Chapter 18- Gases: Hints and Answers 10/13/10
Ideal Gas Law
1. A car tire gauge is used to fill a tire to a gauge pressure of 32 lb/in2 on a cold morning when the
temperature is -10°C. Remember that gauge pressure is [(absolute pressure) – (atmospheric pressure)].
a) What would the tire gauge read when the tire has been heated up to 35°C?
a) It is useful to remember that atmospheric pressure is 14 lb/in2. We want to set-up the ratio P1/T1= P2/T2,
but we need absolute pressures and temperatures, so: (32+14)/(-10+273)=(∆Px+14)/(35+273) ∆Px
represent the gauge pressure reading =40 lb/in2.
b) Why can we use any unit of absolute pressure in a PV/T ratio, but not gauge pressures?
c) Why must we use only the Kelvin scale for temperature in a PV/T ratio problem?
b&c) If P is proportional to T, they must approach zero together and that only happens with scales that
reflect a true “absolute zero”…The “0” in gauge pressure and Celsius temp. are not “real” zeroes.
2. Practice using the ideal gas law:
a) In chemistry one learns that one mole of any gas occupies 22.4 liters of volume at standard temperature
and pressure (STP), which means 1 atm. pressure and 0 ºC. Verify this statement.
a) Done in class…plug into PV=nRT, taking care to match units, and solve for V.
b) How far apart are the molecules in (a) on average? Assume each molecule sits at the center of a cube.
b) Dividing 22.4 x10-3 m3 by Avogadro’s # gives the average volume occupied by one particle, then taking
the cube root gives the linear dimensions of the cube which is a good approximation to the average
separation between the particles…I got ~3.3 nm.
c) Compare the separation of molecules in a gas to: (i) the size of the molecule themselves (~0.1 nm) and
(ii) the separation of molecules in a solid (~0.3 nm).
c) Using the value obtained in (b) one can see that the separation is ~33x larger than the size of the
particles and ~11x larger than the separation in solids.
d) In state-of-the-art vacuum systems, pressures as low as 1.00 X 10-9 Pa are being attained. Calculate the
number of molecules in a 1.00 m3 vessel at this pressure if the temperature is 27°C.
d) Using PV=NkT N=PV/kT= (10-9)/(1.38x10-23)(300)=2.4 x1011 molecules. This is what we would call
a pretty good “vacuum”!
3. An automobile tire is inflated with a volume of air originally at 10°C and normal atmospheric pressure.
During the process, the air is compressed to 28% of its original volume and the temperature is increases to
40 ºC.
a) What is the tire pressure?
a) (PatmVo/283)= (P1 )(0.28Vo/313) P1 =3.95 atm
b) After the car is driven at high speed, the tire air temperature rises to 85°C and the interior volume of the
tire increases by 2%. (You should understand that “increases by 2%” means: Vnew= 1.02Vold). What is the
new tire absolute pressure in pascals (Pa)?
b) (P1V1/313)= (P2)(1.02V1/358) P2 =4.43 atm=4.65 x 105 Pa.
c) What would a standard pressure gauge read? 3.64 x 105 Pa.
4. Dalton's law of partial pressures states: The total pressure of a mixture of gases is equal to the sum of the
partial pressures of gases making up the mixture.
a) Prove Dalton’s law. Start with the Ideal Gas Law and show that the total pressure P in a container filled
with a mixture of several ideal gases is P = PI + P2 + P3 +… , where PI, P2, etc. are the pressures that each
gas would exert if it alone filled the container (these individual pressures are called the partial pressures of
the respective gases).
a) PV=Ntotal kT, but since Ntotal=(∑Ni) P=( NI + N2 + N3 +…)kT/V=( NI kT/V) + (N2 kT/V) + (N3 kT/V)
+…). Each term is the pressure that each gas would exert if alone in the container (Pi), hence P=(∑Pi).
b) Which would be denser at the same pressure, dry air or moist air? Would sound travel faster in dry air or
in moist air?
b) This was also a question in the Speed of Sound lab. The basic concept, made clear by Dalton’s law, is
that the water molecules in the air displace air molecules. Water molecules (H2O) have less mass than air
molecules (N2 &O2), so the density decreases.
5. A vertical cylinder of cross-sectional area A is fitted with a tight-fitting, frictionless
piston of mass m. There are n moles of an ideal gas in the cylinder at a temperature T.
a) Determine the height h at which the piston is in equilibrium under its own weight.
