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\noindent {\bf PHYSICS 1020}
\begin{center}
{\Large \bf Homework set 2\\}
{\large 28 August 1996\\}
\end{center}
\bigskip
\begin{description}
\item[{\bf [2.1]}] 
Is the atomic theory known, for certain, to be true?  \\
Answer:\\
No; a theory, as all scientific principles, is never certain. It is permanently
subject to testing and experimental verification. Even if a  theory is found to
agree with all of the available observational data, and all of its predictions
have been found to be ``correct'', i.e. borne out by all observations until
today, it my yet be
disproved by tomorrow's more precise observations. It is, in fact, the attempt
to disprove the existing theories which is the purpose of most scientific
experimentation.
\item[{\bf [2.2]}]
Carbon atoms are about 25\% lighter than oxygen atoms (the ratio of their 
weights is 3 to 4). What is the weight ratio of the carbon and oxygen that 
go into the formation of carbon monoxide? Answer the same question for carbon
dioxide.\\
Answer:\\
Carbon monxyde has the chemical composition CO, i.e. there is one carbon
atom for every oxygen atom; therefore the weight ratio of carbon to oxygen 
in this compound is just the weight ratio of the atoms, i.e.  3 to 4.
In carbon dioxide, ${\rm CO_2}$, there are two oxyen atoms per carbon atom, so the carbon 
to oxygen weight ratio in carbon dioxide is 3 to 8.
\item[{\bf [2.4]}] 
Which of these is a pure {\sl compound}, which is an {\sl element}, 
and which is neither:
(a) pure water, (b) oxygen gas, (c) liquid mercury, (d) ${\rm H_2SO_4}$, 
(e) U, (f) air,\\
 (g) He, (h) carbon dioxide, (i) ${\rm H_2}$, 
(j) polluted water? \\
Answer:
\begin{description}
\item [(a)] pure water is a pure {\sl compound} (${\rm H_2O}$)
\item [(b)] oxygen gas  is a pure {\sl element} (${\rm O_2}$)
\item [(c)] liquid mercury  is a pure {\sl element} (${\rm Hg}$)
\item [(d)] ${\rm H_2SO_4}$ (sulphuric acid) is a pure {\sl compound}
\item [(e)] U (Uranium) is a pure {\sl element}
\item [(f)] air is a {\sl mixture} of gases (elements and compounds):
 oxygen, nitrogen, carbon dioxide,
water vapor, argon, and others
\item [(g)] He (Helium) is a pure {\sl element} -- a ``noble gas''
\item [(h)] carbon dioxide  is  a pure {\sl compound} (${\rm CO_2}$) 
\item [(i)] ${\rm H_2}$ is  a pure {\sl element} -- hydrogen 
\item [(j)] polluted water is a {\sl mixture} of many things
\end{description}
\item[{\bf [2.6]}] 
Helium is an inert gas, meaning that it does not readily enter into chemical
reactions with other substances. List five other substances that you would
expect to also be inert gases.
  \\
Answer:\\
All the elements in the same column in the periodic table (``in the same
group'') have similar chemical properties. Therefore we expect all the elements
in the same group as Helium to share the poperty of being chemically inert.
These elements are the ``noble gases'' helium, neon, argon, krypton, xenon,
radon.
\newpage
\item[{\bf [2.8]}]  
In a typical large coal-fed electrical
generating plant, a ton of coal is burned every ten seconds. About how many tons
of carbon dioxide enter the atmosphere every hour from such a plant?
  \\
Answer:\\
Let's assume (as in exercise 2.7) that coal is nearly pure carbon, and that oxygen
and carbon have the same weight. Since one ton of coal (i.e. carbon) is burned
every ten seconds, 3 tons of carbon dioxide (${\rm CO_2}$) enter the atmosphere
every ten seconds. One hour (3600 seconds) has 360 ten-second intervals,
and thus the total amount of carbon dioxide emitted into the atmosphere per hour
= 360 times 3 = 1080 tons.
Using the correct mass ratio (3:4) between carbon and oxygen, one finds that 
the total mass of carbon dioxide is (4+4+3)/3 times that of carbon.
Therefore the total amount of carbon dioxide emitted per hour = (11/3) times 360
= 1320 tons.
 
\item[{\bf [2.14]}] 
If air is put into a sealed container that is then compressed (reduced in volume)
what do you predict will happen to the air pressure on the container walls?
Explain this from a microscopic view.\\
Answer:\\
Reducing the volume will increase the density, i.e. the number of molecules 
per unit volume will increase. In other words, the space available for each 
molecule's motion will be smaller than before.
As a consequence, the number of collisions per unit time of molecules with the 
container wall will be higher then before. Since pressure is due to the force
exerted by the molecules due to their collisions with the walls, an increased 
number of collisions will appear as higher pressure.

\item[{\bf [2.15]}] 
If air is put into a sealed container and heated, what do you predict will
happen to the air pressure on the container walls? Explain this from a
microscopic view.
  \\
Answer:\\
Heating the air in the container will increase the average kinetic energy of
random motion of the gas molecules in the container. As a consequence they will
hit the container walls with higher speeds; this will appear as 
 increased pressure.
\item[{\bf [2.21]}] 
Making estimates: The smallest dust particle visible to the unaided human eye
measures about 0.05 mm across. About how many atoms across is this -- in other
words, if we line up atoms side by side, about how many would it take to make
a line  of atoms 0.05 mm long?
  \\
Answer:\\
The diameter of an atom is about $10^{-10}{\rm m}$, and the dust particle
measures 0.05 mm = $5 \times 10^{-5}{\rm m}$ across. Therefore the number of 
atoms needed to be lined up to make a length equal to the lenght of the dust
particle is 
$${{5 \times 10^{-5}{\rm m}} \over {10^{-10}{\rm m}}} = 5\times 10^5 
= 500 000.$$
It takes half a million atoms.
\newpage
\item[{\bf [2.22]}]
Making estimates: Referring to the preceding exercise: Assume the small dust
particle is shaped like a cube. 
About how many atoms does it contain?
  \\
Answer:\\
There are $5 \times 10^5$ atoms across each of the three sides (dimensions)
of the cube. The volume of a  cube = the ``cube'' (i.e. the third power) of the 
length of its side. Therefore the number of atoms in the cube is equal to the 
third power (``cube'') of the number of atoms across one side, i.e.
$$= (5\times 10^5)^3 = 5^3 \times 10^{5\times 3} = 125 \times 10^{15} = 1.25 \times 10^{17}$$

\item[{\bf [2.25]}] 
Gasoline is a hydrocarbon fuel. What are the main compounds created when
gasoline burns in a car engine?
  \\
Answer:\\
A hydrocarbon is a compound consisting essentially of hydrogen H and carbon C.
Burning means forming a new compound with oxygen. The compounds formed will be
water and carbon dioxide (${\rm H_2O ~ and ~ CO_2}$). 
\end{description}
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