David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 4

Work Done on a Gas

Work done on a gas

Pressure work

To apply the first law of thermodynamics to an ideal gas, we need to know the work done on the gas by the environment. Because of the pressure in a gas, work is done when the volume of the gas changes.

Consider a gas in a sealed container, which exerts a force on the walls of the container. If the volume of the container changes, work is done, since this force is applied to a moving wall.

A schematic of a gas sealed in a rectangular container, drawn as a light cyan box with blue dots for the gas molecules. Red force arrows point straight out of every wall --- left, top, bottom, and right --- because the pressure of the gas pushes outward everywhere on the container. The right-hand wall is a movable grey wall. A dashed line marks where that wall started, and a short teal arrow above it shows the wall being displaced outward by a small amount delta r perpendicular. Force applied over a displacement means work is done as the volume of the container changes.

Derivation: Work done on a gas (optional)

For any small area $A$ of the container wall, the force exerted by the gas on that area is in the direction perpendicular to the wall and outward, and is

\[F_{\mathrm{gc},\perp} = pA,\]

where the subscript $\mathrm{gc}$ stands for “gas on container.”

A close-up of one container wall. The gas, drawn as blue dots on a light cyan field, fills the left half; the wall is a grey vertical slab on the right, with the outside beyond it. A small yellow rectangle on the wall marks one patch of wall area A. A red force arrow starts at the patch and points outward, perpendicular to the wall, labeled F sub g c perpendicular equals p times A: the force of the gas on the container, proportional to the patch area. A dashed reference arrow below points the same way, marking the outward normal as the positive perpendicular direction.

By Newton’s third law, the component of the inward force exerted by the container on the gas is opposite the gas’s force:

\[F_{\mathrm{cg},\perp} = - F_{\mathrm{gc},\perp} = - p A.\]

The same close-up of one container wall, now showing the Newton's-third-law pair acting on the patch of area A. The grey arrow pointing outward, away from the gas, is the force of the gas on the container, labeled F sub g c perpendicular. The red arrow pointing inward, back into the gas, is the force of the container on the gas, labeled F sub c g perpendicular equals minus p times A. The two arrows are equal in length and opposite in direction. Because the positive perpendicular direction, marked by a dashed reference arrow, is outward, the inward force on the gas is negative.

If the wall undergoes a small displacement $\Delta r_{\perp}$ in the direction perpendicular to the wall, then the work done on the gas by the wall is

\[W = F_{\mathrm{cg},\perp} \, \Delta r_{\perp} = - p A \, \Delta r_{\perp}.\]

A wall of area A sliding outward. The gas is a light cyan region with blue dots on the left; a dashed line marks where the wall started, and the grey wall now stands a short distance to the right of it. The thin band between the dashed line and the wall is shaded darker cyan and labeled delta V: it is the extra volume the gas gained. A dimension arrow across the band gives its width, the perpendicular displacement delta r perpendicular, and a dimension arrow along the wall gives its area A, so delta V equals A times delta r perpendicular. A red arrow at the wall, labeled F sub c g, points back into the gas: the force of the container on the gas is opposite to the displacement, so the work done on the gas is negative.

The product $A \, \Delta r_{\perp}$ is the gas’s change in volume $\Delta V$.

Sign convention for work

We find that the work done on the gas for a small volume change $\Delta V$ is

\[W = - p \, \Delta V.\]

It’s important to get the sign right in this equation!

  • If the wall moves outward, then the gas expands ($\Delta V > 0$). Since the displacement is opposite the inward force of the container on the gas, the work done on the gas is negative ($W < 0$).
  • Conversely, if the wall moves inward, the gas is compressed ($\Delta V < 0$), and the work done on the gas is positive ($W > 0$).

Work is the area under a $pV$ curve

We can calculate the total work done in a process by adding up the work done over many small volume changes. In the limit of $\Delta V \to 0$, the work done on the gas is the integral

\[\boxed{W = -\int_{V_i}^{V_f} p \, dV}\]

Remember that the integral $\int p \, dV$ is the signed area between the graph of $p(V)$ and the $V$-axis. So we have

\[\boxed{W = -\left(\text{area under the curve in a } pV \text{ diagram}\right).}\]

Work in a constant volume process

In a constant volume process, because the volume does not change, no work is done on the gas. The $pV$ curve for a constant volume process is a vertical line with no area below the curve.

A pressure versus volume diagram showing a constant-volume process. Two faint dashed hyperbolas are reference isotherms, the lower one at temperature T sub 1 and the upper one at higher temperature T sub 2. A vertical process arrow runs upward from state 1 on the lower isotherm to state 2 on the upper isotherm. Dashed guide lines run from the two states to the axes: both states drop to the same volume, labeled V sub 1 equals V sub 2, and they mark the lower pressure p sub 1 and the higher pressure p sub 2. The pressure increases while the volume stays the same. There is no area under a vertical path, so the work is zero and all of the added heat increases the internal energy.