Don’t forget atmospheric pressure.
a) Pgas=(Patm +mg/A), and Vgas=Ah nRT=(Patm +mg/A)(Ah)h=nRT/( PatmA+mg)
b) What is the value for h if n ~ 0.20 mol, T ~ 400 K, A ~ 0.0080 m2, and m ~ 20.0 kg?
b) 66 cm
6. On a day when the atmospheric pressure is 1 atm and the temperature is 20 ºC, a diving bell in the shape
of a cylinder 4 m tall and closed at the upper end is lowered into the water to aid in the construction of the
underground foundation of a bridge tower. The water inside the bell rises to within 1.5 m from the top and
the temperature drops to 8 ºC.
a) Find the air pressure inside the bell.
?
a) Let A be the cross-section of the cylinder. Then (Patm A4/293)= (P1 )(1.5A/281)
 P1 =2.56 atm=258 kPa
4m
b) How far below the water surface is the bottom of the bell located?
b) Here you have to recall Pascal’s principle: Pb=Pa+ghh~16 m. This is not to the bottom of the bell,
but to the water level inside the bell, so add 2.5m to get ~18.5 m.
c) To what minimum pressure must the air in the bell be raised to expel the water that entered?
c) You must raise the pressure inside the bell to the pressure value at the bottom of the bell. Again using
Pascal’s principle, Pb=Pa+gh=258 x103 +2.5(103)(9.8)=283 kPa.
Point of interest: During the building of the Brooklyn Bridge in NY the diving bell had to descend to
unprecedented depths. When workers began to die on the return to the surface, the dangers of a sudden
depressurization on the human body (a condition now called “the bends”) became obvious.
7. By volume, air is composed of approximately 78% nitrogen (N2), 21% oxygen (02), and 1% other gases.
Ignore the 1% other gases.
a) Verify that the molecular mass of air is ~ 29 g/mole.
a) M of N2=28 and M of O2=32 M of air=28(.78) + 32(.21) ~29 g/mole
b) Find the mass of a cubic meter of air at standard conditions (1.00 atm, 0.0 ºC).
b) At STP one mole occupies 22.4 x10-3 m3 of volume, so (1/22.4 x10-3)=(n/1) n=44.6 moles. At 29
g/mole mass=1.29 kg
c) Given this result, calculate the lifting force on a helium-filled balloon with a volume of 1.00 m3 at a
pressure of 1.00 atm. Hint: Recall that, according to Archimedes’ principle, the buoyant force is equal to
the weight of the displaced fluid.
c) M or He= 4 g/mole, so the mass 1 m3 =44.6 x4=178 g, and the weight is 1.75 N. The displaced fluid is
air so the buoyant force= 1.29 x 9.8=12.6 N. Subtract the weight of the helium to get the lifting force=
12.6-1.75~11 N
d) Show that a helium-filled balloon has 92.6% the lifting force of a similar hydrogen-filled balloon.
c) M or H2= 2 g/mole, so the mass 1 m3 =44.6 x4=89 g, and the weight is 0.87 N. Lifting force of the
hydrogen balloon=12.6-0.87~12 N(11/12)x100 ~92%
7B. Atmospheric pressure is due to the weight of the air. It is therefore reasonable that atmospheric
pressure should decrease as you rise above the Earth’s surface. In general pressure is proportional to the
density of the air, P=C where C is a constant (if temperature and “g” are assumed to remain unchanged
within the range). Use this fact and Pascal’s law in differential form to prove that atmospheric pressure
decrease exponentially with height above the surface. What does the constant “C” here represent?
h
The Kinetic Theory of Gases
8A. Show that the following versions of the ideal gas law are all equivalent: i) PV=nRT ; ii)PV=NkT; iii)
PM=RT. Make sure to define the terms in the equations.
8. In a 20 sec interval, 500 stones strike a glass window of area 0.70 m2 at an angle of 45º to the window
surface. Each stone has a mass of 5 g and a speed of 8 m/s.
a) If the collisions are elastic, find the average force and pressure on the window.
a) Using the formula derived below: F= 1.41 N and P= 2.0 Pa.
b) Derive a general expression for the pressure generated when N particles with speed v and mass m strike
a surface A at an angle ø from the perpendicular to the surface in a time interval ∆t.
b) From Newton’s 2nd law: F=∆p/∆t= N(2mvsinø)/∆t, and P= F/A=N(2mvsinø)/A∆t
9. For equal 5.0 liters containers of Argon gas (40 g/mole) and Helium gas (4.0 g/mole) at 1 atm and 300 K
compare the following: a) total translational kinetic energy; b) average kinetic energy of the molecules; c)
rms speed of the molecules; c) momentum of molecules at the rms speed.
