Chapter 6, Problem 6.1P

Ten mL of pure liquid water in a cylinder with a movable piston is heated at a constant pressure of 1 atm from an initial temperature of 80°C. The temperature of the system is monitored, and the following behavior is observed:

(a) What is happening in steps AB, BC, and CD? What is the temperature corresponding to the horizontal portion of the curve?

(b) Estimate the volume occupied by the water at points B and C. (Assume the vapor follows the ideal-gas equation of state.)

Interpretation Introduction

(a)

Interpretation:

Identify the process in AB, BC, CD areas and calculate horizontal zone temperature.

Concept introduction:

Energy is transferred into a system when it is heated. The system will change depending on the energy it receives. This can happen through increase in temperature. A heating curve is called a plot of the temperature versus time.

Answer to Problem 6.1P

• AB − Water liquid heating zone (temperature increase)
• BC − Vaporization of water (liquid to gas phase)
• CD − water vapor (gas phase) heating zone

Horizontal portion temperature = 100°C

Explanation of Solution

AB Step

As heat is absorbed, the temperature of the liquid begins to increase this is due to the increase in the kinetic energy of the molecules of liquid. For standard atmospheric pressure, the rise in temperature takes place until it reaches to 100°C. With increasing temperature, the volume of the liquid remains constant due to little expansion for liquids.

BC Step

At this point, the heat is consumed to begin vaporization of liquid. This temperature is known as the bubble point temperature. Due to change in the previous state, the temperature remains constant and water molecules are moving from liquid to vapor phase. At point C, the last drop of liquid gets evaporated.

CD Step

The temperature increases above 100 °C, when heating steam boils all the liquid into vapor. This cause increase in the volume. Step B shows transition of liquid to vapor. The boiling point of water is 100 °C (liquid here is water) and pressure is 1 atm.

Interpretation Introduction

(b)

Interpretation:

Volume occupied by water at point B and C should be calculated.

Concept introduction:

The ideal gas equation is represented as follows:

$PV=nRT$

Here, $P$ = Gas pressure

$V$ = Gas volume

$n$ = number moles

$R$ = the gas constant

$T$ = temperature in Kelvin

Answer to Problem 6.1P

At point B, the water is present as liquid only thus, the volume will be 10 mL

At point C $=17.\text{0 L}$

Explanation of Solution

At point B, the water is present as liquid only thus, the volume will be 10 mL

$\begin{array}{l}{\rho }_{{\text{H}}_{\text{2}}\text{O}}=1.\text{0 g/mL}\\ M_{{\text{H}}_{\text{2}}\text{O}}=18.01\text{6 g/mol}\end{array}$

The number of moles in 10 mL of liquid water is calculated as follows:

$\begin{array}{l}n=\frac{V\rho }{M}\\ =\frac{1\text{0 mL}\times \frac{\text{1}\text{.0 g}}{\text{mL}}}{\frac{\text{18}\text{.016 g}}{\text{mol}}}\\ =0.555\text{ mol}\end{array}$

The volume of vapor can be calculated using the ideal gas equation as follows:

$V=\frac{nRT}{P}$

Here, number of moles is 0.555 mol at 1 atm and ${100}^0\text{C}$. Converting temperature into Kelvin thus,

${100}^0\text{C=}\left(\text{100+273}\right)\text{ K=373 K}$

Putting the values,

$=\frac{\text{0}\text{.555 mol×}\frac{\text{0}\text{.0826 L}\text{.atm}}{\text{mol}\text{.K}}\text{×373 K}}{\text{1 atm}}$

$=17.\text{0 L}$

Thus, at point C, the volume occupied by the water occupies is 17 L.

At point C, $=17.0\text{ L}$ the water is present in vapor from.