Physics · Thermodynamics

Reversible isothermal work of an ideal gas

Q: What does this relationship describe?

$W_{\mathrm{by}}=nRT\ln\frac{V_f}{V_i}$ describes work delivered by ideal gas along a reversible isothermal piston path at fixed amount. Expansion is positive and compression negative. T is constant absolute temperature; $V_i$ and $V_f$ are accessible gas volumes in the same unit. Under the alternative convention, work on the gas has the opposite sign. The logarithmic result is path dependent: identical endpoints can produce different work under finite external load or free expansion. An isothermal ideal state comparison has $\Delta U=0$, but actual heat still follows from the full process balance. One isothermal branch alone does not determine engine efficiency.

Q: Where does the formula come from, and what does its structure tell us?

Start with actual piston work $\delta W_{\mathrm{by}}=p_{\mathrm{ext}}\,\mathrm dV$. In the frictionless reversible limit, external load differs only infinitesimally from equilibrium gas pressure. You may then replace $p_{\mathrm{ext}}$ with $p=\frac{nRT}{V}$. At fixed n and constant T, factor nRT out of the integral; $\int\frac{\mathrm dV}{V}$ gives $\ln\frac{V_f}{V_i}$. The dimensionless structure explains why equal volume ratios at equal n,T give equal reversible work. Quasistatic states alone do not guarantee reversibility with friction. A finite pressure jump still requires actual external load; vacuum supplies no load work even though accessible gas volume increases during expansion.

Q: What does a checked numerical calculation look like?

At 0.300 mol and 360 K, gas expands reversibly from 4.00 to 12.00 L. $nRT=897.9619628\,\mathrm J$ and $\ln3$ give $W_{\mathrm{rev}}=986.5120470\,\mathrm J$. Absolute endpoint pressures are 224490.4907 and 74830.16356 Pa. For the closed ideal process without external-energy change or other work, $Q_{\mathrm{rev}}=W_{\mathrm{rev}}$ and $\Delta U=0$. The work area lies between rectangles $p_f\Delta V=598.6413\,\mathrm J$ and $p_i\Delta V=1795.9239\,\mathrm J$. A second check gives $\frac{Q_{\mathrm{rev}}}{T}=2.740311242\,\mathrm{J\,K^{-1}}$, matching $nR\ln3$. Identical endpoints under constant external load pf instead deliver only 598.6413 J; do not transfer the logarithmic formula unchecked to that finite pressure step.

Q: Under what conditions may this relationship be used?

Require an ideal gas, fixed amount, constant absolute temperature and a mechanically reversible piston path without dissipative loads. Volume means space actually accessible to the gas. Additionally equating Q with Wgas requires a closed system, unchanged bulk kinetic and external potential energy, and absence of other work modes. Real gases can have volume-dependent internal energy even at equal temperature. Leakage and open flow installations need different balances. A merely slow piston can remain irreversible because of friction. Calculate an irreversible path with actual $p_{\mathrm{ext}}$; the reversible equilibrium curve is then only a comparison, not an automatically traversed work path.

Q: Which common mistake changes the result?

Isothermal does not automatically mean reversible. For free expansion in an isolated rigid assembly, $Q=W=\Delta U=0$ and ideal endpoint temperatures are equal. $nRT\ln\frac{V_f}{V_i}$ would only be work on a different reversible comparison path. Another error reverses signs: $V_f<V_i$ gives negative work by gas but positive work on gas. The logarithm also needs a dimensionless ratio; both volumes may be in litres when units cancel. Individual litres in pV with pascals instead require conversion to m³. Use natural logarithms and avoid rounding intermediate values before integration and the final energy-balance check.

For reversible isothermal expansion of an ideal gas, the work done by the gas depends logarithmically on the volume ratio.

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Formula

Calculate

W_{G a s} = n RT ln (\frac{V _{f}}{V _{i}})

W in J; n in mol; R in J/(mol·K); T in K; volume ratio dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
WGas	work done by the gas on its surroundings	J
$n$	amount of gas	mol
$R$	molar gas constant	J/(mol·K)
$T$	constant absolute temperature	K
$Vi,Vf$	initial and final volumes in the same unit	m³ or L, provided the same unit is used consistently

LaTeX

W_{Gas}=nRT\ln\!\left(\frac{V_f}{V_i}\right)

QUANTA STUDYquanta-study.de

Normalized reversible pressure 1/u between volume factors 1 and 3, above constant external load 1/3 with a smaller work area. — For n=0.300mol,T = 360 K and volume ratio 3: reversible 986.51 J; constant external load 598.64 J. The horizontal line is external load, not an equilibrium gas path.

Isothermal work calculator

Find the work and heat of an ideal gas in a reversible isothermal expansion or compression, or the final volume from the work.

Calculation model

Find

Fixed amount of substance n

Amount of the enclosed gas, e.g. 1 mol.

Absolute equilibrium temperature T

Constant absolute temperature during the process, e.g. 300 K; °C is converted.

Positive initial volume Vi

Volume before the process, e.g. 10 L.

Positive final volume Vf

Volume afterwards, e.g. 20 L; smaller than at the start means compression.

Filled with example values. Type your own and the result updates straight away.

Result: Work by gas on surroundings Wby

Wby = 1.729 kJ

With n = 1 mol, T = 300 K, Vi = 10 L and Vf = 20 L, Wby = 1.729 kJ.

