Chapter 16 Webassign

Keep in mind that there are only three types of problems in Chapter 16: thermal expansion, ideal gas

law, or kinetic theory. It should be fairly simple to identify, from the information provided in the

problem, into which of these three categories a problem belongs.

Problem 1: This is a simple thermal expansion problem, i.e. you can use the expression

L = LI T to calculate L.

Note that you are given Li in km, a is 11 x 10-6 C-1 (from the table in Chap 16…) and T is the

difference between the two given temperatures. Your answer for L will be in km.

How do we compensate for the expansion of a pipeline? You cannot use expansion “gaps”, as we do

with bridges or railroad tracks, but we have to ensure the pipeline will not expand horizontally. So we

create vertical “loops” (or we could use horizontal loops that are oriented perpendicular to the

pipeline… i.e. if the pipeline is north-south, the loops could be east-west) in the pipe that can expand

when the temperature increases, keeping the ends of the pipeline fixed.

Problem 2: This problem is similar to Problem 1, but instead of the length of a pipeline you are given

the radius of the eyeglass frame. Using the initial and final radius (i.e. the radius of the lens is the final

radius of the frame…) you can determine r. You are given for the frame material. So you can

determine T. Use this and the given initial temperature to find the final temperature of the eyeglass

frame.

Problem 3: While this problem is very wordy, you should recognize it as a typical “a lot of words, but

very simple” problem. It describes a practical use for thermal expansion: a valve that is heated by hot

water, which causes a spring to expand and close the valve (if the water is too hot.)

If you carefully pick through the words, you should see that you are given:

initial length of the spring coeff of thermal expansion for the spring T for the spring

and you are asked to calculate L for the spring.

Problem 4: In this problem you are given a steel container which is filled with a liquid. Both the

container and the liquid are then warmed, causing both to expand. You are then asked to find the

volume of the liquid that spills.

The key to this problem is to recognize that the liquid and the steel container both expand. If we label

the initial volume of the container, which is also the initial volume of the liquid, as “V1”, the final volume

of the container as “V2“ and the final volume of the liquid as “V3”, we can write:

volume spilled = V3 – V2 V2 = V1 + Vsteel V3 = V1 + Vliquid

where the latter two expressions are simply expressing that the final volume of each is the initial

volume plus the increase in volume.

Now we can use our thermal expansion expressions for the steel and the liquid:

Vsteel = V1 3 T and Vliquid = V1 T

where = 11 x 10-6 OC-1 and the value of is given in the problem.

Combine these equations algebraically by eliminating V3 V2 Vsteel and Vliquid by substitution.

Simplify the result and then make one simple calculation of the volume spilled.

Problem 5: This one is similar to Problem 4, i.e a metal container filled with a liquid is warmed and

some of the liquid spills. But this problem is sneakier. But a good picture, which will help us see how

to organize our information, will be very useful.

Note the picture describes three situations:

1. The cylinder is cool and is completely filled with turpentine.

2. The cylinder (and turpentine) is warmed, and some turpentine spills.

3. The cylinder is cooled, and the level of turpentine is below the top of the cylinder.

We can draw all three of these, label the given information, and label the information we are asked to

find.

Part (a) is very similar to Problem 4: you can write the volume of the “warm container” as

V2 = V1 + V where V = V1 3 T

Similarly, you can write the volume of the “warm turpentine” as

V3 = V1 + V where V = V1 T

And finally, the volume spilled is V4 = V3 – V2 If you solve this algebraically, you should get exactly

the same expression for V4 that you got for “volume spilled” in Problem 4. That is, do the algebra once

( in Problem 4) and it will save you can use the result here also.

What about Part (b)? This is the tricky part. The problem asks you to find the distance I have labeled

“h” in the picture above. Fortunately the picture allows us to see what we’re working with. And we can

write down some simple ideas:

The turpentine does not fill the cylinder in Picture C because some of the turpentine is “missing”.

The “missing” turpentine is the volume that spilled.

We can define the fraction of turpentine that spilled, which is the fraction that is missing.

Using this logic:

fraction of turpentine that spilled = amount that spilled / total turpentine

In Picture B, this means

fraction of turpentine that spilled = V4 / V3

We can then write, for Picture C:

fraction of turpentine missing = h / H

These two fractions must be equal, from the logic presented above. So…

h / H = V4 / V3

You are given H, and you found V4 to answer Part a. You can now calculate V3, which is just the total

volume of the expanded turpentine (see the expressions above.) And then you can calculate h…

which is the answer to Part (b).

