18.7.5 Constant-Temperature Process

The last process we wish to look at for now is one that takes place at a constant temperature. A constant-temperature process is called an isothermal process. An isothermal process is one for which $T_{f} = T_{i}$ . Because $p V = n R T$ , a constant-temperature process in a closed system (constant $n$ ) is one for which the product $p V$ doesn’t change. Thus

p_{f} V_{f} = p_{i} V_{i}

(18.22)

in an isothermal process.

One possible isothermal process is illustrated in FIGURE 18.15a, where a piston is being pushed down to compress a gas. If the piston is pushed slowly, then heat energy transfer through the walls of the cylinder keeps the gas at the same temperature as the surrounding liquid. This is an isothermal compression. The reverse process, with the piston slowly pulled out, would be an isothermal expansion.

The figure labeled ("a") consists of two parts labeled before and after that illustrate the change in volume and pressure at constant temperature T.

The figure labeled (b) illustrates a pressure versus volume graph.

Full Alternative Text

Representing an isothermal process on the $p V$ diagram is a little more complicated than the two previous processes because both $p$ and $V$ change. As long as $T$ remains fixed, we have the relationship

p = \frac{n R T}{V} = \frac{constant}{V}

(18.23)

The inverse relationship between $p$ and $V$ causes the graph of an isothermal process to be a hyperbola. As one state variable goes up, the other goes down.

The process shown as $1 \to 2$ in FIGURE 18.15b represents the isothermal compression shown in Figure 18.15a. An isothermal expansion would move in the opposite direction along the hyperbola.

The location of the hyperbola depends on the value of $T$ . A lower-temperature process is represented by a hyperbola closer to the origin than a higher-temperature process. FIGURE 18.15c shows four hyperbolas representing the temperatures $T_{1}$ to $T_{4}$ , where $T_{4} > T_{3} > T_{2} > T_{1}$ . These are called isotherms. A gas undergoing an isothermal process moves along the isotherm of the appropriate temperature.

EXAMPLE 18.10 Compressing air in the lungs

An ocean snorkeler takes a deep breath at the surface, filling his lungs with $4.0 L$ of air. He then descends to a depth of $5.0 m$ . At this depth, what is the volume of air in the snorkeler’s lungs?

MODEL At the surface, the pressure in the lungs is $1.00 atm$ . Because the body cannot sustain large pressure differences between inside and outside, the air pressure in the lungs rises—and the volume decreases—to match the surrounding water pressure as he descends.

SOLVE The ideal-gas law for a sealed container is

V_{2} = \frac{p_{1}}{p_{2}} \frac{T_{2}}{T_{1}} V_{1}

Air is quickly warmed to body temperature as it enters through the nose and mouth, and it remains at body temperature as the snorkeler dives, so $T_{2} / T_{1} = 1$ . We know $p_{1} = 1.00 atm = 101, 300 Pa$ at the surface. We can find $p_{2}$ from the hydrostatic pressure equation, using the density of seawater:

\begin{matrix} p_{2} & = & p_{1} + ρ g d = 101, 300 Pa + (1030 kg / m^{3}) (9.80 m / s^{2}) (5.0 m) \\ = & 151,800 Pa \end{matrix}

With this, the volume of the lungs at a depth of 5.0 m is

V_{2} = \frac{101, 300 Pa}{151, 800 Pa} \times 1 \times 4.0 L = 2.7 L

REVIEW The air inside your lungs does compress—significantly—as you dive below the surface.

EXAMPLE 18.11 A multistep process

A gas at 2.0 atm pressure and a temperature of $200 ° C$ is first expanded isothermally until its volume has doubled. It then undergoes an isobaric compression until it returns to its original volume. First show this process on a $p V$ diagram. Then find the final temperature ( $in ° C$ ) and pressure.

MODEL The final state of the isothermal expansion is the initial state for an isobaric compression.