Working · rearrange for Wby, substitute, evaluate

1. Rearrange for Wby
$W_{\rm by}=nRT\ln\frac{V_f}{V_i}$
2. Substitute values (in SI units)
- Vi = 10 L = 0.01 m³
- Vf = 20 L = 0.02 m³
$1\cdot 8.31446\cdot 300\cdot\ln\frac{0.02}{0.01}=1728.94$
3. Result
Wby = 1.729 kJ

Quick tips

Benchmark: if 1 mol at 300 K doubles its volume, it does about 1.73 kJ of work. Halving the volume costs exactly as much.
Sign check: if Vf is larger than Vi the logarithm is positive, the gas does work and takes in the same amount of heat. In a compression both reverse.

Assumptions and limits

Closed ideal gas sample with fixed composition and amount, constant positive absolute temperature, reversible frictionless mechanically quasistatic piston path. Accessible volumes positive. Wby positive in expansion, Won=−Wby. Q=Wby only with volume work alone and unchanged bulk kinetic/external potential energy. Irreversible work uses actual external load, not endpoints alone. n=0 is only a formal empty-sample limit of the forward model.

Derivation & background

Actual piston work is $W_{\mathrm{by}}=\int p_{\mathrm{ext}}\,\mathrm dV$ . For the mechanically reversible frictionless ideal-gas path at fixed n,T, replace pext with $p=\frac{nRT}{V}$ ; integration gives $nRT\ln\frac{V_f}{V_i}$ . Quasistatics alone does not exclude friction. Expansion gives positive and compression negative work by gas; work on gas has the opposite sign. Isothermal irreversible endpoints do not determine this logarithmic work.

Exam blueprint

Validity range

Reversible isothermal mechanically quasistatic frictionless piston path for ideal gas at fixed n with positive accessible Vi,Vf. T in kelvin, volume ratio dimensionless. Actual irreversible work uses pext; slow evolution alone does not guarantee reversibility.

Derivation steps

During isothermal expansion ideal internal energy stays unchanged, yet the gas can deliver work. The logarithmic formula describes a particular reversible path. Identical initial and final volumes alone do not determine actual work.

1Specify a closed ideal gas sample with fixed amount and positive constant temperature.
2Actual volume work uses the external load: $\delta W_{\mathrm{by}}=p_{\mathrm{ext}}\,\mathrm dV$ .
3Only in the frictionless mechanically reversible limit may $p_{\mathrm{ext}}$ be replaced by $p=\frac{nRT}{V}$ .
4Factor out the constant product: $W_{\mathrm{rev}}=nRT\int_{V_i}^{V_f}\frac{\mathrm dV}{V}$ .
5Integration gives $W_{\mathrm{rev}}=nRT\ln\frac{V_f}{V_i}$ ; expansion is positive, compression negative.
6Equal ideal-gas endpoint temperatures imply $\Delta U=0$ . With no external-energy change or other work, $Q=W_{\mathrm{by}}$ ; continue using actual external load for irreversible work.

Rearrangements

Volume ratio

V_f/V_i=exp(W_Gas/(nRT))

W_gas is positive for expansion under this convention.

Task variant

One mole of ideal gas at 300 K expands reversibly and isothermally from 10 to 20 L. Calculate work by the gas and work on the reversible return at the same temperature.

One mole of ideal gas at 300 K expands reversibly and isothermally from 10 L to 20 L.
$W_{\mathrm{by}}=RT\ln2\approx1.73\,\mathrm{kJ}$ .

Check: the reversible return halves the final volume back to 10 L, giving $W_{\mathrm{by}}\approx-1.73\,\mathrm{kJ}$ : equal magnitude, opposite sign.

Common mistakes

Using a variable with the wrong meaning or unit, thereby exceeding the model’s domain.

Match each variable to its definition, make units consistent and state the assumption being used.

Exam context

Problems in Thermodynamics assess formula choice, units, model assumptions and a brief result check.

Formula cluster

Thermodynamics

Isothermal gas work connects the ideal gas law with the first law and forms one part of reversible heat-engine processes.

ΔU = Q − W

First Law of Thermodynamics

pV = nRT

Ideal Gas Law

η = 1 − T_k/T_w

Carnot Efficiency

Worked example

One mole of ideal gas at 300 K expands reversibly and isothermally from 10 L to 20 L. $W_{\mathrm{by}}=RT\ln2\approx1.73\,\mathrm{kJ}$ . Check: the reversible return halves the final volume back to 10 L, giving $W_{\mathrm{by}}\approx-1.73\,\mathrm{kJ}$ : equal magnitude, opposite sign.

Applications

Analyze thermodynamic paths, heat engines and the distinction between work and heat.

How do you calculate with Work in Isothermal Expansion of an Ideal Gas?

Here is how to work through a typical Work in Isothermal Expansion of an Ideal Gas (Wgas=nRT ln(Vf/Vi)) task step by step:

1
Task
One mole of ideal gas at 300 K expands reversibly and isothermally from 10 to 20 L. Calculate work by the gas and work on the reversible return at the same temperature.
Solution path
1. One mole of ideal gas at 300 K expands reversibly and isothermally from 10 L to 20 L.
2. $W_{\mathrm{by}}=RT\ln2\approx1.73\,\mathrm{kJ}$ .
Check: the reversible return halves the final volume back to 10 L, giving $W_{\mathrm{by}}\approx-1.73\,\mathrm{kJ}$ : equal magnitude, opposite sign.

Understand and apply the equation

Why gas can do work without changing temperature

Gas pushes a piston outward and lifts a load, delivering energy as work. If temperature nevertheless remains constant, energy must enter during an appropriate isothermal expansion. The first law explains this heat transfer; the gas equation supplies the pressure curve for the reversible model path.[1][2]

Isothermal specifies the temperature condition. Reversible additionally specifies an ideal limiting process without dissipative losses and with vanishingly small driving differences. Rapid expansion against a fixed external pressure can reach the same endpoints without delivering the same work.[1]

The logarithmic work formula requires a closed ideal gas sample of fixed composition, positive constant absolute temperature and reversible volume work. Work is positive when the gas delivers it to the surroundings. This sign convention remains unchanged during compression.[1]

Which work sign applies to reversible isothermal expansion and compression?