Problem 6: Here you are told a story about air in a tire. The air is a gas… this is an ideal gas law

problem.

Draw three pictures:

Picture 1 Picture 2 Picture 3

cool air outside the tire air in the warm tire air in the hot tire

Label the appropriate information: p V T n for each picture. You are not given n, so simply

label them as

n1 = ? n2 = ? n3 = ?

You are given a value of p1 (i.e. one atmosphere…). You are given T1, T2 and T3. Convert them to

Kelvin when you label them on your picture.

You are asked to find p2, so label p2 = ? on your Picture 2. And you are asked to find p3, so

label p3 = ? on your Picture 3.

All three volumes are unknown, so you can label all three as = ?

Now… write the ideal gas law for Picture 1, and write the ideal gas law for Picture 2. You can also

write

n1 = n2 and V2 = B V1

where B is the given percentage (i.e. you are told how the volume in Picture 2 relates to the

original volume in Picture 1.)

Now… divide the two ideal gas law expressions, and eliminate n2 and V2 by substitution. You

should be able to simplify the resulting expression so that the only unknown is p2. Solve for it and

calculate it.

Solving for p3 is similar. Write the ideal gas law for Picture 2 and Picture 3. Also write:

n2 = n3 and V3 = 1.02 V2

Divide the two ideal gas law equations, and eliminate n3 and V3 by substitution. Simplify and solve

for p3.

Problem 7: This problem tells you the volume and pressure of a tank of helium. The options for this

chapter are rather limited: this has to be an IGL problem.

In the problem, all of the helium is initially in the tank but later most of the helium is in the balloons

(while some of the helium remains in the tank.) The problem provides the pressure of the helium in

the balloons. You have to recognize that the tank can only fill balloons as long as the helium in the

tank is at a higher pressure than that of the balloons (so the helium will flow into the balloon.)

Therefore, the final pressure of the helium in the tank (when it can no longer fill balloons) is the same

as the given pressure of the helium in the balloons.

This is a multi-state IGL problem. Draw three pictures:

1. Tank in the initial state, with all of the helium in it. Label p V T n.

2. Balloons... how many? Call it “N”. Label the volume of all balloons as V2, and label p T n.

3. Tank in final state, with some helium remaining. Label p V T n.

Now... write IGL for each of the three pictures. You can assume that all three temperatures are the

same. Recognize that the connection between the pictures is that the total amount of helium (i.e. the

actual molecules, or the number of moles) is the same initially and finally. So you can write:

helium in tank initially = helium in balloons + helium remaining in tank

or

n1 = n2 + n3

Use the IGL for each of the three to replace each n, i.e. replace each n with its pV/RT. You should

find that the only unknown in the equation is V2, the volume of all the balloons. Get it... and since you

know the volume of each balloon, you can get the number of balloons.

Problem 8: This problem asks you to find the temperature of the air (i.e. a gas) inside a hot air

balloon. Considering the topics from Chapter 16, it should be obvious (because there’s really no

option) that this is an IGL problem.

Draw two pictures: one with the balloon on the ground and filled with cool air (i.e. air identical to the

outside air) and one with the balloon floating while filled with hot air. Note that the problem mentions

the density of the outside air, which is the same as the density of the air in the balloon in your first

picture. For each picture, label density, pressure and temperature. Note that T1 is given and T2 is

unknown. The second density is unknown.

But both pressures are given! This is because the balloon is open to the outside air, so the pressure

of the air inside the balloon is the same as the pressure outside. Essentially, as the air in the balloon

is heated some of the molecules will leave through the open bottom to keep the pressure equal to that

of the outside. The temperature increases, the density decreases, pressure remains the same.

Write the density version of the IGL for each of the two pictures.

Now... the balloon is floating in air. And the air is a fluid. And when an object is floating in a fluid, we

have a buoyancy problem! (Yes, a little review is in order.)

For your floating balloon, draw a FBD. The only two forces are mg down, and B up. Write the

force equation: B - mg = 0

Replace B with fluid Vdispl g . Then you can eliminate g from this equation.

What is fluid? The “fluid” the balloon floats in is the outside air, so fluid is just the given density of

the outside air.

What is Vdispl? You can assume that the given volume of the balloon is essentially the volume of the

air displaced by the balloon (i.e. we ignore the volume of the cargo, assume it is small compared to

the balloon itself.)

What is “m”? It is the given mass of the cargo, plus the mass of the air inside the balloon. So you can

replace m with

mcargo + 2 V

Now... you have three equations: your force equation and your two IGL equations. Count unknowns in

these three equations. Solve for the temperature in the second picture.