Gas
Ktotal=3NkT/2=3PV/2
Kaverage= 3kT/2
vrms=(3RT/M)1/2
prms=m vrms
-21
Argon
758 J
6.21 x10 J
433 m/s
2.9 x10-23 kg m/s
Helium
758 J
6.21 x10-21 J
1370 m/s
9.1 x10-24 kg m/s
d) Reconsider the problem in the case the volumes were different, for example if the volume of Argon
were 5.0 lt but the volume of Helium were 10 lt. Which answers above would change?
d) Only the total K value of the Helium would change and it would be twice the value.
10. Use the kinetic molecular theory to explain the following:
a) Alcohol taken internally makes you feel warmer. Yet when it is rubbed on your body, it lowers body
temperature. Explain the latter effect. Alcohol molecules take energy from skin to evaporate…
b) Explain why the temperature of the liquid decreases when the container’s pressure is reduced, and how
this makes it possible to freeze water. At lower pressures the faster molecules in a liquid can evaporate
more easily. This leaves slower molecules behind cooling the liquid. If the pressure is reduced enough, the
liquid can cool to freezing temp.
c) Cook books often give special instructions for cooking at high altitudes. Explain why. At higher
altitudes the atmospheric pressure is lower and water boils at a lower temperature, thus affecting the
cooking of food…
11. The escape velocity is the speed that a mass must have to escape from the gravitational attraction of a
planet. In mechanics we found that vescape= (2GM/R)1/2 where M and R are the mass and radius of the
planet and G is the universal gravitational constant (G=6.67x10-11 Nm2/kg2).
a) Show that a mass m must have kinetic energy equal to (mgR) to escape from the surface of a planet of
radius R and acceleration due to gravity g.
a) K=mv2/2 = m(2GM/R)/2=mR(GM/R2). The term in parenthesis is “g” so Kescape=mgR and vesc=
vesc=√(2gR).
b) If the planet in question is earth, at what temperature would the average nitrogen molecule (28 g/mole)
be able to escape earth’s gravitational pull? How about hydrogen molecules (2.0 g/mole)?
b) Since vesc=√(2gR) and RE =6.4x106 m  vesc=11.2 km/s for earth’s surface. Now let vesc =
vrms=√(3RT/M)T= vesc2M/3R. Tesc for N2 = 1.4 x 105 K and for H2= 104 K, both of which are pretty high
temperatures so we are likely to hang on to these gases in our atmosphere.
c) The moon’s g=1.63 m/s2 and its radius is 1740 km. Do these values explain why the moon has almost no
atmosphere?
c) For the moon’s surface vesc=2.4 km/s. These speeds are quite a bit lower and it would be a lot easier for
gas molecules to escape from the moon than from the earth.
Van der Waals' Equation of State
12. Explain the rationale for the correction terms in the van der Waals equation.
a) What do the a and b constant represent? What is meant by the statement that these are empirical
constants?
a) Read the text book…the general idea is that “b” reflects the volume of the gas molecules themselves and
that “a” reflects the intermolecular forces between gas molecules…but these factors are determined
experimentally not theoretically and it is very difficult to figure exactly what these factors represent.
b) For oxygen the constant b is measured to be 31.8 cm3/mol. Assuming a spherical shape, estimate the
diameter of the molecule. ~0.50 nm
13. Explain the concept of the mean free path. Read the text book…
a) The mean free path is inversely proportional to the number density of the particles in the gas and also
inversely proportional to the cross-sectional area of the particles. Why do these make sense?
a) Conceptually these make sense: the more particles and the larger their cross-section the more likely
there are to be collisions, reducing the average distance the particles travel between them…
b) Surprisingly the mean free path does not depend on the speed of the particles. Why is that?
b) This is less obvious, as it seems that more speed would mean that the particles could travel longer
distances between collisions, but further thought leads to the conclusion that more speed also means more
collisions and the effect of the random speed of the particles cancels out in the mean-free-path formula.
The faster molecules, however, will take less time to travel the same distance as the slower molecules
c) At room temperature, the average speed of an air molecule is several hundred meters per second.
Given these large average speeds, why does it take the odor of perfume (or other smells) several minutes to
travel across the room?
c) The random average speeds of the air molecules is irrelevant, the perfume molecules will diffuse in
accordance with their mean-free-times and that is a much slower process…
13B. Consider a gas at STP. If the diameter of the particles is 0.5 nm, determine the mean-free-path. What
additional information would you need to find and the mean-free-time?