W_{\mathrm{by,rev}}=nRT\ln\frac{V_f}{V_i}

Expansion gives positive work by the gas; reversible compression has negative sign.[1]

n is fixed amount, T constant absolute gas temperature, and V actual accessible gas volume. Vi and Vf must use the same unit. A litre ratio is therefore valid, whereas individual litres in pV with pascals require conversion. With work on the gas positive, the same transfer is $W_{\mathrm{on}}=-W_{\mathrm{by}}$ .[1]

From moving piston to logarithm

Pressure decreases during expansion. Multiplying initial pressure by the whole volume change therefore overestimates reversible work. Divide the path into small volume steps, add the contributions $p\,\mathrm dV$ , and take the integral.[1]

\begin{aligned}\delta W_{\mathrm{by}}&=p_{\mathrm{ext}}\,\mathrm dV\\W_{\mathrm{by,rev}}&=\int_{V_i}^{V_f}\frac{nRT}{V}\,\mathrm dV\\&=nRT\ln\frac{V_f}{V_i}\end{aligned}

In the reversible frictionless limit, internal pressure and external load differ only infinitesimally.[1]

External load determines actual piston work delivered outward. Only mechanical quasistatics without dissipative loads permits replacing it by equilibrium gas pressure. Ideal equation of state and constant T give $p(V)=\frac{nRT}{V}$ ; integrating $\frac1V$ gives the natural logarithm. Quasistatic means slow enough for defined states, but alone does not guarantee reversibility in the presence of friction.[1]

Equal endpoints, different work areas

QUANTA STUDYquanta-study.de

The solid curve is the reversible working pressure in Example 1; the dashed line is the constant external pressure in Example 2. Area beneath the respective working pressure gives about 986.5 J or 598.6 J. The second line is not an equilibrium path of gas pressure. Numerically,

1\,\mathrm{kPa}\cdot1\,\mathrm L=1\,\mathrm J

.[1]

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Heat replaces delivered energy

\begin{aligned}\Delta U&=0\\Q&=W_{\mathrm{by}}\\\Delta S_{\mathrm{gas}}&=nR\ln\frac{V_f}{V_i}\end{aligned}

This concerns fixed n of an ideal gas in a closed system without bulk kinetic or external potential-energy change and without additional work modes.[2][3]

For fixed composition, ideal U depends only on T, not volume. During reversible isothermal expansion the gas continually absorbs heat to replace delivered work. Adiabatic expansion is therefore not an interchangeable path. Entropy is a state property: the same change can occur along an irreversible path with different heat and additional generation. The first law of thermodynamics checks this process balance using the same work convention throughout.[2][3]

Three examples with different paths

Triple volume in a reversible comparison

Given: $n=0.300\,\mathrm{mol}$ , $T=360\,\mathrm K$ , $V_i=4.00\,\mathrm L$ , $V_f=12.00\,\mathrm L$ . Closed ideal gas in a reversible isothermal frictionless piston process.
Find: Find work, heat and endpoint pressures.

Approach: The comparison path is reversible and isothermal. $W_{\mathrm{rev}}=nRT\ln\frac{V_f}{V_i}$ . Find energy scale and ratio.
$\begin{aligned}nRT&=897.9619628\,\mathrm J\\{} \frac{V_f}{V_i}&=3\end{aligned}$
Evaluate the integral.
$W_{rev}=nRT\ln3=986.5120470\,\mathrm J$
Convert litres for pressure calculation.
$\begin{aligned}p_i&\approx224490.4907\,\mathrm{Pa}\\p_f&\approx74830.16356\,\mathrm{Pa}\end{aligned}$

Result: The gas delivers about 986.51 J of work and absorbs the same heat. $\Delta U=0$ ; absolute pressure falls to one third.

Check: The area lies between the final-pressure rectangle of 598.64 J and the initial-pressure rectangle of 1795.92 J. Also, $\frac{Q_{\mathrm{rev}}}{T}=2.740311242\,\mathrm{J\,K^{-1}}=nR\ln3$ . The geometric and entropy checks agree.

The same endpoints against constant external load

Given: The same amount and endpoint states as above. Constant external pressure $p_{\mathrm{ext}}=p_f=74830.16356\,\mathrm{Pa}$ . A 360 K heat bath establishes the same final temperature. No other work; external energies agree initially and finally.
Find: Calculate actual work and total entropy change.

Approach: Actual constant external load determines the work. $W_{\mathrm{Last}}=p_{\mathrm{ext}}(V_f-V_i)$ . Integrate actual load rather than reversible gas-pressure curve.
$W_{step}=p_{ext}(V_f-V_i)=598.6413085\,\mathrm J$
Equal ideal endpoint T gives $\Delta U=0$ .
$\begin{aligned}Q_{step}&=W_{step}\\{} \Delta S_{bath}&=-\frac{Q_{\mathrm{step}}}{360}\end{aligned}$
Add gas state change and bath.
$\Delta S_{total}=2.740311242-1.662892524=1.077418718\,\mathrm{J/K}$

Result: The finite step delivers about 598.64 J instead of 986.51 J and generates 1.07742 J/K total entropy.

Check: Gas endpoints and ΔSgas are identical. A well-defined T at every transient instant is not asserted; constant load determines work and the complete endpoint balance determines Q.

In the following compression, internal energy of the closed ideal gas remains unchanged. Work input and heat rejection must therefore balance.[1][2]

Compression: where does the input work go?