Problem 9: This problem presents an air bubble exhaled by a diver at a depth below the ocean

surface. Note that you know the pressure at this depth... we did this last semester in Chap 15:

absolute pressure at depth = po + gd where po = 101,300 Pa

You can calculate the pressure at the depth where the bubble is exhaled. You can also assume that

the pressure of the air inside the bubble is equal to the pressure of the surrounding water (because if

the pressures were different, the air bubble would either expand or be compressed.)

The problem mentions the temperature, volume and pressure of the air in the bubble... so this has to

be an IGL problem. Draw two pictures: one of the bubble at depth, and one with the bubble near the

surface. You can assume that the pressure near the surface is just po. Label p V n T for each of

the two pictures. Recognize that all of the air remains trapped in the bubble as it rises, so n is the

same for each of the two pictures.

Write the IGL for each of the two pictures. Divide the equations. You should find that n and R

cancel and that you’re left with one equation and one unknown... the final volume.

(Note: the initial volume is given in cm3 and the final volume is asked for in cm3. Do not convert

units... if you enter V1 in cm3, you should find the calculation for V2 comes out in cm3 also.)

Problem 10: You are given a tank of oxygen, along with pressure readings and mass of the oxygen

(which is related to the number of moles.) The problem tells you the temperature of the oxygen is

constant, and you can assume the volume is also constant. What kind of problem is this? It’s a gas

and they mention pressure, volume and temperature. You guessed it: ideal gas law.

Draw two pictures: one with the tank at the initial pressure, and one with the tank at the final

pressure. Note that they give you gauge pressures in “atm”. You must convert these to absolute

pressure, but don’t work too hard at it! Just add ONE atm and leave your pressures in “atm” (i.e. there

is no reason to convert the units of pressure.)

Label p V n T for each picture. Also label m, the mass of the gas, for each picture. Write the IGL

for each picture. Divide the equations... you should be able to cancel V R T, so your equation

includes only the ratio of pressures and the ratio of “n’s”.

Multiply each n by M, the molar mass of O2. DO NOT use a number... instead, recognize that M

times n is just the mass of the gas. So now you have an equation that should include the ratio of the

two pressures and the ratio of the two masses.

Calculate m2. Now that you know the original mass and the mass remaining... answer the question.

Problem 11: This problem gives you the volume, temperature and pressure of a gas... but then it

asks you for the kinetic energy of the gas and for one molecule. This is a “kinetic theory” problem.

Take a look at the notes... “kinetic theory” is all about how the parameters of the gas (p V T) tell us

something about the actual motion of the molecules (i.e. speed and kinetic energy.) The connection is

pretty simple, algebraically.

There are two boxes on Page 11 of the Chapter 16 notes. The first box shows the end result of the

kinetic theory derivation: the average translational kinetic energy of one gas molecule. Note that this is

a simple calculation: all you need is k (which is given in the notes) and the temperature (in Kelvin, of

course.)

The total kinetic energy of all the gas molecules is simply the average KE of one times “N”, the

number of molecules. Note that the result is:

NkTK total 2

3

Recognize that the right side includes “NkT”, which is the same as the right side of the third version

of the ideal gas law, pV = NkT. Which means you can replace NkT for this gas with pV for the gas.

So now:

pVK total 2

3

And now you can calculate the answer for Part a with just the given information. Note that the

pressure and volume are given in atm and L, but the problem asks for the energy in Joules. You will

need to convert the pressure and volume into Pascals and m3 to get the final answer in Joules.

Problem 12: You are given the diameter of a balloon, the temperature of the helium inside it, and the

pressure of the helium. And you are asked to find the number of helium atoms in the balloon.

Recognize that Part (a) is a simple ideal gas law question. I would suggest first finding the number of

moles of helium, and then use Avagadro’s number to get the number of atoms. To do this:

Use the given diameter to find the volume, where the volume of a sphere is D3 / 6

Your volume will be in cm cubed… convert to liters (recall that 1000 cc = 1 liter)

Convert your temperature to Kelvin!

Now you can use p V = n R T where R = 0.08206 atm-L / mol-K to find the number of moles.

For Part (b), use the expression, also from kinetic theory (also in the notes), for the average kinetic

energy of a single atom (or gas molecule… atoms are molecules for the noble gases.) You should

only need your temperature, in Kelvin, and Boltzmann’s constant k = 1.38 x 10-23 J/K

For Part (c), use the expression, also from kinetic theory (also in the notes), for vrms . The molar mass

for helium is 4 grams/mol, which is 4 x 10-3 kg/mol. Be sure to use R = 8.314 J/mol-K for this

calculation (i.e. standard units.)

