1 Practice Problems


Chapter 1
Introduction
Problem 1.1
Consider a glass container, half-full of water and half-full of air, at rest on a labo-
ratory table. List some similarities and di erences between the liquid (water) and
the gas (air).
Solution
Similarities
1. The gas and the liquid are comprised of molecules.
2. The gas and the liquid are uids.
3. The molecules in the gas and the liquid are relatively free to move about.
4. The molecules in each uid are in continual and random motion.
Di erences
1
2 CHAPTER 1. INTRODUCTION
1. In the liquid phase, there are strong attractive and repulsive forces between
the molecules; in the gas phase (assuming ideal gas), there are minimal forces
between molecules except when they are in close proximity (mutual repulsive
forces simulate collisions).
2. A liquid has a de nite volume; a gas will expand to ll its container. Since
the container is open in this case, the gas will continually exchange molecules
with the ambient air.
3. A liquid is much more viscous than a gas.
4. A liquid forms a free surface, whereas a gas does not.
5. Liquids are very di cult to compress (requiring large pressures for small com-
pression), whereas gases are relatively easy to compress.
6. With the exception of evaporation, the liquid molecules stay in the container.
The gas molecules constantly pass in and out of the container.
7. A liquid exhibits an evaporation phenomenon, whereas a gas does not.
Comments
Most of the di erences between gases and liquids can be understood by considering
the di erences in molecular structure. Gas molecules are far apart, and each
molecule moves independently of its neighbor, except when one molecule approaches
another. Liquid molecules are close together, and each molecule exerts strong
attractive and repulsive forces on its neighbor.
Problem 1.2
In an ink-jet printer, the ori ce that is used to form ink drops can have a di-
ameter as small as 3 × 10 6 m. Assuming that ink has the properties of water,
does the continuum assumption apply?
Solution
The continuum assumption will apply if the size of a volume, which contains enough
molecules so that e ects due to random molecular variations average out, is much
smaller that the system dimensions. Assume that 104 molecules is su cient for
averaging. If is the length of one side of a cube that contains 104 molecules and
is the diameter of the ori ce, the continuum assumption is satis ed if

ż 1

The number of molecules in a mole of matter is Avogadro s number: 6 02 × 1023
The molecular weight of water is 18, so the number of molecules ( ) in a gram of
3
water is
µ Å› µ Å›
6 02 × 1023 molecules mole
=
mole 18 g
molecules
= 3 34 × 1012
g
The density of water is 1 g/cm3 so the number of molecules in a cm3 is 3 34 × 1012
The volume of water that contains 104 molecules is
104 molecules
Volume =
3 34 × 1012 molecules
cm3
= 3 0 × 10 19 cm3
Since the volume of a cube is 3, where is the length of a side
p
3
= 3 0 × 10 19 cm3
= 6 2 × 10 7 cm
= 6 2 × 10 9 m
Thus
6 2 × 10 9 m
=
3 0 × 10 6 m
= 0 0021

Since ż 1 the continuum assumption is quite good.

4 CHAPTER 1. INTRODUCTION


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