Work is one of the primary ways that energy can be transferred between a system and its surroundings. In general, work occurs whenever a force acts through a displacement. If a force resists motion, energy must be expended to move an object against that force. Conversely, if a force causes motion, the system doing the pushing can transfer energy to its surroundings.
Although work is often introduced in the context of moving objects, the same idea appears in many different physical systems. A gas can perform work by expanding against an external pressure, a battery can perform electrical work by moving charge through a potential difference, and a spring can store energy that is released as mechanical work. In each case, work is calculated by multiplying a generalized force by a generalized displacement.
Because work represents a transfer of energy, it is measured in joules (J). One joule is the amount of work required to move an object one meter against a force of one newton:
\[ 1\ \mathrm{J} = 1\ \mathrm{N\,m} \]
The common forms of work encountered in chemistry are all variations on this same idea: force multiplied by the displacement through which that force acts.
The simplest form of work is mechanical work. If a constant force acts through a displacement, the work is given by
\[ w = F\Delta x \]
where \(F\) is the magnitude of the force and \(\Delta x\) is the displacement. This expression can be used to calculate the work required to push a box across a floor, lift an object against gravity, or pull an object with a rope.
In thermodynamics, one of the most important forms of work is pressure-volume work. When a gas expands against an external pressure, the pressure acts as the generalized force and the change in volume acts as the generalized displacement:
\[ w = -P_{\mathrm{ext}}\Delta V \]
The negative sign indicates that an expanding system transfers energy to the surroundings by performing work. Pressure-volume work plays a central role in the study of gases and thermodynamic processes.
Electrical systems provide another important example. When a quantity of charge moves through a potential difference, electrical work is performed:
\[ w = Vq \]
where \(V\) is the voltage difference and \(q\) is the amount of charge transferred. Batteries, fuel cells, and electrochemical cells all perform electrical work in this manner.
Many other forms of work can be described using the same force-times-displacement framework. Stretching a rubber band or compressing a spring involves a tension force acting through a displacement, while expanding the surface of a liquid involves surface tension acting through a change in surface area. Although the details differ, each example follows the same general principle: work equals a generalized force multiplied by a generalized displacement.
A 2.54 kg mass is moved 2.0 m along a horizontal surface by applying a constant force of 3.0 N. Calculate the work performed by the force.
First, identify the appropriate equation:
\[ w = F\Delta x \]
The force and displacement are given as
\[ F = 3.0\ \mathrm{N} \]
\[ \Delta x = 2.0\ \mathrm{m} \]
Substituting these values into the equation gives
\[ w = \left(3.0\ \mathrm{N}\right) \left(2.0\ \mathrm{m}\right) \]
\[ w = 6.0\ \mathrm{N\,m} \]
Since \(1\ \mathrm{N\,m}=1\ \mathrm{J}\),
\[ w = 6.0\ \mathrm{J} \]
Therefore, the force performs 6.0 J of work while moving the object.
Note: Although the problem states that the mass is 2.54 kg, the mass does not appear in the work equation. Because the force is already known, only the force and displacement are needed to calculate the work performed.
A gas expands against a constant external pressure of \(1.00\ \mathrm{atm}\). The volume increases from \(11.2\ \mathrm{L}\) to \(22.4\ \mathrm{L}\) at \(0^\circ\mathrm{C}\). Calculate the work performed by the gas during the expansion.
For an expansion against a constant external pressure, the work is given by
\[ w = -P_{\mathrm{ext}}\Delta V \]
First calculate the change in volume:
\[ \Delta V = V_f - V_i = 22.4\ \mathrm{L} - 11.2\ \mathrm{L} \]
\[ \Delta V = 11.2\ \mathrm{L} \]
Substitute the values into the work equation:
\[ w = -\left(1.00\ \mathrm{atm}\right) \left(11.2\ \mathrm{L}\right) \]
\[ w = -11.2\ \mathrm{L\,atm} \]
The unit \(\mathrm{L\,atm}\) is a unit of energy and can be converted to joules using
\[ 1\ \mathrm{L\,atm} = 101.325\ \mathrm{J} \]
Therefore,
\[ w = -11.2\ \mathrm{L\,atm} \left( \frac{101.325\ \mathrm{J}} {1\ \mathrm{L\,atm}} \right) \]
\[ w = -1135\ \mathrm{J} \]
\[ w = -1.14\times10^3\ \mathrm{J} \]
Therefore, the gas performs \(1.14\times10^3\ \mathrm{J}\) of work on the surroundings. The negative sign indicates that energy leaves the system as the gas expands.
Note: The temperature is not needed for this calculation because the external pressure is constant and the work depends only on \(P_{\mathrm{ext}}\) and the change in volume. In later chapters, the temperature becomes important when calculating reversible expansion work using the Ideal Gas Law.
