Work

Work (w) is the transfer of energy that causes organized, uniform motion in the surroundings.

This concept describes a process where the chaotic, random actions of individual particles give rise to an organized, macroscopic change. Consider the gas inside a cylinder with a movable piston. While each individual gas molecule zips around randomly, their countless collisions against the inner face of the piston produce a net, directional force.

This steady, outward force is the statistical result of the chaotic impacts. If this net force is sufficient to push the piston, the energy of the system has been used to move a macroscopic object over a distance. This uniform movement of the piston is the hallmark of work and is fundamentally different from the random, chaotic molecular motion associated with heat. This connection between a collective force and its resulting motion is the basis of work in chemistry.

Work: From Random Motion to Organized Motion Initial State (V 1 , T 1 ) V₁ Net upward force from random collisions System performs work (w > 0) Final State (V 2 , T 2 ) V₂ Uniform motion of piston (Δx) The system's internal energy decreased from U₁ to U₂, resulting in a lower final temperature. Thus, T₂ < T₁.


Work is formally defined as the energy transferred when a force acts on an object over a distance. The familiar equation from physics is:

\[ w = F_{\mathrm{ext}}~d \] In thermodynamics, we are almost always concerned with the change between an initial and a final state. Therefore, it is more precise to describe this movement as a displacement, or a change in position, represented by the symbol Δx. This gives us the more formal version of the work equation:

\[ w = F_{\mathrm{ext}}~\Delta x \] where w is work, Fext is the opposing external force, and Δx is the displacement (xfinal − xinitial). For a simple, one-way process like a piston moving, the distance (d) and the displacement (Δx) are the same. We use the Δx notation because it creates a clear parallel to the pressure-volume work done by a gas.

In chemistry, there are several types of work, such as the ordered movement of electrons in a battery (electrical work). However, the most common type we will study is the work associated with a change in volume. This is called pressure-volume work or PV-work.

Pressure-Volume Work

PV-work occurs when a gas expands or is compressed against an external pressure. Consider a gas confined in a cylinder with a movable piston.

  • Expansion (Work done by the system): If the gas inside the cylinder expands, it pushes the piston outward against the external pressure of the surroundings. The system is doing work on the surroundings.
  • Compression (Work done on the system): If the surroundings exert a greater pressure, the piston moves inward, compressing the gas. The surroundings are doing work on the system.

The equation for calculating PV-work done at a constant external pressure is: \[ w = -P_{\mathrm{ext}} \Delta V \]

where

  • w is the work.
  • Pext is the constant external pressure against which the volume change occurs.
  • ΔV is the change in the system’s volume (Vfinal − Vinitial).

Sign Conventions for Work

The negative sign in the PV-work equation is part of the sign convention, which is always defined from the system’s point of view.

  1. Expansion (System does work and loses energy):
    • The system expands, so ΔV is positive (V_final > V_initial).
    • The system pushes on the surroundings, losing energy. Therefore, w must be negative.
    • w = -P_ext * (+ΔV) = negative value
  2. Compression (System has work done on it and gains energy):
    • The system is compressed, so ΔV is negative (V_final < V_initial).
    • The surroundings push on the system, giving it energy. Therefore, w must be positive.
    • w = -P_ext * (-ΔV) = positive value

A positive value for w means energy has been “deposited” into the system’s energy account, while a negative value means energy has been “withdrawn.”

Units of Work and Energy: The Joule

The standard SI unit for work, and all forms of energy, is the Joule (J). The Joule is a derived unit, defined in terms of the base SI units for mass (kg), length (m), and time (s): \[ 1~\mathrm{J} \equiv 1~\mathrm{kg} ~ \frac{\mathrm{m^2}}{\mathrm{s^2}} \] A perfect illustration of this is the equation for kinetic energy (Ek) which is the energy an object possesses due to its motion. The kinetic energy is a function of an object’s mass (m) and velocity (v): \[ E_{\mathrm{k}} = \frac{1}{2}~m v^2 \] By analyzing the units, we can see how this equation results in Joules:

  • Mass (m) has SI units of kg
  • Velocity (v) has SI units of m s−1

\[ \text{Units of } E_{\mathrm{k}} = (\mathrm{kg}) (\mathrm{m~s^{-1}})^2 = \mathrm{kg} ~ \mathrm{m^2} ~ \mathrm{s^{-2}} \equiv \mathrm{J} \]

Practice

What is the kinetic energy (in J) of a typical baseball, with a mass of 145 g, moving at a velocity of 37.6 m s–1?