Work in a constant pressure process

In a constant pressure process, the $pV$ curve is a horizontal line, and the region under the curve is a rectangle.

A pressure versus volume diagram showing a constant-pressure expansion. Two faint dashed hyperbolas are reference isotherms, the lower at temperature T sub 1 and the upper at higher temperature T sub 2. A horizontal process arrow at a fixed pressure, labeled p sub 1 equals p sub 2, runs rightward from state 1 on the lower isotherm to state 2 on the upper isotherm, so the volume increases from V sub 1 to V sub 2 at constant pressure. Dashed guide lines drop from both states to the volume axis. The rectangular area under the path, of height p and width the change in volume, is hatched in purple; this area, p times delta V, is the magnitude of the work.

We can calculate the work as minus the signed area under this line. Note that the area has a sign depending on the direction of the process: the area is positive for expansion and negative for compression.

A pressure versus volume diagram showing a constant-pressure expansion. A horizontal process arrow at fixed pressure p runs rightward from state 1 at volume V sub 1 to state 2 at the larger volume V sub 2, so the volume increases at constant pressure. Dashed guide lines drop from each state to the axes. The rectangular region under the path, of height p and width the change in volume, is hatched in purple; because the volume increases, this signed area is positive and the work done on the gas is negative.

Expansion: the area under the curve is positive, so $W < 0$.

A pressure versus volume diagram showing a constant-pressure compression, the mirror image of the expansion beside it. A horizontal process arrow at the same fixed pressure p runs leftward from state 1 at volume V sub 1 to state 2 at the smaller volume V sub 2, so the volume decreases at constant pressure. Dashed guide lines drop from each state to the axes. The rectangular region under the path is hatched in purple and has the same shape as before, but because the process runs toward smaller volume the signed area is negative and the work done on the gas is positive.

Compression: the area under the curve is negative, so $W > 0$.

When calculating the work done in a constant pressure process, $p$ is constant, so we can pull it out of the integral:

\[W = -\int_{V_i}^{V_f} p \, dV = -p \int_{V_i}^{V_f} dV = -p \left( V_f - V_i \right) = -p \, \Delta V\]

The work done on a gas in a constant pressure process is

\[\boxed{W = -p \, \Delta V,}\]

which is minus the area of the rectangle under the $pV$ curve.

Work in an isothermal process

For an isothermal process, we can calculate the work by finding the area under the isotherm given by $p = nRT/V$ between the initial and final volumes.

A pressure versus volume diagram showing an isothermal expansion. A single hyperbola p equals C over V represents an isotherm at constant temperature. The gas expands along the isotherm from state 1, at volume V sub 1 and high pressure p sub 1, to state 2, at the larger volume V sub 2 and lower pressure p sub 2; that stretch of the curve is drawn thicker, with arrowheads pointing from 1 to 2. Dashed guide lines run from each state to the axes. The area under the curve between the two volumes is hatched in purple; it equals the integral of p dV, the magnitude of the work. Because temperature is constant, the internal energy change is zero, so the heat added equals the work done by the gas.

Plugging this into the expression for work, we get

\[W = -\int_{V_i}^{V_f} p \, dV = -\int_{V_i}^{V_f} \frac{nRT}{V} \, dV = -nRT \int_{V_i}^{V_f} \frac{1}{V} \, dV = -nRT \left[ \ln V \right]_{V_i}^{V_f}\]

We find that the work done on the gas in an isothermal process is

\[\boxed{W = -nRT \ln\left(\frac{V_f}{V_i}\right).}\]

Example: Work along two paths

A sample of $0.15\ \mathrm{mol}$ of nitrogen gas is in a sealed container with a movable piston. The gas initially occupies volume $V_1 = 1.6\ \mathrm{L}$ at a pressure of $p_1 = 220\ \mathrm{kPa}$. Consider the following two processes:

  • (A) the gas expands at constant pressure to a volume of $V_2 = 2.2\ \mathrm{L}$, then is cooled at constant volume until it reaches pressure $p_2 = 160\ \mathrm{kPa}$.
  • (B) the gas is cooled at constant volume to the same final pressure $p_2 = 160\ \mathrm{kPa}$, then expands at constant pressure to the final volume $V_2 = 2.2\ \mathrm{L}$.

Try parts (a) and (b) yourself first:

  • (a) Draw a $pV$ diagram showing both paths (A) and (B).
  • (b) Do these two states fall on the same isotherm?
  • (c) Calculate the work done on the gas along each path.
  • (d) How much work would be done in an isothermal process connecting these same two states?
Solution

(a) Path A runs right along $p_1$, then down to state 2. Path B runs down to $p_2$, then right to state 2. The diagram also shows the isotherm through both states, found in part (b) and used in part (d).