Given: $n=0.200\,\mathrm{mol}$ , $T=320\,\mathrm K$ , $V_i=6.00\,\mathrm L$ , $V_f=3.00\,\mathrm L$ . Closed ideal gas, reversible isothermal compression, no changes in external energies.
Find: Find work by the gas, work on the gas and heat.

Approach: the volume ratio is below one, making its natural logarithm negative. For the unchanged ideal gas, equal temperature means equal internal energy.
$\begin{aligned}W_{\mathrm{by}}&=nRT\ln\frac{V_f}{V_i}\\W_{\mathrm{on}}&=-W_{\mathrm{by}}\\Q&=W_{\mathrm{by}}\quad(\Delta U=0)\end{aligned}$
Substitute: both volumes can remain in litres within the ratio.
$\begin{aligned}W_{\mathrm{by}}&=(0.200)(8.314462618)(320)\ln\frac{3.00}{6.00}\,\mathrm J\\&\approx-368.841\,\mathrm J\\W_{\mathrm{on}}&\approx+368.841\,\mathrm J\\Q&\approx-368.841\,\mathrm J\end{aligned}$

Result: About 369 J of work is done on the gas, and it rejects the same energy as heat.

Check: Check: $Q-W_{\mathrm{by}}=0$ . On the reversible return at the same temperature, the logarithm takes the reciprocal argument and work reverses sign. Without heat rejection this compression would not be isothermal.

Isothermal does not guarantee logarithmic work

For free expansion into vacuum, $p_{\mathrm{ext}}=0$ and $W_{\mathrm{by}}=0$ . An isolated rigid assembly also has $Q=0$ ; ideal equilibrium endpoints have equal T. Gas entropy nevertheless increases. The real evolution can be temporarily inhomogeneous, so an equilibrium p(V) curve is not its actual work path. Entropy change along a reversible path can compare these same endpoints without assigning heat transfer to the actual free-expansion path.[2][3]

Real gases need an appropriate equation of state and can have volume-dependent U even at fixed T. Leakage, flow installations, piston friction and additional electrical work change the balance. For the same initial state, prescribed expansion and one heat bath, the reversible comparison gives maximum delivered work; comparing paths requires the same boundary conditions.[1][2]

How do you infer volume ratio from reversible work?

\begin{aligned}\frac{V_f}{V_i}&=\exp\left(\frac{W_{rev}}{nRT}\right)\\{} T&=\frac{W_{rev}}{nR\ln(\frac{V_f}{V_i})}\end{aligned}

Temperature inversion requires volume ratio different from 1 and the reversible model path.[1]

For small relative change $\varepsilon=\frac{V_f-V_i}{V_i}$ , $\ln(1+\varepsilon)\approx\varepsilon$ . Then $nRT\varepsilon$ approximates $p_i\Delta V$ . For larger expansion this initial-pressure rectangle exceeds actual reversible work because p falls. The natural logarithm is dimensionless; taking the logarithm of an isolated dimensional litre value is invalid.[1]

Five exercises with solutions

Calculate · Foundation

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work $W_{\mathrm{by}}$ done by the gas.

Show hint

$\frac{V_f}{V_i}$ <1.

Learning goal: Calculate reversible compression work with the selected sign convention.

Show solution and calculation

Approach: use the natural logarithm of the volume ratio for the reversible isothermal path.
Compression makes the logarithm negative.
Work on the gas is about +1062.78 J; work by the gas and heat are negative.

Check: equal ideal-gas endpoint temperatures imply $\Delta U=0$ , so heat and work must cancel in the energy balance.

\begin{aligned}W_{\mathrm{by}}&=nRT\ln\frac{V_f}{V_i}\\&=(0.45)(8.314462618)(310)\ln\frac25\,\mathrm J\\&\approx-1062.775873\,\mathrm J\\W_{\mathrm{on}}&=-W_{\mathrm{by}}\\Q&=W_{\mathrm{by}}\end{aligned}

Calculate · Application

A reversible path has $W=nRT\ln4$ . What is $\frac{V_f}{V_i}$ ?

Show hint

Apply the exponential.

Learning goal: Infer volume ratio from a logarithmic work expression.

Show solution and calculation

Approach: divide the work equation by the positive product of amount, gas constant and temperature, then exponentiate.
Volume ratio is four.

Check: the reversible return at the same temperature uses the reciprocal ratio and gives opposite work; net work vanishes.

\begin{aligned}\frac{W}{nRT}&=\ln\frac{V_f}{V_i}=\ln4\\\frac{V_f}{V_i}&=\exp(\ln4)=4\\W_{\mathrm{return}}&=nRT\ln\frac14=-W\end{aligned}

Interpret · Application

Ideal gas freely expands in an isolated rigid assembly from V to 2V. What are Q,W and ΔU?

Show hint

Actual path is not reversible.

Learning goal: Distinguish heat, work and internal energy in free isolated expansion.

Show solution and calculation

Approach: inspect actual transfer mechanisms.
The isolated rigid outer boundary allows neither heat nor work; free expansion into vacuum likewise supplies no volume work.
Internal energy and, in the ideal model, endpoint temperature therefore stay unchanged.
Entropy nevertheless increases.

Check: reversible comparison work belongs to a different process and must not be used here.

\begin{aligned}Q&=0\\W&=0\\\Delta U&=Q-W=0\\T_f&=T_i\\\Delta S&=nR\ln\frac{2V}{V}=nR\ln2>0\end{aligned}

Interpret · Challenge

An ideal gas expands at constant temperature with $nRT=600\,\mathrm J$ , from 2.0 L to 4.0 L. Path A is reversible and quasistatic; path B proceeds against constant external pressure 150 kPa and ends at equilibrium. Compare work done by the gas and heat supplied. Macroscopic kinetic energies vanish in the initial and final states.