Solution

Using the kinetic energy equation

\[ \begin{align*} E_{\mathrm{k}} &= \dfrac{1}{2}~mv^2 \\[1.5ex] &= \dfrac{1}{2} \left ( 14\bar{5}~\mathrm{g} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \left ( \dfrac{37.\bar{6}~\mathrm{m}}{\mathrm{s}} \right )^2 \\[1.5ex] &= 10\bar{2}.49~\mathrm{kg~m^{2}~s^{-2}} \\[1.5ex] &= 102~\mathrm{J} \end{align*} \]

Calculating Work in Joules

Just as with kinetic energy, when calculating pressure-volume (PV) work, we must ensure our units result in Joules. The most direct way to achieve this is to use SI units for pressure and volume from the start:

  • Pressure (P) in Pascals (Pa)
  • Volume (V) in cubic meters (m3)

If you use these units, the result of the PΔV calculation will automatically be in Joules.

Handling Common, Non-SI Units

In practice, especially in chemistry, pressure is often given in atmospheres (atm) and volume in liters (L). While convenient, multiplying these units results in an energy unit called the liter-atmosphere (L atm).

Since thermodynamic laws (like the First Law) require energy terms to be combined, work and heat must be in the same unit. Because heat is typically expressed in Joules, it is essential to convert work from L atm to Joules. This is achieved using the exact conversion factor: \[ 1~\mathrm{L ~ atm} = 101.325~\mathrm{J} \] Therefore, you can calculate work using the given units of L and atm, and then apply this conversion factor to get the final answer in Joules, ensuring it can be correctly used in further thermodynamic calculations.

An older, non-SI unit for energy is the calorie (cal), originally defined as the energy needed to raise one gram of water by one degree Celsius. The calorie is now formally defined in relation to the Joule: 1 cal = 4.184 J (exactly). The “Calories” reported on food labels (Cal, with a capital C) are actually kilocalories (1 Cal = 1000 cal = 1 kcal).

Visualizing Irreversible Work

The single-step expansion we discussed (w = -P_ext * ΔV) is fast, simple, and happens against a constant external pressure. This type of process is called irreversible. But is it the most efficient way for a system to perform work?

Consider the expansion of a gas in a cylinder. To get the most work out of the expansion, we would want to push against the highest possible pressure at every step. The highest possible pressure is one that is just infinitesimally less than the gas’s own internal pressure. If we could manage this delicate balance throughout the entire expansion, we would achieve a reversible process.

The Path Matters: Reversible vs. Irreversible

A reversible process is a theoretical pathway that proceeds in an infinite number of tiny steps. At every moment, the system is perfectly in equilibrium with its surroundings.

  • Irreversible Expansion (One Big Step): Imagine a piston held in place by a large weight. If we suddenly remove the weight, the external pressure drops instantly to a lower constant value, and the gas expands rapidly against this new pressure. The work done is w = -P_ext * ΔV, where Pext is this final, constant external pressure.

  • Reversible Expansion (Infinite Tiny Steps): Now, imagine the piston is held down by a pile of very fine sand. If we remove one grain of sand at a time, the external pressure decreases by a tiny amount, and the gas expands by a tiny amount. We repeat this process an infinite number of times. At every step, the external pressure is perfectly matched to the internal pressure of the gas (P_ext ≈ P_gas).

Graphical View of Work

This difference is easiest to see on a PV graph, where the work done is the area under the curve.

The figure above presents a direct, side-by-side comparison of the work done when a gas expands from the same initial state to the same final state via two different pathways.

1. The Left Panel: Irreversible Path

This panel illustrates a sudden, single-step expansion. Imagine the gas is held at a high pressure (P1) and the external pressure is instantly dropped to the final pressure (P2). The gas then expands against this new, constant external pressure.

  • The path is shown by the dashed red line.
  • The work done, wirrev, corresponds to the shaded red rectangular area. This area is calculated as w = -P_ext * ΔV, where P_ext is the low, final pressure.

2. The Right Panel: Reversible Path

This panel shows the ideal, slow expansion where the external pressure is always perfectly matched to the gas’s internal pressure. As the gas expands, the pressure decreases gradually along the solid blue curve.

  • The work done, wrev, is the entire shaded blue area under the curve.

The Conclusion

By visually comparing the two panels, the conclusion is immediate and clear: the shaded area for the reversible path is significantly larger than the area for the irreversible path. This graphically demonstrates the fundamental principle of this topic:

The maximum work a system can do on its surroundings is achieved through a reversible pathway.

|wrev| > |wirrev|

The single, fast, irreversible step is less efficient at getting work done because the expansion happens against a much lower average pressure compared to the slow, controlled, reversible process.

