Work is energy transferred when a system undergoes a displacement against a resisting force. For gases, the most common form of work is pressure-volume work, often called \(pV\) work. This occurs when a gas expands or contracts against an external pressure.
The infinitesimal amount of work associated with a small change in volume is
\[ dw = -P_{\mathrm{ext}}\,dV \]
The negative sign reflects the sign convention used in thermodynamics. When a gas expands, \(dV > 0\), so \(w < 0\) because the system is doing work on the surroundings. When a gas is compressed, \(dV < 0\), so \(w > 0\) because the surroundings are doing work on the system.
If the external pressure remains constant during the expansion, it can be removed from the integral:
\[ w = -\int_{V_1}^{V_2} P_{\mathrm{ext}}\,dV \]
\[ w = -P_{\mathrm{ext}} \int_{V_1}^{V_2} dV \]
\[ w = -P_{\mathrm{ext}} (V_2-V_1) \]
\[ w = -P_{\mathrm{ext}}\Delta V \]
This expression is commonly used for irreversible expansions in which the gas expands against a constant opposing pressure.
A reversible expansion represents a limiting case in which the gas is always in mechanical equilibrium with its surroundings. At every point along the pathway,
\[ P_{\mathrm{ext}} = P_{\mathrm{gas}} \]
For an ideal gas,
\[ P = \frac{nRT}{V} \]
If the temperature remains constant, the work becomes
\[ w = -\int_{V_1}^{V_2} \frac{nRT}{V}\,dV \]
Since \(n\), \(R\), and \(T\) are constant during an isothermal process,
\[ w = -nRT \int_{V_1}^{V_2} \frac{1}{V}\,dV \]
\[ w = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
This expression gives the maximum amount of work that can be obtained from an expansion between two specified states.
Because the pressure decreases continuously during the expansion, a reversible expansion always produces more work than an irreversible expansion connecting the same initial and final states.
A constant-volume (or isochoric) process is one in which the volume of the system does not change:
\[ dV = 0 \]
Since pressure-volume work requires a change in volume,
\[ dw = -P_{\mathrm{ext}}\,dV = 0 \]
and therefore
\[ w = 0 \]
The First Law (\(dU\ =\ dq\ +\ dw \)) then simplifies to
\[ dU = dq \]
Under constant-volume conditions, all heat transferred into or out of the system changes the internal energy.
The amount of heat required to change the temperature of a system depends on its heat capacity. This is defined as follows: \[ C = \frac{dq}{dT} \] At constant volume, since dw = 0, and dU = dq, the heat capacity is defined by
\[ C_V = \left( \frac{\partial U} {\partial T} \right)_V \]
(See the Partial Derivatives Module for help in understanding partial derivatives.) For a small temperature change,
\[ dU = C_V\,dT \]
Since \(dU=dq\) for a constant-volume process,
\[ dq = C_V\,dT \]
If the heat capacity is approximately constant over the temperature range of interest,
\[ q = C_V\Delta T \]
and
\[ \Delta U = C_V\Delta T \]
These relationships make constant-volume processes particularly convenient because the heat transferred can be determined directly from the temperature change.
Big picture: The work done by a gas depends strongly on the pathway followed. Constant-pressure expansions, reversible expansions, and constant-volume processes all involve different amounts of work, even when they connect similar states. At constant volume no pressure-volume work is performed, making the connection between heat flow and internal energy especially simple.
Calculate the work done by \(1.00\ \mathrm{mol}\) of an ideal gas expanding isothermally at \(298\ \mathrm{K}\) from \(10.0\ \mathrm{L}\) to \(24.4\ \mathrm{L}\) against a constant external pressure of \(1.00\ \mathrm{atm}\).
For an expansion against a constant external pressure,
\[ w = -P_{\mathrm{ext}}\Delta V \]
First calculate the volume change:
\[ \Delta V = V_2 - V_1 \]
\[ \Delta V = 24.4\ \mathrm{L} - 10.0\ \mathrm{L} \]
\[ \Delta V = 14.4\ \mathrm{L} \]
Substituting into the work expression:
\[ w = -(1.00\ \mathrm{atm}) (14.4\ \mathrm{L}) \]
\[ w = -14.4\ \mathrm{atm\cdot L} \]
Converting to joules:
\[ w = -14.4\ \mathrm{atm\cdot L} \left( \frac{ 101.325\ \mathrm{J} } { 1\ \mathrm{atm\cdot L} } \right) \]
\[ w = -1.46\times10^3\ \mathrm{J} \]
Therefore,
\[ w = -1.46\ \mathrm{kJ} \]
The negative sign indicates that the gas does work on the surroundings as it expands.
Calculate the work done by \(1.00\ \mathrm{mol}\) of an ideal gas expanding reversibly and isothermally at \(298\ \mathrm{K}\) from \(10.0\ \mathrm{L}\) to \(24.4\ \mathrm{L}\).
For a reversible isothermal expansion of an ideal gas,
\[ w = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
Substituting the known values:
\[ w = -(1.00\ \mathrm{mol}) \left( 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) (298\ \mathrm{K}) \ln \left( \frac{24.4}{10.0} \right) \]
\[ w = -(2477.6\ \mathrm{J}) \ln(2.44) \]
\[ w = -2.21\times10^3\ \mathrm{J} \]
Therefore,
\[ w = -2.21\ \mathrm{kJ} \]
Notice that the magnitude of the reversible work exceeds the magnitude of the work calculated for the constant-pressure expansion:
\[ |w_{\mathrm{rev}}| > |w_{\mathrm{irrev}}| \]
This occurs because the pressure of the gas gradually decreases as the gas expands. The gas therefore continues to push against the surroundings with a larger pressure throughout much of the expansion than it would if it expanded against a constant external pressure.
Important: A reversible expansion always produces the maximum possible amount of work between a specified initial and final state. Although perfectly reversible processes do not occur in the real world, they represent an important limiting ideal case that establishes the upper limit on the work obtainable from a thermodynamic process.
Choose the correct answer for the randomly generated work problem.
Big picture: Work depends on how a process occurs, not simply on the initial and final states. Constant-pressure, reversible, and constant-volume processes all involve different amounts of work, making the thermodynamic pathway a crucial part of solving First Law problems.