Show hint

For A integrate gas pressure; for B integrate constant external pressure. For this ideal gas, ΔU vanishes for both isothermal paths.

Learning goal: Compare path-dependent work for the same isothermal state change.

Show solution and calculation

Approach: find the area under the respective working pressure over the volume range.
Reversible work is about 415.89 J; constant-external-pressure work is 300 J.
With $\Delta U=Q-W$ and $\Delta U=0$ , supplied heat equals work on each path.

Check: final gas pressure is indeed 150 kPa. Equal endpoints and temperature therefore do not imply equal work or heat.

\begin{aligned}W_A&=nRT\ln\frac{V_f}{V_i}=600\ln2\,\mathrm J\approx415.89\,\mathrm J\\{}W_B&=p_{\mathrm{ext}}(V_f-V_i)=(150\,000)(0.0020)\,\mathrm J=300\,\mathrm J\\{}Q_A&=W_A\\{}Q_B&=W_B\\{}p_f&=\frac{600\,\mathrm J}{0.0040\,\mathrm{m^3}}=150\,\mathrm{kPa}\end{aligned}

[1][2]

Error analysis · Application

An ideal gas expands reversibly and isothermally from 1.0 L to 10.0 L with $nRT=200\,\mathrm J$ . A solution uses base-ten logarithm and obtains $W=200\,\mathrm J$ . Correct the value and explain why an arbitrary logarithm does not satisfy the same formula. W denotes work done by the gas.

Show hint

The integral of $\frac1V$ is ln V. Another logarithm requires a conversion factor; its argument must be dimensionless.

Learning goal: Connect the natural logarithm to the work integral.

Show solution and calculation

Approach: obtain work by integrating reversible gas pressure.
This produces the natural logarithm of the volume ratio, giving about 460.52 J.
A base-ten logarithm would require an additional factor ln 10.

Check: exponentiating the correct work divided by nRT recovers volume ratio 10; using 200 J would instead give only e.

\begin{aligned}W&=\int_{V_i}^{V_f}\frac{nRT}{V}\,\mathrm dV=nRT\ln\frac{V_f}{V_i}\\&=200\ln10\,\mathrm J\approx460.52\,\mathrm J\\\ln x&=(\ln10)\log_{10}x\\\exp\!\left(\frac{W}{nRT}\right)&=10\end{aligned}

[1]

Compare work areas meaningfully

For one mole at 300 K and doubled volume, reversible work remains 1728.943896 J, about 1.73 kJ. In a p–V plot, area under the reversible isotherm means work, not stored internal energy. A rectangular external-load path and curved reversible path need common axis scaling; otherwise the graphic conveys a false work ordering.[1]

In a heat-engine cycle one isothermal branch is only part of total work. Sum every branch using one sign convention to obtain the net result. One branch alone determines neither machine efficiency nor required driving power without timing information.[2]

What to carry into your next calculation

Reversible isothermal volume work is $W_{\mathrm{by}}=nRT\ln\frac{V_f}{V_i}$ . This formula describes a particular path. A closed ideal gas has unchanged internal energy at equal temperature; with the stated energy-balance assumptions, $Q=W_{\mathrm{by}}$ .[1][2]

Sources for this explanation

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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Quanta. Work in Isothermal Expansion of an Ideal Gas. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/isotherme-arbeit-ideales-gas

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Frequently asked questions about Work in Isothermal Expansion of an Ideal Gas

What does this relationship describe?+

$W_{\mathrm{by}}=nRT\ln\frac{V_f}{V_i}$ describes work delivered by ideal gas along a reversible isothermal piston path at fixed amount. Expansion is positive and compression negative. T is constant absolute temperature; $V_i$ and $V_f$ are accessible gas volumes in the same unit. Under the alternative convention, work on the gas has the opposite sign. The logarithmic result is path dependent: identical endpoints can produce different work under finite external load or free expansion. An isothermal ideal state comparison has $\Delta U=0$ , but actual heat still follows from the full process balance. One isothermal branch alone does not determine engine efficiency.

Where does the formula come from, and what does its structure tell us?+

Start with actual piston work $\delta W_{\mathrm{by}}=p_{\mathrm{ext}}\,\mathrm dV$ . In the frictionless reversible limit, external load differs only infinitesimally from equilibrium gas pressure. You may then replace $p_{\mathrm{ext}}$ with $p=\frac{nRT}{V}$ . At fixed n and constant T, factor nRT out of the integral; $\int\frac{\mathrm dV}{V}$ gives $\ln\frac{V_f}{V_i}$ . The dimensionless structure explains why equal volume ratios at equal n,T give equal reversible work. Quasistatic states alone do not guarantee reversibility with friction. A finite pressure jump still requires actual external load; vacuum supplies no load work even though accessible gas volume increases during expansion.

What does a checked numerical calculation look like?+

At 0.300 mol and 360 K, gas expands reversibly from 4.00 to 12.00 L. $nRT=897.9619628\,\mathrm J$ and $\ln3$ give $W_{\mathrm{rev}}=986.5120470\,\mathrm J$ . Absolute endpoint pressures are 224490.4907 and 74830.16356 Pa. For the closed ideal process without external-energy change or other work, $Q_{\mathrm{rev}}=W_{\mathrm{rev}}$ and $\Delta U=0$ . The work area lies between rectangles $p_f\Delta V=598.6413\,\mathrm J$ and $p_i\Delta V=1795.9239\,\mathrm J$ . A second check gives $\frac{Q_{\mathrm{rev}}}{T}=2.740311242\,\mathrm{J\,K^{-1}}$ , matching $nR\ln3$ . Identical endpoints under constant external load pf instead deliver only 598.6413 J; do not transfer the logarithmic formula unchecked to that finite pressure step.

