18.1.1 State Variables

The parameters used to characterize or describe a macroscopic system are known as state variables because, taken all together, they describe the state of the macroscopic system. You met some state variables in earlier chapters: volume, pressure, mass, mass density, and thermal energy. We’ll soon introduce several new state variables.

One important state variable, the mass density, is defined as the ratio of two other state variables:

ρ = \frac{M}{V} (mass density)

(18.1)

In this chapter we’ll use an uppercase $M$ for the system mass and a lowercase $m$ for the mass of an atom. TABLE 18.1 is a short list of mass densities.

TABLE 18.1 Densities of materials

Substance	$ρ (kg / m^{3})$
Air at STP^*	1.29
Ethyl alcohol	790
Water (solid)	920
Water (liquid)	1000
Aluminum	2700
Copper	8920
Gold	19,300
Iron	7870
Lead	11,300
Mercury	13,600
Silicon	2330

^* $T = 0 ° C, p = 1 atm$

A system is said to be in thermal equilibrium if its state variables are constant and not changing. As an example, a gas is in thermal equilibrium if it has been left undisturbed long enough for $p$ , $V$ , and $T$ to reach steady values.

EXAMPLE 18.1 The mass of a lead pipe

A project on which you are working uses a cylindrical lead pipe with outer and inner diameters of 4.0 cm and 3.5 cm, respectively, and a length of 50 cm. What is its mass?

SOLVE The mass density of lead is $ρ_{lead} = 11, 300 kg / m^{3}$ . The volume of a circular cylinder of length $l$ is $V = π r^{2} l$ . In this case we need to find the volume of the outer cylinder, of radius $r_{2}$ , minus the volume of air in the inner cylinder, of radius $r_{1}$ . The volume of the pipe is

V = π r_{2}^{2} l - π r_{1}^{2} l = π (r_{2}^{2} - r_{1}^{2}) l = 1.47 \times 10^{- 4} m^{3}

Hence the pipe’s mass is

M = ρ_{lead} V = 1.7 kg