In the previous example, we converted \(\mathrm{L\cdot atm}\) to joules using the relationship
\[ 1\ \mathrm{L\cdot atm} = 101.325\ \mathrm{J} \]
An alternative way to obtain this conversion factor is by using the Ideal Gas Law constant, \(R\). The gas constant can be written in several equivalent forms:
\[ R = 8.314\ \frac{\mathrm{J}} {\mathrm{mol\cdot K}} \]
\[ R = 0.08206\ \frac{\mathrm{L\cdot atm}} {\mathrm{mol\cdot K}} \]
Since both expressions represent the same physical constant, their ratio can be used as a conversion factor:
\[ \frac{ 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} }{ 0.08206\ \mathrm{L\cdot atm\ mol^{-1}\ K^{-1}} } = 101.3\ \frac{\mathrm{J}} {\mathrm{L\cdot atm}} \]
Notice that the units of mol and K cancel, leaving only \(\mathrm{J/(L\cdot atm)}\). This gives exactly the conversion factor needed to convert pressure-volume work into joules.
\[ -11.2\ \mathrm{L\cdot atm} \left( \frac{8.314\ \mathrm{J}} {0.08206\ \mathrm{L\cdot atm}} \right) = -1.14\times10^3\ \mathrm{J} \]
Physical Chemistry Tip: The value of \(R\) is not a single number—it depends on the units being used. One of the most common mistakes in physical chemistry is choosing the wrong numerical value for \(R\). Always examine the units required by the problem before selecting which form of the gas constant to use.
Consider \(1.00\ \mathrm{mol}\) of an ideal gas expanding isothermally from \(11.2\ \mathrm{L}\) to \(22.4\ \mathrm{L}\) at \(0^\circ\mathrm{C}\). In this example, the external pressure is adjusted so that it is always equal to the internal pressure of the gas. (This is the case for a reversible expansion, which will be more fully explained in Chapter 3 - The First Law.)
Because the pressure is not constant, we cannot simply use \(w=-P_{\mathrm{ext}}\Delta V\). Instead, the pressure changes as the volume changes, so the work must be calculated using an integral:
\[ w = -\int_{V_i}^{V_f} P_{\mathrm{ext}}\,dV \]
Since the external pressure is equal to the internal pressure of the ideal gas, we can calculate the pressure at each volume using the Ideal Gas Law:
\[ P_{\mathrm{ext}} = P_{\mathrm{int}} = \frac{nRT}{V} \]
As the gas expands, \(V\) increases. Because \(P=nRT/V\), the pressure decreases during the expansion. This changing pressure is why calculus is required.
Substitute \(P_{\mathrm{ext}}=\frac{nRT}{V}\) into the work integral:
\[ w = -\int_{V_i}^{V_f} \frac{nRT}{V}\,dV \]
Since \(n\), \(R\), and \(T\) are constant during an isothermal expansion, they can be moved outside the integral:
\[ w = -nRT \int_{V_i}^{V_f} \frac{1}{V}\,dV \]
\[ w = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
Now substitute the given values:
\[ n = 1.00\ \mathrm{mol} \]
\[ T = 273.15\ \mathrm{K} \]
\[ V_i = 11.2\ \mathrm{L} \qquad V_f = 22.4\ \mathrm{L} \]
Use \(R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}\) so that the work is calculated directly in joules:
\[ w = -\left(1.00\ \mathrm{mol}\right) \left(8.314\ \frac{\mathrm{J}}{\mathrm{mol\cdot K}}\right) \left(273.15\ \mathrm{K}\right) \ln\left(\frac{22.4\ \mathrm{L}}{11.2\ \mathrm{L}}\right) \]
\[ w = -\left(2271\ \mathrm{J}\right)\ln(2.00) \]
\[ w = -1.57\times10^3\ \mathrm{J} \]
Therefore, the gas performs \(1.57\times10^3\ \mathrm{J}\) of work on the surroundings. The negative sign indicates that energy leaves the gas as it expands.
Important comparison: In the constant-pressure example, the work was \(-1.14\times10^3\ \mathrm{J}\). Here, the magnitude of the work is larger because the gas expands against a pressure that starts high and gradually decreases, rather than expanding against the final pressure for the entire process.
One mole of electrons carries a charge of \(96484\ \mathrm{C}\), known as the Faraday constant. Suppose this charge moves through a potential difference of \(1.10\ \mathrm{V}\), which is the standard cell potential of a Daniell cell. Calculate the electrical work performed.
Electrical work is calculated using
\[ w = Vq \]
where \(V\) is the potential difference and \(q\) is the amount of charge transferred.
The values given are
\[ V = 1.10\ \mathrm{V} \]
\[ q = 96484\ \mathrm{C} \]
Substituting these values into the work equation gives
\[ w = \left(1.10\ \mathrm{V}\right) \left(96484\ \mathrm{C}\right) \]
\[ w = 106132\ \mathrm{V\cdot C} \]
Since one volt is one joule per coulomb,
\[ 1\ \mathrm{V} = 1\ \frac{\mathrm{J}}{\mathrm{C}} \]
the coulombs cancel, leaving joules:
\[ w = 106132\ \mathrm{V\cdot C} \left( \frac{1\ \mathrm{J}} {1\ \mathrm{V\cdot C}} \right) \]
\[ w = 1.06\times10^5\ \mathrm{J} \]
Therefore, moving one mole of electrons through a potential difference of \(1.10\ \mathrm{V}\) performs \(1.06\times10^5\ \mathrm{J}\) of electrical work.
Connection to chemistry: The quantity \(96484\ \mathrm{C\,mol^{-1}}\) is called the Faraday constant, \(F\). Because electrochemical cells move electrons through a potential difference, electrical work is often written as
\[ w = nFV \]
where \(n\) is the number of moles of electrons transferred. This relationship will become important when studying electrochemistry and the thermodynamics of galvanic cells.
Calculate the work for the randomly generated problem below. Enter only the numerical value.
Big picture: Mechanical work, pressure-volume work, and electrical work all follow the same fundamental idea: a generalized force acting through a generalized displacement. The specific variables change from one physical system to another, but the underlying concept of energy transfer remains the same.