Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV

Core Intuition

Intuition Why does the

path matter for work? Work is not a state function — it depends on how you get from state A to state B, not just the endpoints. Imagine lifting a heavy book onto a shelf using a spring balance:

Reversible (slow, careful): You raise it infinitely slowly, so at every instant the upward force you apply exactly balances gravity (only an infinitesimal excess to keep it moving). No jerks, no energy dissipated — you could reverse the process by an infinitesimal change and get all the energy back. Maximum useful work is exchanged.
Irreversible (sudden): You yank the book up fast against a single constant resisting force. It's jerky, energy is dissipated (heat/sound), and you cannot recover it by an infinitesimal reversal.

For gases: reversible expansion means the external pressure $P_{\text{ext}}$ equals the gas pressure $P_{\text{gas}}$ at every moment (differing only by an infinitesimal $dP$ ) — the system is always in quasi-equilibrium. Irreversible expansion means $P_{\text{ext}}$ is finite and noticeably less than $P_{\text{gas}}$ — the gas does a "burst" of work against a lower external pressure and dissipates the difference.

Definitions

Derivation: Reversible Isothermal Work

WHY isothermal? Temperature constant $\Rightarrow$ for an ideal gas, $PV = nRT$ holds throughout.

Step 1: Start with the work integral $w = -\int_{V_1}^{V_2} P_{\text{ext}} \, dV$

Step 2: For reversible process, $P_{\text{ext}} = P_{\text{gas}}$ Because the gas is always in quasi-equilibrium, the external pressure equals the gas pressure (the infinitesimal $dP$ contributes nothing to the integral): $P_{\text{ext}} = P_{\text{gas}} = \frac{nRT}{V} \quad \text{(ideal gas law)}$

WHY can we substitute $P_{\text{gas}}$ ? Only in reversible processes! In irreversible, $P_{\text{ext}} \neq P_{\text{gas}}$ .

Step 3: Substitute into the integral $w = -\int_{V_1}^{V_2} \frac{nRT}{V} \, dV$

Step 4: Factor out constants (T is constant in isothermal) $w = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV$

Step 5: Integrate $w = -nRT \left[ \ln V \right]_{V_1}^{V_2} = -nRT (\ln V_2 - \ln V_1)$

Step 6: Simplify using log properties $\boxed{w_{\text{rev}} = -nRT \ln\left(\frac{V_2}{V_1}\right)}$

WHAT does this mean?

If $V_2 > V_1$ (expansion): $\ln(V_2/V_1) > 0 \Rightarrow w < 0$ (work done by gas)
If $V_2 < V_1$ (compression): $\ln(V_2/V_1) < 0 \Rightarrow w > 0$ (work done on gas)
Logarithmic dependence: doubling volume from 1→2 L does the same work as 10→20 L (same ratio).

Derivation: Irreversible Expansion

WHAT'S DIFFERENT? External pressure $P_{\text{ext}}$ is constant throughout — think of a gas expanding against atmospheric pressure suddenly.

Step 1: Work integral $w = -\int_{V_1}^{V_2} P_{\text{ext}} \, dV$

Step 2: $P_{\text{ext}}$ is constant $w = -P_{\text{ext}} \int_{V_1}^{V_2} dV$

Step 3: Integrate $w = -P_{\text{ext}} (V_2 - V_1) = -P_{\text{ext}} \Delta V$

WHY is this less work than reversible? In reversible expansion, the gas pushes against a pressure that decreases smoothly as volume increases (because $P_{\text{ext}} = P_{\text{gas}} = nRT/V$ tracks the gas). In a constant- $P_{\text{ext}}$ irreversible expansion, the gas pushes against a lower constant pressure the whole time. Less resistance $\Rightarrow$ less work extracted.

Visual Representation

HOW TO READ THE DIAGRAM:

Blue curve (reversible): Area under the $PV$ isotherm from $V_1$ to $V_2$ . This is $w_{\text{rev}}$ — maximum work.
Red rectangle (irreversible): Area under constant $P_{\text{ext}}$ line. This is $w_{\text{irrev}}$ — always smaller.
The gap: Represents "lost" work due to irreversibility.

