Example 20.6 Boiling Water
Suppose 1.00 g of water vaporizes isobarically at atmospheric pressure (1.013 × 10⁵ Pa). Its volume in the liquid state is Vi = Vliquid = 1.00 cm³, and its volume in the vapor state is Vf = Vvapor = 1671.00 cm³. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the steam and the surrounding air; imagine that the steam simply pushes the surrounding air out of the way.
SOLVE IT
Conceptualize Notice that the temperature of the system does not change. There is a phase change occurring as the water evaporates to steam.
Categorize Because the expansion takes place at constant pressure, we categorize the process as isobaric. We will use equations developed in the preceding sections, so we categorize this example as a substitution problem.
Use the equation for an isobaric process to find the work done on the system as the air is pushed out of the way:
W = -P(Vf - Vi)
= -(1.013 × 10⁵ Pa)(1.67 × 10⁻³ m³ - 1.00 × 10⁻⁶ m³)
= -169.07 J
Use the latent heat equation and the latent heat of vaporization for water to find the energy transferred into the system by heat:
Q = mLv = (1.00 × 10⁻³ kg)(2.26 × 10⁶ J/kg) = 2260 J
Use the first law to find the change in internal energy of the system:
ΔEint = Q + W = 2260 J + W = 2090.93 kJ
MASTER IT
As you can see in the example, the volume in the gaseous state is higher than that in the liquid state at boiling. As a good approximation, we can ignore the initial volume occupied by the liquid. Now with 1.00 g of ethyl alcohol (molar mass = 46.1 g), assuming the gas is ideal, what is the change in internal energy as it vaporizes at its boiling point?
ΔEint = 185.11
Review the example problem and check your calculation for Vf.