For a gas expanding from V1 to V2:

  1. Irreversible Work: The work is the rectangular area defined by the constant, final external pressure (P_ext) and ΔV.
  2. Reversible Work: The work is the entire area under the curved path that the gas follows as its internal pressure drops during the expansion.

As you can see from a typical PV diagram, the area under the curve for the reversible path is significantly larger than the rectangular area for the irreversible path. This leads to a critical conclusion:

The maximum amount of work that can be done by a system during an expansion is achieved through a reversible pathway.

The Equation for Reversible Work

Because the pressure is constantly changing during a reversible expansion, we cannot use the simple -PΔV formula. Calculating the area under this curve requires calculus. For an ideal gas at a constant temperature (an isothermal reversible process), the work is given by the equation: \[ w_{\mathrm{rev}} = -nRT \ln \left( \frac{V_{\mathrm{final}}}{V_{\mathrm{initial}}} \right) \] where

  • wrev is the reversible work, typically in Joules
  • n is the number of moles of gas
  • R is the ideal gas constant (≈8.314 J mol−1 K−1)
  • T is the absolute temperature in Kelvin
  • Vfinal is the final volume of the gas
  • Vinitial is the initial volume of the gas

This equation is fundamental for calculating the theoretical maximum work for gas expansions and is a benchmark for thermodynamic efficiency.

Chemical Reactions and Work

While the image of a piston in a cylinder is a useful model, pressure-volume work is a fundamental aspect of many common chemical reactions. For chemical reactions, the primary source of pressure-volume work is the production or consumption of gases and can be signified by the change in moles of the gas (Δngas) of the reaction.

This is because the volume occupied by one mole of gas is vastly larger than the volume of the same amount of a liquid or solid. Therefore, any change in the number of moles of gas results in a significant change in the system’s total volume, forcing the system to push against the constant pressure of the atmosphere.

If a reaction produces more moles of gas than it consumes, the system expands. If it consumes more moles of gas than it produces, the system contracts (is compressed).

The stoichiometric coefficients in a balanced equation are dimensionless ratios. When calculating the actual change in moles of gas for a reaction, we interpret these coefficients as moles. This gives Δngas units of mol, which is needed for calculating the work done (w = −PΔV) or total energy changes (ΔH, ΔU).

Δngas is positive

Consider the following reaction.

\[ \begin{align*} \mathrm{CaCO_3(s)} \longrightarrow \mathrm{CaO(s)} + \mathrm{CO_2}(g) \end{align*} \]

Here, the reaction starts with 0 moles of gas but ends with 1 mole of gas. The change in moles of gas (Δngas) is

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{CO_2(g)}) \bigl ] ~ - ~ \bigr [ 0~\mathrm{mol} \bigl] \\[1.5ex] &= \,\bigr [ 1~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 0~\mathrm{mol} \bigl ] \\[1.5ex] &= 1~\mathrm{mol} \end{align*} \] Since Δngas is positive, the system expands and performs PV-work on the surroundings. This constitutes work done by the system, meaning the value of work (w) is negative.

Δngas is negative

Consider the following reaction.

\[ \begin{align*} \mathrm{N_2(g)} + 3~\mathrm{H_2(g)}\longrightarrow 2~\mathrm{NH_3(g)} \end{align*} \]

Here, the reaction starts with 4 moles of gas and ends with 2 moles of gas. The change in moles of gas is

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{NH_3(g)}) \bigl ] ~ - ~ \bigr [ n(\mathrm{N_2(g))} + n(\mathrm{H_2(g)}) \bigl] \\[1.5ex] &= \,\bigr [ 2~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 1~\mathrm{mol} + 3~\mathrm{mol} \bigl ] \\[1.5ex] &= \,\bigr [ 2~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 4~\mathrm{mol} \bigl ] \\[1.5ex] &= -2~\mathrm{mol} \end{align*} \] Since Δngas is negative, the system compresses and PV-work is performed on the system. This means the value of work (w) is positive.

Δngas is zero

The change in moles of gas can also be zero.

\[ \begin{align*} \mathrm{H_2(g)} + \mathrm{Cl_2(g)}\longrightarrow 2~\mathrm{HCl(g)} \end{align*} \]

Here, the reaction starts with 2 moles of gas and ends with 2 mole of gas. The change in moles of gas is

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{HCl(g)}) \bigl ] ~ - ~ \bigr [ n(\mathrm{H_2(g))} + n(\mathrm{Cl_2(g)}) \bigl] \\[1.5ex] &= \,\bigr [ 2~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 1~\mathrm{mol} + 1~\mathrm{mol} \bigl ] \\[1.5ex] &= \,\bigr [ 2~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 2~\mathrm{mol} \bigl ] \\[1.5ex] &= 0~\mathrm{mol} \end{align*} \]

When the change in the number of moles of gas is zero, the volume change (ΔV) for the reaction is considered negligible at constant temperature and pressure. Consequently, the pressure-volume work (w) is approximately zero.