Basically a formula plugger…From the gas law N/V=P/kT=1.01x105/1.38 x10-23(273)=2.68
x1025lmean=V/Nπd2√2= 34 nm. To find the mean-free-time t=lmean/vave) you need the molecular mass of
the gas so you can determine the average speed of the particles vave=1.60√(RT/M).
Maxwell-Bolztmann Distribution
14A. Describe the difference between a normal (Bell-shaped) distribution and the Maxwell-Bolztmann
distribution. Explain why one is symmetric and the other not. In a Bell-shape distribution both extremes of
the distribution are equally likely and this leads to symmetry, but in the Max-Boltz distribution there is the
natural cut-off of “0” speed at the low end of the distribution, but no such definite cut-off at the high end,
hence the asymmetry….If you look at the pictures in the text book you will note that the M-B distribution
gets more symmetric as the average speed gets larger.
14B. Describe how the basic Maxwell-Bolztmann distribution graph of “Nv vs. v” would change if: (a) the
temperature of the gas sample increases; (b) if the number of particles in the sample of the gas increases;
(c) the sample is made of a different gas with more massive particles.
a) The graph would flatten and stretch out, since the speeds would be on average higher, and it would look
more symmetric. But the area under the graph must be the same because it represents the total number of
particles.
b) The area under the graph would increase but the speed distribution would be the same.
c) The area under the graph would be the same but speed values of the distribution would decrease, so the
graph would look compressed and taller.
15. Twenty particles, each of mass m and confined to a volume V, have the following speeds: two have
speed v; three have speed 2v; five have speed 3v; four have speed 4 v; three have speed 5v; two have speed
6v;one has speed 7v. What is the probability of each speed in the sample?
Also find (a) the average speed, (b) the rms speed, (c) the most probable speed, (d) the pressure they exert
on the walls of the vessel, (e) the average kinetic energy per particle, and (f) the “temperature” of this
group of particles.
The probability of each speed in the sample would be the ratio of the no. of particles with a particular
speed divided by the total no. of particles: for speed “v” the probability is 2/20=0.1, etc…if you add all
the probabilities you should get “1”.
a) vave=[2(v) + 3(2v) +…]/20=3.65v; (b) vrms={[2(v)2 + 3(2v)2 +…]/20}1/2 =3.99v; (c) vmp=3v; (d)
PV=(2N/3)(mvrms2/2)P=106mv2/V ; (e) Kave=(mvrms2/2)= 7.96mv2; (f) T=2Kave/3k=(mvrms2/3k)
16. A vessel contains 1.00 X 104 oxygen molecules at 500 K. (a) Make a quantitative graph of the Maxwell
speed distribution function versus speed with points at speed intervals of 100 m/s. (b) Determine the most
probable speed from this graph. (c) Calculate the average and rms speeds for the molecules and label these
points on your graph. (d) From the graph, estimate the fraction of molecules with speeds in the range 300
m/s to 600 m/s.
This takes a bit of time and requires using the Max-Boltzmass distribution function to calculate Nv values
for 10 to 12 velocities by 100 m/s jumps (100, 200, 300 m/s etc..), and plotting them in a graph of Nv vs. v
that will look like the one in the text-book. From the graph you can find answers to (b) & (d). You can
determine vrms (624 m/s) and vave (575 m/s) and vmp (510 m/s ) from the formulas.
17. This problem will guide you in deriving the formulas for the rms [vrms=1.73√(kBT/m)], average
[vave=1.60√(kBT/m)], and most popular [vmp=1.41√(kBT/m)] speeds of the molecules of a gas at a
temperature T.
a) To find the most probable speed is relatively simple. Note that vmp corresponds to the point where the
slope of the speed distribution curve (dNv/ dv) is zero.
a) Take the derivative of the M-B function and set it equal to zero. It is convenient to simplify the formula
to Nv= NAv2e-bv^2, where b=(m/2kT) and A=(4/√π)(m/2kT)3/2 and are constants in this problem. So,
d(Av2e-bv^2)/dv=0. Then solve for v…
b) To derive the other formulas requires setting up a definite integral and looking up the solution to this
integral in an integral table. For the average and the rms speeds you need to realize that the average value
1 
n
of any power of v (vn) is the solution to the integral: v ave
  v n N v dv . The solution to these integrals
N 


2
2
3 
1
for powers “3” and “4” are found in integral tables to be:  x 3eax dx  2 , and  x 4 eax dx  2
.