Worked Examples

Worked example Example 1: Reversible Isothermal Expansion

Problem: 2 moles of ideal gas at 300 K expand reversibly and isothermally from 5 L to 15 L. Calculate work done. Given: $n = 2 \, \text{mol}$ , $T = 300 \, \text{K}$ , $V_1 = 5 \, \text{L}$ , $V_2 = 15 \, \text{L}$ , $R = 8.314 \, \text{J mol}^{-1}\text{K}^{-1}$ . Step 1: Apply reversible work formula $w_{\text{rev}} = -nRT \ln\left(\frac{V_2}{V_1}\right)$ WHY this formula? Reversible + isothermal + ideal gas.

Step 2: Substitute values $w_{\text{rev}} = -(2)(8.314)(300) \ln\left(\frac{15}{5}\right)$ $= -4988.4 \ln(3)$ Step 3: Calculate $\ln(3)$ $\ln(3) \approx 1.0986$

Step 4: Final answer $w_{\text{rev}} = -4988.4 \times 1.0986 = \boxed{-5480 \, \text{J} = -5.48 \, \text{kJ}}$

WHY negative? Work done by the gas (expansion).

Worked example Example 2: Irreversible Expansion Against Atmosphere

Problem: Same gas (2 mol, 300 K) expands from 5 L to 15 L against constant atmospheric pressure $P_{\text{ext}} = 1 \, \text{atm} = 101.325 \, \text{kPa}$ . Calculate work.

Step 1: Apply irreversible work formula $w_{\text{irrev}} = -P_{\text{ext}} \Delta V$

Step 2: Calculate $\Delta V$ $\Delta V = 15 - 5 = 10 \, \text{L} = 0.01 \, \text{m}^3$

WHY convert to m³? To match SI units with kPa.

Step 3: Substitute $w_{\text{irrev}} = -(101.325 \, \text{kPa})(0.01 \, \text{m}^3)$ $= -1013.25 \, \text{J} = \boxed{-1.01 \, \text{kJ}}$

COMPARISON: Reversible gave −5.48 kJ, irreversible only −1.01 kJ. Reversible extracts 5.4× more work!

Worked example Example 3: Two-Step Irreversible Expansion

Problem: Same gas expands from 5 L to 10 L against 2 atm, then 10 L to 15 L against 1 atm. Total work?

Step 1: Work in each step $w_1 = -P_{\text{ext},1} \Delta V_1 = -(2 \, \text{atm})(5 \, \text{L}) = -10 \, \text{L·atm}$ $w_2 = -P_{\text{ext},2} \Delta V_2 = -(1 \, \text{atm})(5 \, \text{L}) = -5 \, \text{L·atm}$

WHY separate? Different $P_{\text{ext}}$ in each step — this is exactly the "path-dependent" $P_{\text{ext}}(V)$ case; we integrate piecewise.

Step 2: Total work $w_{\text{total}} = w_1 + w_2 = -10 - 5 = -15 \, \text{L·atm}$

Step 3: Convert to joules $1 \, \text{L·atm} = 101.325 \, \text{J}$ $w_{\text{total}} = -15 \times 101.325 = \boxed{-1520 \, \text{J} = -1.52 \, \text{kJ}}$ INSIGHT: Still less than reversible (−5.48 kJ) but more than single-step irreversible (−1.01 kJ). More steps closer to equilibrium $\Rightarrow$ more work — showing irreversible work genuinely depends on the $P_{\text{ext}}$ path.

Common Mistakes

Active Recall Questions

#flashcards/chemistry

What is the formula for reversible isothermal work for an ideal gas?

w_{\text{rev}} = -nRT \ln(V_2/V_1)

What is the formula for irreversible expansion work against a constant external pressure?

w_{\text{irrev}} = -P_{\text{ext}} \Delta V

Why is reversible work always greater in magnitude than irreversible work for expansion?

Reversible expansion works against a pressure that tracks the gas (

P_{\text{ext}}=P_{\text{gas}}=nRT/V

), while irreversible expansion works against a lower external pressure, so less total work is extracted.

In the equation

w = -\int P_{\text{ext}} dV

, why do we use

P_{\text{ext}}

and not

P_{\text{gas}}

Work depends on the pressure the gas pushes against (external), not its own internal pressure. Only in reversible processes do these become equal at each instant.

Does irreversible work depend only on endpoints and

P_{\text{ext}}

Only when

P_{\text{ext}}

is constant. In general, irreversible work depends on the whole

P_{\text{ext}}(V)

path, so you must integrate

-\int P_{\text{ext}}\,dV

For an expansion from 5 L to 15 L, is the work positive or negative?

Negative (work done BY the gas on surroundings).

