Critical Temperature and Pressure

An introduction to the critical temperature and the critical pressure of a working substance

Contents

Overview
Page Comments

Overview

Definitions

Working substance (WS) = the WS is used as the carrier for heat energy. The state of the WS is defined by the values of its properties, e.g. pressure, volume, temperature, internal energy, enthalpy. These properties are also sometimes called functions of state.

<br/>

Key facts

The critical temperature ( $\inline T_C$ ) of a WS is the temperature at and above which vapours of that WS cannot be liquefied, no matter how much pressure is applied.

The critical pressure ( $\inline P_C$ ) of a WS is the pressure required to liquefy the WS at its critical temperature.

<br/>

Constants

$T_{C,air} = -220^\circ F$

$P_{C,air} = 573 \; psi$

$T_{C,water} = 706^\circ F$

$P_{C,water} = 3200 \; psi$

In order to introduce the critical temperature and the critical pressure of a working substance (WS), consider the pressure ( $\inline P$ ) - volume ( $\inline V$ ) plot diagramed in Figure 1, which highlights the change in volume with the application of pressure for a fixed mass of gas, found at different constant temperatures. The various curves thus obtained (also called isotherms) are represented in Figure 1 in blue.

Consider for example the $\inline T_1$ isotherm. When initially found at low pressures (point $\inline A_1$ ), the WS is gaseous. In addition, over the $\inline A_1-B_1$ interval, the WS generally exhibits the characteristics of a gas. At point $\inline B_1$ , the WS is found at saturation, as any slight increase in pressure will result in a change from the vapour state to the liquid state. Afterwards, over the $\inline B_1-C_1$ interval, the WS can be found as a mix of both vapour and liquid states. Also over the $\inline B_1-C_1$ interval, the pressure is virtually constant, while the volume is decreasing. At point $\inline C_1$ , the WS is found entirely in a liquid state. From point $\inline C_1$ , the graph becomes almost vertical, indicating that a significant further application of pressure leads to a very little change in volume (as expected, as liquids are virtually incompressible).

Consider now the $\inline T_2$ isotherm. As can be seen from Figure 1, the $\inline T_2$ isotherm follows a path which is very similar to that of the $\inline T_1$ isotherm. However, the $\inline B_2-C_2$ interval is smaller than the corresponding $\inline B_1-C_1$ interval. This indicates that the properties of the liquid and gas states of the WS are becoming increasingly similar, leading to a point where they will coincide. This point is indeed reached for the $\inline T_C$ isotherm, which does not show any horizontal discontinuity.

Note: The locus of all the $\inline A$ and $\inline B$ pairs corresponding to all the isotherms constructs the gray curve highlighted in Figure 1.

The $\inline T_C$ temperature is called the critical temperature, while the corresponding pressure is called the critical pressure ( $\inline P_C$ ). As can be seen from Figure 1, the critical temperature of a WS is the temperature at and above which vapours of that WS cannot be liquefied, no matter how much pressure is applied. The critical pressure is, in turn, the pressure required to liquefy the WS at its critical temperature.

Every WS has its own characteristic critical temperature and pressure. For example, for air: $\inline T_{C,air}=-220^\circ F$ , and $\inline P_{C,air}=573 \; psi$ , while for water: $\inline T_{C,water}=706^\circ F$ , and $\inline P_{C,water}=3200 \; psi$ .

Now consider the same process as in Figure 1, but this time plotted on a temperature ( $\inline T$ ) - entropy ( $\inline S$ ) diagram (see Figure 2A). The various curves obtained for the different constant pressures (also called isobars) are represented in blue.

Consider for example the $\inline P_1$ isobar. The enthalpy corresponding to this constant pressure, $\inline H_1$ , can be calculated as the area under the $\inline P_1$ isobar, limited by $\inline T_1$ and $\inline S_1$ (the blue shaded area in Figure 2B).

However, this area can also be calculated as:

$H_1 = T_1 S_1 - \mathcal{A}_{ABC}$

(1)

where $\inline \mathcal{A}_{ABC}$ is the light blue shaded area in Figure 2B.

Now imagine another isobar, $\inline P_2$ . Similarly, the corresponding enthalpy $\inline H_2$ can be calculated as the area under the $\inline P_2$ isobar, limited by $\inline T_2$ and $\inline S_2$ (the blue shaded area in Figure 2C).