VISUALIZE FIGURE 18.16 shows the process. As the gas expands isothermally, it moves downward along an isotherm until it reaches volume $V_{2} = 2 V_{1}$ . The gas is then compressed at constant pressure $p_{2}$ until its final volume $V_{3}$ equals its original volume $V_{1}$ . State 3 is on an isotherm closer to the origin, so we expect to find $T_{3} < T_{1}$ .

The figure illustrates a pressure versus volume graph with two curves at different temperatures.

FIGURE 18.16 Full Alternative Text

SOLVE $T_{2} / T_{1} = 1$ during the isothermal expansion and $V_{2} = 2 V_{1}$ , so the pressure at point 2 is

p_{2} = p_{1} \frac{T_{2}}{T_{1}} \frac{V_{1}}{V_{2}} = p_{1} \frac{V_{1}}{2 V_{1}} = \frac{1}{2} p_{1} = 1.0 atm

We have $p_{3} / p_{2} = 1$ during the isobaric compression and $V_{3} = V_{1} = \frac{1}{2} V_{2}$ , so

T_{3} = T_{2} \frac{p_{3}}{p_{2}} \frac{V_{3}}{V_{2}} = T_{2} \frac{\frac{1}{2} V_{2}}{V_{2}} = \frac{1}{2} T_{2} = 236.5 K = - 36.5 ° C

where we converted $T_{2}$ to 473 K before doing calculations and then converted $T_{3}$ back to $° C$ . The final state, with $T_{3} = - 36.5 ° C$ and $p_{3} = 1.0 atm,$ is one in which both the pressure and the absolute temperature are half their original values.

CHAPTER 18 CHALLENGE EXAMPLE Depressing a piston

A large, 50.0-cm-diameter metal cylinder filled with air supports a 20.0 kg piston that can slide up and down without friction. The piston is 100.0 cm above the bottom when the temperature is $20 ° C$ . An 80.0 kg student then stands on the piston. After several minutes have elapsed, by how much has the piston been depressed?

MODEL The metal walls of the cylinder are a good thermal conductor, so after several minutes the gas temperature—even if it initially changed—will return to room temperature. The final temperature matches the initial temperature. Assume that the atmospheric pressure is 1 atm.

VISUALIZE FIGURE 18.17 shows the cylinder before and after the student stands on it. The volume of the cylinder is $V = A h$ , and only $h$ changes.

The figure consists of two parts that illustrate a cylinder filled with gas and a piston.

FIGURE 18.17 Full Alternative Text

SOLVE The ideal-gas law for a sealed container is

\frac{p_{2} A h_{2}}{T_{2}} = \frac{p_{1} A h_{1}}{T_{1}}

Because $T_{2} = T_{1}$ , the final height of the piston is

h_{2} = \frac{p_{1}}{p_{2}} h_{1}

The pressure of the gas is determined by the mass of the piston (and anything on the piston) and the pressure of the air above. In equilibrium,

p = p_{atmos} + \frac{M g}{A} = {\begin{array}{l} 1.023 \times 10^{5} Pa & piston only \\ 1.063 \times 10^{5} Pa & piston and student \end{array}

where we used $p_{atmos} = 1 atm = 1.013 \times 10^{5} Pa$ and $A = π r^{2} = 0.196 m^{2}$ . The final height of the piston is

h_{2} = \frac{1.023 \times 10^{5} Pa}{1.063 \times 10^{5} Pa} \times 100.0 cm = 96.2 cm

The question, however, was by how much the piston is depressed. This is $h_{1} - h_{2} = 3.8 cm$ .

REVIEW Neither the piston nor the student increases the gas pressure to much above 1 atm, so it’s not surprising that the added weight of the student doesn’t push the piston down very far.

18.7.5 Constant-Temperature Process

FIGURE 18.15 A constant-temperature (isothermal) process.

EXAMPLE 18.10 Compressing air in the lungs

EXAMPLE 18.11 A multistep process

FIGURE 18.16 A $p V$ diagram for the process of Example 18.11.

CHAPTER 18 CHALLENGE EXAMPLE Depressing a piston

FIGURE 18.17 The student compresses the gas.