0
0
8a a
2a
You need to define “x” appropriately for each
 case.
b) This problem is simpler than it seems; it mostly requires symbol manipulations. Since the variable of
integration is “v”, use the same simplificationsas in (a): Nv= Av2e-bv^2,
where b=(m/2kT) and
3/2
A=(N4/√π)(m/2kT) and are constants of integration. The integral set-up is trivial. The solution to the
integrals are found in table and given above. For the regular average n=1 and you get v3 in the integral,
so the first of the given definite integrals applies. For the rms average n=2 and you get v4 in the integral,
so the second of the given definite integrals applies. The rest is algebra…really! Conceptually the most
important thing is to understand the validity of the averaging-integral expression given in the problem.
Molar Heat Capacities
18. Review the concept of molar heat capacity for gases at constant volume and answer these questions:
a) Why does a diatomic gas have larger thermal energy content per mole than a monatomic gas at the same
temperature? Explain what is meant by a degree of freedom.
Review your notes and read the text-book…Basically, the theory says that diatomic molecules have
rotational as well as translational kinetic energy…each different mode of carrying energy is called a
“degree of freedom”…
b) Inspecting the magnitudes of CV (and later also for CP) for the diatomic and polyatomic gases we find
that the value increases with increasing molecular mass. Give a qualitative explanation of this observation.
19. A 2.0-liter container has a center partition that divides it into two equal volumes, as shown. The walls
of the container are well insulated but heat can flow through the partition.
a) If the left side contains 2 moles of H2 gas at 200 K and the right side contains 1 mole
2 mole
1 mole
of O2 gas at 300 K, what’s the final equilibrium temperature?
H2
O2
b) If we replace the hydrogen (H2) with 2 moles of helium (He) on the left side,
what would be the final equilibrium temperature?
c) Suppose that instead of letting heat flow through the partition we had simply removed it and allowed the
gases to mix. Would the results of the questions above be different? Would there be any other important
differences that you can think of?
a)Both are diatomic so the Cv’s are the same and cancel out233 K
b) One is monatomic and one is diatomic so the Cv’s do not cancel245 K
c) The equilibrium temperatures would be the same but in (c) the molecules would be all mixed up and the
system would be in a more disordered state. We would say that the entropy of the system has increased,
and we’ll be studying this important concept in a later chapter.
Work done by expanding or contracting gases
18. Show that Work= ∫PdV for expanding gases and that the unit is the joule. The following problems all
ask you to find the work done by an expanding gas.
P(atm)
4
I
A
Work=∫F·dx . For a gas F=PA and, if the volume is increasing, A is
3
moving outward by an amount dxWork=∫PAdx=∫PdV. Units: Pa·m =
(N/m2)m3=Nm=Joule.
2
B
F
19. A gas expands from Initial to Final state along three possible paths
as indicated in the figure. Calculate the work in joules done by the gas
along the paths IAF, IF, and IBF. In order to get answer in joules you must
2
4 V(lit)
3
change units to Pa and m . It’s relatively easy to find the area under the graph along the chosen path:
WIAF=PIA ∆VIA = 4(1.01 x 105) (2 x 10-3)=808 J; WIF = (3/2)(1.01 x 105) (2 x 10-3)+202= 505 J;
WIBF=PBF ∆VBF = (1.01 x 105) (2 x 10-3)=202 J;
20. Helium with an initial volume of 1.00 liter and an initial pressure of 10.0 atm expands to a final volume
of 1.00 m3. The relationship between pressure and volume during expansion is PV=constant.
a) Draw the PV graph and determine the final pressure. Pfinal=0.010 atm
b) Determine the value of the constant. (10.1 x 105)(10-3)=1010 J
P
c) Determine the work done by the helium during expansion.
10 atm
d) According to the gas law the “constant” here must be “nRT”.
Not to scale
Derive the general work expression for a gas expanding at
constant temperature T.
c&d)The area under this graph requires an integral:
10-3 m3
V
Work=∫PdV=nRT ∫dV/V=nRT ln(Vfinal/Vinit)= 1010 ln(103)=6980 J
P
21. A sample of ideal gas is expanded to twice its original volume of
1.0 m3 in a quasi-static process for which P= V2, with = 5.0 atm/m6,
as shown. How much work was done by the expanding gas?
Here also we need to integrate to find the work:
Work=∫PdV= V2dV=Vf3 – Vi3)= (5.05 x 105/3)(8-1)=11.8 x 105 J.
P=v2
V
1 m3
2 m3