Problem 13: This problem presents two types of gas molecules at the same temperature and asks

for the average kinetic energy for each... but only provides one answer box! This is because kinetic

theory is simple: the average kinetic energy of a molecule only depends on the temperature of the

gas, which means all molecules (regardless of the type) at the same temperature have the same

average kinetic energy.

Calculate Part a the same way you did Part b in Problem 11.

To find the rms speed of each molecule, use the expression provided on Page 13 of the notes:

M

RTvrms

3 where M is the molar mass of the gas.

Since helium and argon are both noble gases, their molecules are the same as their atoms and their

molar masses are simply the atomic molar masses: 4 grams/mol for helium and 40 grams/mol for

argon.

Watch your units!! See Page 13 on the Chapter 16 notes.

Problem 14: Oh this problem... kind of a long story, right? But you know what that means...

The story is essentially that:

Three astronauts each exhale 1.09 kg of CO2 each day for a week.

Each mol of CO2 is recycled and creates one mol of O2 and one mol of CH4 .

The methane (i.e. CH4) from the week is stored in a tank.

Now can you...

Calculate the mass of CO2 they exhale in one week?

Calculate the number of moles of CO2 exhaled in one week?

Calculate the number of moles of CH4 created in one week? (This one is very simple!)

Now you have n T and V for the methane... can you get the pressure?

Problem 15: This is a thermal expansion problem. The idea is that each horizontal section of the

bridge, in the first picture, expands when the temperature increases. The final length of each section

is illustrated by the diagonal section in the second picture.

First things first: you only need to consider one side (either the left or the right section) of the bridge,

because they are identical.

Draw a picture of a triangle:

The horizontal side is the “cooler” length of one section of the bridge; label this L1.

The vertical side is “y”, which you are asked to find.

The diagonal is the “warmer”, i.e. expanded, length of the section: label this L2.

Now you can write Pythagorean theorem for L1 L2 and y. You can also write that:

L2 = L1 + L and L = L1 T

Eliminate L algebraically and simplify the resulting expression. Then eliminate L2 algebraically,

and simplify the resulting expression to solve for y. You should then be able to make one simple

calculation for y.

Problem 16: This problem includes three pictures and a story. The story goes:

Once there was an open cylinder filled with air. Then a piston was placed on top of the air and the air

was compressed slightly. Then a guy stood on the piston and the air was compressed some more.

Then the air was heated up and the piston (with the man) rose higher.

You will need four pictures. They only gave you the first three. Draw all four. The fourth one

should have the man and piston but should be at the same height as the piston in Picture 2.

It should be obvious this is an ideal gas law problem, since it deals with the volume, pressure and

temperature of the air inside the cylinder. Label p V n T for each of the four pictures. Most of these

are unknown, and that’s okay. Notice that since they give you the radius and depth of the cylinder,

you could calculate the volume of Picture 1, but you don’t need to.

The trick to this problem is the pressure for Pictures 2, 3 and 4. To determine the pressure of the air in

the cylinder, draw a FBD for the piston. What forces act on the piston? mg down, po A down (i.e.

pressure of the outside air times the area of the piston) and pA up (where this “p” is the pressure

of the air inside.) The forces must be balanced so

For Picture 2: p2 A = po A + mg

Solve this for p2 = ... and calculate p2. Then do the same for Picture 3; notice that the only

difference is that in Picture 3 the mass will be that of the piston and man. Do the same for Picture 4...

although you should realize now that pressure in Picture 4 is the same as in Picture 3 (because it is

related to the mass of the piston and man.)

Now write the ideal gas law for each picture. To get the height of the piston in Picture 2:

Divide the IGL equations for Pictures 1 and 2,

You should be able to cancel n T and R

Now you have only the ratio of p1 and p2 with that of V 1 and V2.

Replace V1 and V2 with AH for each picture (area times height.)

They give you the height in Picture 1; calculate the height in Picture 2.

Repeat the steps above but for the IGL equation for Picture 1 and Picture 3. Now you have the height

of the piston for Pictures 2 and 3... and you can answer Part (a).

For Part (b), use the IGL equations for Picture 2 and Picture 4. These two pictures have the same

volume, so when you divide the IGL equations you can cancel V n and R. Only the ratios of

pressure and temperature will remain. You have both pressures and the temperature for Picture 2...

so you can calculate the temperature in Picture 4.