18.7.3 Constant-Volume Process

Because a gas completely fills its container, the volume of the gas is the volume of the container. Many important gas processes take place in a container of constant, unchanging volume often referred to as a rigid container. A constant-volume process is called an isochoric process, where iso is a prefix meaning “constant” or “equal” while choric is from a Greek root meaning “volume.” An isochoric process is one for which

V_{f} = V_{i}

(18.19)

For example, suppose that you have a gas in the closed, rigid container shown in FIGURE 18.11a. Warming the gas with a Bunsen burner will raise its pressure without changing its volume. This process is shown as the vertical line $1 \to 2$ on the $p V$ diagram of FIGURE 18.11b. A constant-volume cooling, by placing the container on a block of ice, would lower the pressure and be represented as the vertical line from 2 to 1. Any isochoric process appears on a $p V$ diagram as a vertical line.

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

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

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EXAMPLE 18.7 A constant-volume gas thermometer

A constant-volume gas thermometer is placed in contact with a reference cell containing water at the triple point. After reaching equilibrium, the gas pressure is recorded as 55.78 kPa. The thermometer is then placed in contact with a sample of unknown temperature. After the thermometer reaches a new equilibrium, the gas pressure is 65.12 kPa. What is the temperature of this sample?

MODEL The thermometer’s volume doesn’t change, so this is an isochoric process.

SOLVE The temperature at the triple point of water is $T_{1} = 0.01 ° C = 273.16 K$ . The ideal-gas law for a closed system is $p_{2} V_{2} / T_{2} = p_{1} V_{1} / T_{1}$ . The volume doesn’t change, so $V_{2} / V_{1} = 1$ . Thus

\begin{matrix} T_{2} & = & T_{1} \frac{V_{2}}{V_{1}} \frac{p_{2}}{p_{1}} = T_{1} \frac{p_{2}}{p_{1}} = (273.16 K) \frac{65.12 kPa}{55.78 kPa} \\ = & 318.90 K = 45.75 ° C \end{matrix}

The temperature must be in kelvins to do this calculation, although it is common to convert the final answer to $° C$ . The fact that the pressures were given to four significant figures justified using $T_{K} = T_{C} + 273.15$ rather than the usual $T_{C} + 273$ .

REVIEW $T_{2} > T_{1}$ , which we expected from the increase in pressure.