Pressure versus volume diagram, with volume in liters on the horizontal axis and pressure in kilopascals on the vertical axis; the axes start at 1.4 liters and 140 kilopascals rather than at zero. State 1 is at 1.6 liters and 220 kilopascals (upper left); state 2 is at 2.2 liters and 160 kilopascals (lower right). Three paths connect them, each with arrows pointing from state 1 toward state 2. Path A, a solid blue line, runs horizontally right from state 1 at 220 kilopascals to 2.2 liters, then straight down to state 2. Path B, a dashed red line, runs straight down from state 1 to 160 kilopascals, then horizontally right to state 2. The isothermal path, a dotted teal curve, bends between the two, below path A and above path B.

(b) Two states lie on the same isotherm if they have the same temperature. Converting to SI units,

\[\begin{aligned} V_1 &= 1.6\times 10^{-3}\ \mathrm{m^3}, & p_1 &= 220\times 10^{3}\ \mathrm{Pa}, \\ V_2 &= 2.2\times 10^{-3}\ \mathrm{m^3}, & p_2 &= 160\times 10^{3}\ \mathrm{Pa}. \end{aligned}\]

The container is sealed, so $n$ stays fixed. Solving the ideal gas law $pV = nRT$ for the temperature gives $T = pV/nR$, so

\[\begin{aligned} T_1 &= \frac{p_1 V_1}{nR} = \frac{(220\times 10^{3}\ \mathrm{Pa})(1.6\times 10^{-3}\ \mathrm{m^3})}{(0.15\ \mathrm{mol})(8.314\ \mathrm{J/(mol\cdot K)})} = 282\ \mathrm{K}, \\ T_2 &= \frac{p_2 V_2}{nR} = \frac{(160\times 10^{3}\ \mathrm{Pa})(2.2\times 10^{-3}\ \mathrm{m^3})}{(0.15\ \mathrm{mol})(8.314\ \mathrm{J/(mol\cdot K)})} = 282\ \mathrm{K}. \end{aligned}\]

The two temperatures match, so both states lie on the same isotherm, at $T = 282\ \mathrm{K}$.

(c) No work is done on the gas during a constant-volume step, since $\Delta V = 0$. Each path does work only in its constant-pressure step, where $W = -p\,\Delta V$.

Path A expands at $p_1$:

\[\begin{aligned} W_A &= -p_1 \left(V_2 - V_1\right) \\ &= -(220\times 10^{3}\ \mathrm{Pa})(2.2\times 10^{-3}\ \mathrm{m^3} - 1.6\times 10^{-3}\ \mathrm{m^3}) = \boxed{-132\ \mathrm{J}}. \end{aligned}\]

Path B expands at $p_2$:

\[\begin{aligned} W_B &= -p_2 \left(V_2 - V_1\right) \\ &= -(160\times 10^{3}\ \mathrm{Pa})(2.2\times 10^{-3}\ \mathrm{m^3} - 1.6\times 10^{-3}\ \mathrm{m^3}) = \boxed{-96\ \mathrm{J}}. \end{aligned}\]

Both works are negative because the gas expands along each path. Path A expands at the higher pressure, so the gas does more work on its surroundings: $W_A$ is more negative than $W_B$.

(d) Part (b) showed that both states lie on the isotherm at $T_1 = 282\ \mathrm{K}$. The work done on the gas in an isothermal process is

\[W = -nRT_1 \ln\left(\frac{V_2}{V_1}\right).\]

Since $nRT_1 = p_1 V_1$, we can write this as

\[\begin{aligned} W &= -p_1 V_1 \ln\left(\frac{V_2}{V_1}\right) \\ &= -(220\times 10^{3}\ \mathrm{Pa})(1.6\times 10^{-3}\ \mathrm{m^3}) \ln\left(\frac{2.2\ \mathrm{L}}{1.6\ \mathrm{L}}\right) = \boxed{-112\ \mathrm{J}}. \end{aligned}\]

The isotherm lies between paths A and B on the $pV$ diagram, so its work falls between $W_A$ and $W_B$. The three processes join the same two states but do different amounts of work: work depends on the path.

Path functions vs. state functions

The example above illustrates an important point: the work done on a gas depends on the process connecting the initial and final states. Knowing only the initial and final states is not enough to determine the work done on a gas.

We call a quantity that depends on the path between two states a path function. Both work and heat are path functions.

In contrast, the internal energy of an ideal gas depends only on the temperature, so it is determined by the state of the system. A quantity that only depends on the state of the system is called a state function. The change in a state function (like $\Delta E_{\mathrm{int}}$) only depends on the initial and final states, and not the specific process connecting them.