Under what conditions may this relationship be used?+

Require an ideal gas, fixed amount, constant absolute temperature and a mechanically reversible piston path without dissipative loads. Volume means space actually accessible to the gas. Additionally equating Q with Wgas requires a closed system, unchanged bulk kinetic and external potential energy, and absence of other work modes. Real gases can have volume-dependent internal energy even at equal temperature. Leakage and open flow installations need different balances. A merely slow piston can remain irreversible because of friction. Calculate an irreversible path with actual $p_{\mathrm{ext}}$ ; the reversible equilibrium curve is then only a comparison, not an automatically traversed work path.

Which common mistake changes the result?+

Isothermal does not automatically mean reversible. For free expansion in an isolated rigid assembly, $Q=W=\Delta U=0$ and ideal endpoint temperatures are equal. $nRT\ln\frac{V_f}{V_i}$ would only be work on a different reversible comparison path. Another error reverses signs: $V_f<V_i$ gives negative work by gas but positive work on gas. The logarithm also needs a dimensionless ratio; both volumes may be in litres when units cancel. Individual litres in pV with pascals instead require conversion to m³. Use natural logarithms and avoid rounding intermediate values before integration and the final energy-balance check.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §3.2: Work, Heat, and Internal Energy.

More formulas: Heat and thermodynamics

Q = m \cdot c \cdot \Delta T

Heat Capacity: Q = m·c·ΔT

P = \sigma \cdot A \cdot T^4

Stefan-Boltzmann Law

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

Q = m L

Latent heat during a phase change

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Symbol

Meaning

Unit

WGas

work done by the gas on its surroundings

n

amount of gas

mol

R

molar gas constant

J/(mol·K)

T

constant absolute temperature

Vi,Vf

initial and final volumes in the same unit

m³ or L, provided the same unit is used consistently

Derivation & background

Exam blueprint

Validity range

Derivation steps

1Specify a closed ideal gas sample with fixed amount and positive constant temperature.
2Actual volume work uses the external load: $\delta W_{\mathrm{by}}=p_{\mathrm{ext}}\,\mathrm dV$ .
3Only in the frictionless mechanically reversible limit may $p_{\mathrm{ext}}$ be replaced by $p=\frac{nRT}{V}$ .
4Factor out the constant product: $W_{\mathrm{rev}}=nRT\int_{V_i}^{V_f}\frac{\mathrm dV}{V}$ .
5Integration gives $W_{\mathrm{rev}}=nRT\ln\frac{V_f}{V_i}$ ; expansion is positive, compression negative.
6Equal ideal-gas endpoint temperatures imply $\Delta U=0$ . With no external-energy change or other work, $Q=W_{\mathrm{by}}$ ; continue using actual external load for irreversible work.

How do you calculate with Work in Isothermal Expansion of an Ideal Gas?

Here is how to work through a typical Work in Isothermal Expansion of an Ideal Gas (Wgas=nRT ln(Vf/Vi)) task step by step:

Task

One mole of ideal gas at 300 K expands reversibly and isothermally from 10 to 20 L. Calculate work by the gas and work on the reversible return at the same temperature.

Solution path

One mole of ideal gas at 300 K expands reversibly and isothermally from 10 L to 20 L.
$W_{\mathrm{by}}=RT\ln2\approx1.73\,\mathrm{kJ}$ .

Check: the reversible return halves the final volume back to 10 L, giving $W_{\mathrm{by}}\approx-1.73\,\mathrm{kJ}$ : equal magnitude, opposite sign.

Why gas can do work without changing temperature

Which work sign applies to reversible isothermal expansion and compression?

W_{\mathrm{by,rev}}=nRT\ln\frac{V_f}{V_i}

Expansion gives positive work by the gas; reversible compression has negative sign.[1]

From moving piston to logarithm

\begin{aligned}\delta W_{\mathrm{by}}&=p_{\mathrm{ext}}\,\mathrm dV\\W_{\mathrm{by,rev}}&=\int_{V_i}^{V_f}\frac{nRT}{V}\,\mathrm dV\\&=nRT\ln\frac{V_f}{V_i}\end{aligned}

In the reversible frictionless limit, internal pressure and external load differ only infinitesimally.[1]

Equal endpoints, different work areas

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1\,\mathrm{kPa}\cdot1\,\mathrm L=1\,\mathrm J

.[1]

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Heat replaces delivered energy

\begin{aligned}\Delta U&=0\\Q&=W_{\mathrm{by}}\\\Delta S_{\mathrm{gas}}&=nR\ln\frac{V_f}{V_i}\end{aligned}

This concerns fixed n of an ideal gas in a closed system without bulk kinetic or external potential-energy change and without additional work modes.[2][3]

Three examples with different paths

Triple volume in a reversible comparison

Given: $n=0.300\,\mathrm{mol}$ , $T=360\,\mathrm K$ , $V_i=4.00\,\mathrm L$ , $V_f=12.00\,\mathrm L$ . Closed ideal gas in a reversible isothermal frictionless piston process.
Find: Find work, heat and endpoint pressures.

Approach: The comparison path is reversible and isothermal. $W_{\mathrm{rev}}=nRT\ln\frac{V_f}{V_i}$ . Find energy scale and ratio.
$\begin{aligned}nRT&=897.9619628\,\mathrm J\\{} \frac{V_f}{V_i}&=3\end{aligned}$
Evaluate the integral.
$W_{rev}=nRT\ln3=986.5120470\,\mathrm J$
Convert litres for pressure calculation.
$\begin{aligned}p_i&\approx224490.4907\,\mathrm{Pa}\\p_f&\approx74830.16356\,\mathrm{Pa}\end{aligned}$

Result: The gas delivers about 986.51 J of work and absorbs the same heat. $\Delta U=0$ ; absolute pressure falls to one third.