Practice

Consider the vigorous reaction of zinc metal with hydrochloric acid, a common experiment in a general chemistry lab: \[ \mathrm{Zn(s) + 2~HCl(aq) \longrightarrow H_2(g) + ZnCl_2(aq)} \]

Determine if work is positive, negative, or approximately zero.

Solution

The change in moles of gas is calculated from the stoichiometry of the reaction.

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{H_2(g)}) \bigl ] ~ - ~ \bigr [ 0~\mathrm{mol} \bigl] \\[1.5ex] &= \,\bigr [ 1~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 0~\mathrm{mol} \bigl ] \\[1.5ex] &= 1~\mathrm{mol} \end{align*} \]

Since Δngas is positive, a gas is produced, causing the system to expand and perform work on the surroundings. By convention, work done by the system is negative.

Practice

The combustion of octane in a car engine is the classic example of a chemical reaction generating work. \[ \mathrm{2~C_8H_{18}(g) + 25~O_2(g) \longrightarrow 16~CO_2(g) + 18~H_2O(g)} \]

Determine if work is positive, negative, or approximately zero.

Solution

The change in moles of gas is calculated from the stoichiometry of the reaction.

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{CO_2(g)}) + n(\mathrm{H_2O(g)}) \bigl ] ~ - ~ \bigr [ n(\mathrm{C_8H_{18}(g))} + n(\mathrm{O_2(g)}) \bigl] \\[1.5ex] &= \,\bigr [ 16~\mathrm{mol} + 18~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 2~\mathrm{mol} + 25~\mathrm{mol} \bigl ] \\[1.5ex] &= \,\bigr [ 34~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 27~\mathrm{mol} \bigl ] \\[1.5ex] &= 7~\mathrm{mol} \end{align*} \]

Since Δngas is positive, a gas is produced, causing the system to expand and perform work on the surroundings. By convention, work done by the system is negative.

This is the type of reaction that takes place inside the internal combustion engine in a typical car. A sudden, massive increase in the amount of gas inside the engine cylinder, combined with the heat produced, creates an enormous pressure that forcefully pushes the piston down, doing the mechanical work of moving a car.

Practice

Biological systems can perform PV-work. The metabolic breakdown of glucose is a form of combustion: \[ \mathrm{C_6H_{12}O_6(s) + 6~O_2(g) \longrightarrow 6~CO_2(g) + 6~H_2O(l)} \]

Determine if work is positive, negative, or approximately zero.

Solution

The change in moles of gas is calculated from the stoichiometry of the reaction.

\[ \begin{align*} \Delta n_{\mathrm{gas}} &= \sum n(\mathrm{products})_{\mathrm{gas}} ~ - ~ \sum n(\mathrm{reactants})_{\mathrm{gas}} \\[1.5ex] &= \,\bigr [ n(\mathrm{CO_2(g)}) \bigl ] ~ - ~ \bigr [ n(\mathrm{O_2(g))} \bigl] \\[1.5ex] &= \,\bigr [ 6~\mathrm{mol} \bigl ] ~ - ~ \bigr [ 6~\mathrm{mol}\bigl ] \\[1.5ex] &= 0~\mathrm{mol} \end{align*} \]

In this specific case, the number of moles of gas is the same on both sides and Δngas = 0. Therefore, the PV-work done by our bodies from this process is essentially zero. Almost all of the energy from metabolism is released as heat.

Summary

Work in Thermodynamics

Work (w) is the transfer of energy via organized, uniform motion. In chemistry, the most common form is pressure-volume work:

\[ w = -P_{\mathrm{ext}} \Delta V \]

The negative sign ensures proper sign convention from the system’s point of view:

  • Expansion (ΔV > 0): System does work on surroundings, w < 0 (energy leaves)
  • Compression (ΔV < 0): Surroundings do work on system, w > 0 (energy enters)

Work in Chemical Reactions

For chemical reactions, PV-work arises from changes in the number of moles of gas (Δngas):

  • Δngas > 0: System expands, w is negative
  • Δngas < 0: System contracts, w is positive
  • Δngas = 0: No significant PV-work

The SI unit for work is the Joule (J). When using L and atm, convert using 1 L atm = 101.325 J.