefore absorbs $\Delta Q \approx 5478\ \text{J}$ of heat, every joule of which leaves again as work done on the surroundings during the expansion — none of it raises the gas's internal energy.

Competitive level - isothermal compression with pressure found first

An ideal gas at pressure $3\times10^5\ \text{Pa}$ and volume $2\times10^{-3}\ \text{m}^3$ is compressed isothermally until its volume becomes $0.8\times10^{-3}\ \text{m}^3$ . Find the final pressure, the work done on the gas, and state whether heat is absorbed or released during the compression.

Step 1. Use Boyle's law $P_1V_1 = P_2V_2$ to find the final pressure, since $T$ is unchanged: $P_1V_1 = (3\times10^5)(2\times10^{-3}) = 600\ \text{J}$ (pascal times cubic metre is joule, a useful unit check).

Step 2. Solve for $P_2$ : $P_2 = \dfrac{600}{0.8\times10^{-3}} = 7.5\times10^5\ \text{Pa}$ . The pressure has risen to two and a half times its initial value, consistent with the volume shrinking to $0.4$ of its initial value.

Step 3. Find the work using $W = nRT\ln(V_2/V_1)$ , but here it is cleaner to note $nRT = P_1V_1 = 600\ \text{J}$ directly (no need to separately know $n$ or $T$ ). The volume ratio is $V_2/V_1 = 0.8\times10^{-3}/2\times10^{-3} = 0.4$ , so $\ln(0.4) = -0.916$ . Then $W = 600 \times (-0.916) = -549.6\ \text{J}$ . The work done by the gas is $-550\ \text{J}$ (to three significant figures), meaning $550\ \text{J}$ of work is done on the gas by whatever is compressing it.

Step 4. Determine the heat exchange. Since $\Delta U = 0$ (isothermal), the first law gives $\Delta Q = \Delta W = -550\ \text{J}$ . The negative sign means heat is released by the gas to its surroundings, equal in magnitude to the work done on it — the gas cannot store the compression energy as internal energy, so it must pass every joule straight out as heat. Timing note: in an exam, spot immediately that $nRT$ equals $P_1V_1$ (or $P_2V_2$ ) whenever you already know an initial pressure-volume pair — this saves you from separately hunting for $n$ and $T$ , and is often the fastest route to the work in a Boyle's-law-style question. Budget about 3 minutes for the pressure step and 2–3 minutes for the logarithmic work step; do not spend time re-deriving $nRT = P_1V_1$ from scratch each time.

Common traps

Assuming $\Delta Q = 0$ for an isothermal process. This is exactly backwards — it is the adiabatic process that has $\Delta Q = 0$ . In an isothermal process heat flows freely in or out; it is $\Delta U$ , not $\Delta Q$ , that is zero.
Using $\Delta U = \tfrac{3}{2}nR\Delta T$ -type formulas and forgetting $\Delta T = 0$ . Recomputing $\Delta U$ from a kinetic-theory formula is pointless busywork here; the isothermal shortcut $\Delta U = 0$ applies directly and always, whatever the gas.
Forgetting the natural logarithm in the work formula. Isothermal work is $nRT\ln(V_2/V_1)$ , never $nRT(V_2 - V_1)$ and never $P\Delta V$ — that linear formula belongs only to an isobaric (constant-pressure) process, a different situation entirely.
Dropping or mishandling the sign of the logarithm during compression. When $V_2 < V_1$ , the ratio $V_2/V_1$ is a fraction less than one, and $\ln$ of a fraction is negative — students who forget this report a positive work done by the gas during compression, which is physically wrong.
Confusing the steepness of isothermal and adiabatic curves on a $P$ – $V$ diagram. The adiabatic curve through the same starting point is always steeper than the isothermal curve, because the adiabatic exponent $\gamma$ exceeds 1, making adiabatic pressure fall faster with increasing volume.

You should be able to…

Derive $PV = \text{constant}$ (Boyle's law) for an isothermal process starting from the ideal gas equation of state.
Explain, in terms of internal energy depending only on temperature, why $\Delta U = 0$ and hence $\Delta Q = \Delta W$ for any isothermal process.
Derive and apply $W = nRT\ln(V_2/V_1)$ for the work done during isothermal expansion or compression, including correctly reading the sign.
Use Boyle's law $P_1V_1 = P_2V_2$ to relate initial and final pressures and volumes, including using $P_1V_1$ as a shortcut for $nRT$ .
Distinguish an isothermal process from an adiabatic process by which quantity is held fixed ( $T$ versus $\Delta Q$ ) and identify which curve is steeper on a $P$ – $V$ diagram.
Describe the physical conditions (thermal contact with a reservoir, slow/quasi-static change) needed to keep a real process close to isothermal.

Ask the tutor: Want to practise a few more isothermal work-and-heat problems with different volume ratios, or see how this compares directly to the adiabatic process next in the chapter? Ask the AI tutor for guided isothermal and adiabatic comparison problems.

If the gas is expanding and doing work, where does the energy for that work actually come from, since the internal energy does not change?

It comes entirely from the heat absorbed from the surroundings, not from the gas's own energy store. Picture the gas in contact with a large constant-temperature reservoir, like a water bath. As the gas pushes the piston outward, it would normally cool down (since expansion alone tends to lower a gas's temperature), but the reservoir instantly supplies exactly enough heat to replace what the expansion would have used, keeping $T$ fixed. So the reservoir is effectively the "battery" powering the work, and the gas itself is just a pass-through: heat comes in from the reservoir on one side and an equal amount of work goes out on the other, with $\Delta U$ staying at zero throughout.

Why is the work formula a logarithm here but just P times delta V for other processes?

Because in an isothermal process the pressure itself keeps changing as the volume changes — $P = nRT/V$ is not a constant, so you cannot simply multiply pressure by the volume change. You have to add up (integrate) the work done at every instant, using the true pressure at that instant, and integrating $1/V$ naturally produces a logarithm. In an isobaric process, by contrast, pressure genuinely stays fixed by construction, so the same integral becomes a simple multiplication, $W = P\Delta V$ . The formula you use always has to match how the pressure is actually behaving during that specific process.

Can a real process ever be perfectly isothermal, or is it just an idealisation?

It is an idealisation that real slow processes can approach very closely. True isothermal behaviour needs the gas to be in excellent thermal contact with a reservoir and for the volume change to happen slowly enough (quasi-statically) that heat has time to flow in or out and cancel any tendency for temperature to drift. A gas compressed inside a thin-walled metal cylinder immersed in a large water bath, pushed very slowly, behaves almost perfectly isothermally. A gas compressed suddenly, as in a bicycle pump stroke done quickly, behaves closer to adiabatic instead, because there simply is not enough time for heat to escape.

Is this related to Boyle's law from earlier classes, or is it a completely new law?

It is the very same law, just derived properly this time. In earlier classes Boyle's law, $PV = \text{constant}$ at constant temperature, was usually presented as an experimental rule found by testing gases in the lab. Here, starting from the ideal gas equation $PV = nRT$ , you can see exactly why it must be true: if $n$ and $T$ are both fixed, the right-hand side is a fixed number, so the left-hand side $PV$ has no choice but to stay fixed too. The isothermal process is simply Boyle's law given a proper thermodynamic home, connected to the first law so you can also talk about the heat and work involved, not just the pressure-volume relationship.