The same endpoints against constant external load

Given: The same amount and endpoint states as above. Constant external pressure $p_{\mathrm{ext}}=p_f=74830.16356\,\mathrm{Pa}$ . A 360 K heat bath establishes the same final temperature. No other work; external energies agree initially and finally.
Find: Calculate actual work and total entropy change.

Approach: Actual constant external load determines the work. $W_{\mathrm{Last}}=p_{\mathrm{ext}}(V_f-V_i)$ . Integrate actual load rather than reversible gas-pressure curve.
$W_{step}=p_{ext}(V_f-V_i)=598.6413085\,\mathrm J$
Equal ideal endpoint T gives $\Delta U=0$ .
$\begin{aligned}Q_{step}&=W_{step}\\{} \Delta S_{bath}&=-\frac{Q_{\mathrm{step}}}{360}\end{aligned}$
Add gas state change and bath.
$\Delta S_{total}=2.740311242-1.662892524=1.077418718\,\mathrm{J/K}$

Result: The finite step delivers about 598.64 J instead of 986.51 J and generates 1.07742 J/K total entropy.

Check: Gas endpoints and ΔSgas are identical. A well-defined T at every transient instant is not asserted; constant load determines work and the complete endpoint balance determines Q.

In the following compression, internal energy of the closed ideal gas remains unchanged. Work input and heat rejection must therefore balance.[1][2]

Compression: where does the input work go?

Given: $n=0.200\,\mathrm{mol}$ , $T=320\,\mathrm K$ , $V_i=6.00\,\mathrm L$ , $V_f=3.00\,\mathrm L$ . Closed ideal gas, reversible isothermal compression, no changes in external energies.
Find: Find work by the gas, work on the gas and heat.

Approach: the volume ratio is below one, making its natural logarithm negative. For the unchanged ideal gas, equal temperature means equal internal energy.
$\begin{aligned}W_{\mathrm{by}}&=nRT\ln\frac{V_f}{V_i}\\W_{\mathrm{on}}&=-W_{\mathrm{by}}\\Q&=W_{\mathrm{by}}\quad(\Delta U=0)\end{aligned}$
Substitute: both volumes can remain in litres within the ratio.
$\begin{aligned}W_{\mathrm{by}}&=(0.200)(8.314462618)(320)\ln\frac{3.00}{6.00}\,\mathrm J\\&\approx-368.841\,\mathrm J\\W_{\mathrm{on}}&\approx+368.841\,\mathrm J\\Q&\approx-368.841\,\mathrm J\end{aligned}$

Result: About 369 J of work is done on the gas, and it rejects the same energy as heat.

Isothermal does not guarantee logarithmic work

How do you infer volume ratio from reversible work?

\begin{aligned}\frac{V_f}{V_i}&=\exp\left(\frac{W_{rev}}{nRT}\right)\\{} T&=\frac{W_{rev}}{nR\ln(\frac{V_f}{V_i})}\end{aligned}

Temperature inversion requires volume ratio different from 1 and the reversible model path.[1]

Five exercises with solutions

Calculate · Foundation

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work $W_{\mathrm{by}}$ done by the gas.

Show hint

$\frac{V_f}{V_i}$ <1.

Learning goal: Calculate reversible compression work with the selected sign convention.

Show solution and calculation

Approach: use the natural logarithm of the volume ratio for the reversible isothermal path.
Compression makes the logarithm negative.
Work on the gas is about +1062.78 J; work by the gas and heat are negative.

Check: equal ideal-gas endpoint temperatures imply $\Delta U=0$ , so heat and work must cancel in the energy balance.

\begin{aligned}W_{\mathrm{by}}&=nRT\ln\frac{V_f}{V_i}\\&=(0.45)(8.314462618)(310)\ln\frac25\,\mathrm J\\&\approx-1062.775873\,\mathrm J\\W_{\mathrm{on}}&=-W_{\mathrm{by}}\\Q&=W_{\mathrm{by}}\end{aligned}

Calculate · Application

A reversible path has $W=nRT\ln4$ . What is $\frac{V_f}{V_i}$ ?

Show hint

Apply the exponential.

Learning goal: Infer volume ratio from a logarithmic work expression.

Show solution and calculation

Approach: divide the work equation by the positive product of amount, gas constant and temperature, then exponentiate.
Volume ratio is four.

Check: the reversible return at the same temperature uses the reciprocal ratio and gives opposite work; net work vanishes.

\begin{aligned}\frac{W}{nRT}&=\ln\frac{V_f}{V_i}=\ln4\\\frac{V_f}{V_i}&=\exp(\ln4)=4\\W_{\mathrm{return}}&=nRT\ln\frac14=-W\end{aligned}

Interpret · Application

Ideal gas freely expands in an isolated rigid assembly from V to 2V. What are Q,W and ΔU?

Show hint

Actual path is not reversible.

Learning goal: Distinguish heat, work and internal energy in free isolated expansion.

Show solution and calculation

Approach: inspect actual transfer mechanisms.
The isolated rigid outer boundary allows neither heat nor work; free expansion into vacuum likewise supplies no volume work.
Internal energy and, in the ideal model, endpoint temperature therefore stay unchanged.
Entropy nevertheless increases.

Check: reversible comparison work belongs to a different process and must not be used here.

\begin{aligned}Q&=0\\W&=0\\\Delta U&=Q-W=0\\T_f&=T_i\\\Delta S&=nR\ln\frac{2V}{V}=nR\ln2>0\end{aligned}

Interpret · Challenge

An ideal gas expands at constant temperature with $nRT=600\,\mathrm J$ , from 2.0 L to 4.0 L. Path A is reversible and quasistatic; path B proceeds against constant external pressure 150 kPa and ends at equilibrium. Compare work done by the gas and heat supplied. Macroscopic kinetic energies vanish in the initial and final states.