Why does the reversible isothermal work formula contain

\ln(V_2/V_1)

instead of

(V_2 - V_1)

Because pressure changes with volume as

P = nRT/V

; integrating

1/V

gives the natural logarithm.

What is the key condition for using

w = -nRT \ln(V_2/V_1)

Process must be (1) reversible, (2) isothermal, and (3) ideal gas.

Convert −10 L·atm to joules

−1013.25 J (1 L·atm = 101.325 J)

If a gas expands against vacuum (

P_{\text{ext}} = 0

), what is the work done?

w = 0

(free expansion; no external pressure to push against).

Why is work a path function, not a state function?

Work depends on how the process occurs (reversible vs irreversible, the

P_{\text{ext}}

path, number of steps), not just the initial and final states.

Feynman Explanation

Recall Explain to a 12-year-old

Imagine you have a spring-loaded piston full of air.

Slow and careful way (reversible): You very slowly let the piston move out, matching the push of the air exactly at every instant. Because nothing is jerky and nothing is wasted, the air does the maximum work possible — and you could push it right back with an equally tiny effort.

Fast, rough way (irreversible): You suddenly release the piston against ordinary room air pressure. The air POPS out against that one constant, lower pressure. It does some work, but less, because it isn't carefully matching resistance — it just shoves against a low fixed pressure, and the extra energy gets dissipated.

The logarithm in the reversible formula ( $\ln V_2/V_1$ ) comes from the math of $1/V$ . Doubling the volume always does the same extra work, whether you go from 1→2 cups or 100→200 cups (same ratio).

Mnemonic

Connections

First Law of Thermodynamics — work is one way energy transfers ( $\Delta U = q + w$ )
Isothermal Process — why temperature stays constant and how this relates to $\Delta U = 0$ for ideal gas
PV Diagram — visualizing work as area under the curve
Ideal Gas Law — foundation for deriving reversible work
Entropy and Irreversibility — irreversible processes generate entropy; reversible is the limit
Carnot Cycle — uses reversible isothermal expansions/compressions for maximum efficiency
Free Expansion — limiting case where $P_{\text{ext}} = 0 \Rightarrow w = 0$
Heat Capacity — how energy distributes between $q$ and $w$ in different processes

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho beta, yaha sabse important baat samajhne ki ye hai ki work ek state function nahi hai — matlab kitna work hua ye sirf starting aur ending point pe depend nahi karta, balki raasta (path) kaunsa liya usse bhi depend karta hai. Ek simple picture yaad rakho: agar tum gas ko dheere-dheere, bilkul aaram se expand karne do jahan external pressure gas ke pressure ke almost barabar hota hai (bas ek chhota sa $dP$ ka difference), toh ye reversible process hai aur isme maximum work milta hai. Lekin agar tum ekdum jhatke se, kam external pressure ke against gas ko expand hone do, toh wo irreversible hai aur usme kam work nikalta hai kyunki beech mein energy waste (heat, sound) ho jaati hai.

Ab formula ki taraf aao — reversible isothermal expansion ke liye humne $w = -\int P_{ext}\,dV$ se shuru karke, kyunki reversible mein $P_{ext} = P_{gas} = nRT/V$ hai, integrate karke nikala $w_{rev} = -nRT \ln(V_2/V_1)$ . Yaha do cheezein notice karo: pehli, expansion mein ( $V_2 > V_1$ ) work negative aata hai kyunki gas khud kaam kar rahi hai (IUPAC convention). Dusri, ye logarithmic relationship hai — matlab volume 1→2 double karne mein utna hi work hota hai jitna 10→20 mein, kyunki ratio same hai. Ye insight questions mein bahut kaam aata hai.

Ye topic kyun matter karta hai? Kyunki thermodynamics ka pura foundation isi concept pe khada hai — reversible process humesha maximum work deta hai, aur ye idea aage chal ke entropy, efficiency, aur real engines samajhne mein base banta hai. Exam mein bhi ye pakka aata hai — reversible vs irreversible ka comparison, aur numerical jahan tumhe pata hona chahiye kab kaunsa formula lagana hai. Toh bas yaad rakho: path matters, reversible = maximum work, aur woh log wala formula. Baaki sab isi pe build hoga.

Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV

Core Intuition

Definitions

Derivation: Reversible Isothermal Work

Derivation: Irreversible Expansion

Visual Representation

Worked Examples

Common Mistakes

Active Recall Questions

Feynman Explanation

Mnemonic

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics (Chemical)

Connections