Show hint

For A integrate gas pressure; for B integrate constant external pressure. For this ideal gas, ΔU vanishes for both isothermal paths.

Learning goal: Compare path-dependent work for the same isothermal state change.

Show solution and calculation

Approach: find the area under the respective working pressure over the volume range.
Reversible work is about 415.89 J; constant-external-pressure work is 300 J.
With $\Delta U=Q-W$ and $\Delta U=0$ , supplied heat equals work on each path.

Check: final gas pressure is indeed 150 kPa. Equal endpoints and temperature therefore do not imply equal work or heat.

\begin{aligned}W_A&=nRT\ln\frac{V_f}{V_i}=600\ln2\,\mathrm J\approx415.89\,\mathrm J\\{}W_B&=p_{\mathrm{ext}}(V_f-V_i)=(150\,000)(0.0020)\,\mathrm J=300\,\mathrm J\\{}Q_A&=W_A\\{}Q_B&=W_B\\{}p_f&=\frac{600\,\mathrm J}{0.0040\,\mathrm{m^3}}=150\,\mathrm{kPa}\end{aligned}

[1][2]

Error analysis · Application

An ideal gas expands reversibly and isothermally from 1.0 L to 10.0 L with $nRT=200\,\mathrm J$ . A solution uses base-ten logarithm and obtains $W=200\,\mathrm J$ . Correct the value and explain why an arbitrary logarithm does not satisfy the same formula. W denotes work done by the gas.

Show hint

The integral of $\frac1V$ is ln V. Another logarithm requires a conversion factor; its argument must be dimensionless.

Learning goal: Connect the natural logarithm to the work integral.

Show solution and calculation

Approach: obtain work by integrating reversible gas pressure.
This produces the natural logarithm of the volume ratio, giving about 460.52 J.
A base-ten logarithm would require an additional factor ln 10.

Check: exponentiating the correct work divided by nRT recovers volume ratio 10; using 200 J would instead give only e.

\begin{aligned}W&=\int_{V_i}^{V_f}\frac{nRT}{V}\,\mathrm dV=nRT\ln\frac{V_f}{V_i}\\&=200\ln10\,\mathrm J\approx460.52\,\mathrm J\\\ln x&=(\ln10)\log_{10}x\\\exp\!\left(\frac{W}{nRT}\right)&=10\end{aligned}

[1]

Compare work areas meaningfully

What to carry into your next calculation

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Work in Isothermal Expansion of an Ideal Gas. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/isotherme-arbeit-ideales-gas

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Retain Work in Isothermal Expansion of an Ideal Gas for exams

Create a curated FSRS exam set for Wgas=nRT ln(Vf/Vi): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Reversible isothermal work of an ideal gas

Variables & units

Isothermal work calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermodynamics

Worked example

Applications

How do you calculate with Work in Isothermal Expansion of an Ideal Gas?

Why gas can do work without changing temperature

Which work sign applies to reversible isothermal expansion and compression?

From moving piston to logarithm

Equal endpoints, different work areas

Heat replaces delivered energy

Three examples with different paths

Triple volume in a reversible comparison

The same endpoints against constant external load

Compression: where does the input work go?

Isothermal does not guarantee logarithmic work

How do you infer volume ratio from reversible work?

Five exercises with solutions

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work WbyW_{\mathrm{by}}Wby​ done by the gas.

A reversible path has W=nRTln⁡4W=nRT\ln4W=nRTln4. What is VfVi\frac{V_f}{V_i}Vi​Vf​​?

Ideal gas freely expands in an isolated rigid assembly from V to 2V. What are Q,W and ΔU?

Compare work areas meaningfully

What to carry into your next calculation

Sources for this explanation

Frequently asked questions about Work in Isothermal Expansion of an Ideal Gas

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Work in Isothermal Expansion of an Ideal Gas for exams

Reversible isothermal work of an ideal gas

Variables & units

Isothermal work calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermodynamics

Worked example

Applications

How do you calculate with Work in Isothermal Expansion of an Ideal Gas?

Why gas can do work without changing temperature

Which work sign applies to reversible isothermal expansion and compression?

From moving piston to logarithm

Equal endpoints, different work areas

Heat replaces delivered energy

Three examples with different paths

Triple volume in a reversible comparison

The same endpoints against constant external load

Compression: where does the input work go?

Isothermal does not guarantee logarithmic work

How do you infer volume ratio from reversible work?

Five exercises with solutions

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work WbyW_{\mathrm{by}}Wby​ done by the gas.

A reversible path has W=nRTln⁡4W=nRT\ln4W=nRTln4. What is VfVi\frac{V_f}{V_i}Vi​Vf​​?

Ideal gas freely expands in an isolated rigid assembly from V to 2V. What are Q,W and ΔU?

Compare work areas meaningfully

What to carry into your next calculation

Sources for this explanation

Frequently asked questions about Work in Isothermal Expansion of an Ideal Gas

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Work in Isothermal Expansion of an Ideal Gas for exams

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work $W_{\mathrm{by}}$ done by the gas.

A reversible path has $W=nRT\ln4$ . What is $\frac{V_f}{V_i}$ ?

0.45 mol of ideal gas at 310 K is reversibly and isothermally compressed from 5 to 2 L. Find work $W_{\mathrm{by}}$ done by the gas.

A reversible path has $W=nRT\ln4$ . What is $\frac{V_f}{V_i}$ ?