118 ❯ STEP 4. Review the Knowledge You Need to Score High
find the vapor pressure in a table). Answer A is
correct because it is possible to delay looking up
the vapor pressure of water.- C—Real gases are different from ideal gases
because of two basic factors (see the van der
Waals equation): molecules have a volume, and
molecules attract each other. The molecules’
volume is subtracted from the observed volume
for a real gas (giving a smaller volume), and the
pressure has a term added to compensate for
the attraction of the molecules (correcting for a
smaller pressure). Since these are the only two
directly related factors, answers B and D are
eliminated. The question is asking about volume;
thus, the answer is C. - B—A real gas approaches ideal behavior at
higher temperatures and lower pressures. - D—The partial pressure of any gas is equal
to its mole fraction times the total pressure.
The mole fraction of carbon monoxide is
()++
0.30
0.60 0.30 0.10
=
0.30
1.00
= 0.30, and thepartial pressure of CO is 0.30 × 0.80 atm = 0.24 atm.- C—Using Dalton’s law (PTotal = PA + PB +...),
where PA is unknown and PB is the vapor pres-
sure of water, the partial pressure may be found
by 756 mm Hg – 41 mm Hg = 715 mm Hg. - B—The molar masses of the gases are 2.0 g/mole
for H 2 , 4.0 g/mole for He, and 16 g/mole for
CH 4. Therefore, an 8.0-g sample means 4.0 moles
of H 2 , 2.0 moles of He, and 0.50 mole of CH 4.
The greater the number of moles present, with
volume and temperature being the same, the
greater the pressure in the flask. - B—The molar mass may be obtained by divid-
ing the grams by the number of moles (calcu-
lated from the ideal gas equation). Estimation
works in this case as n =
PV
RT
=
(1.00) (0.250)
(0.1) (400)
. Do
not forget to convert the temperature to kelvin.
11. C—Choice B requires an increase in volume,
not allowed by the problem. Choice C requires
an increase in temperature. Choice A requires a
change in the composition of the gas. Choice D
requires a decrease in the volume.
12. B—Lighter gases effuse faster. The only gas
among the choices that is lighter than methane
is helium. To calculate the molar mass, you
would begin with the molar mass of methane
and divide by the rate difference squared.
13. D—A steel tank will have a constant volume,
and the problem states that the temperature is
constant. Adding gas to the tank will increase
the number of moles of the gas and the pressure
(forcing the argon atoms closer together). A con-
stant temperature means there will be a constant
average speed.
14. C—Deviations from ideal behavior depend on
the size of the molecules and the intermolecu-
lar forces between the molecules. The greatest
deviation would be for a large polar molecule.
Sulfur tetrafluoride is the largest molecule, and
it is the only polar molecule listed.
15. D—The molar mass of gas must be determined.
The simplest method to find the molar mass is
5.47 g
L22.4 L
mol = 123 g/mole (simple factor
label). The molar mass may also be determined
by dividing the mass of the gas by the moles
(using 22.4 L/mole for a gas at STP and using
1 L). If you did not recognize the conditions as
STP, you could find the moles from the ideal gas
equation. The correct answer is the gas with the
molar mass closest to 123 g/mole.- D—The hot air balloon rises because it has a
lower density than the surrounding air. Less
dense objects will float on more dense objects.
In other words, “lighter” objects will float on
“heavier” objects. - D—The densities come from the mass of gas
divided by identical volumes; therefore, the con-
tainer with the greatest mass of gas will have the
greatest density. Each container holds one mole
of gas, which means that each container has the
same number of molecules of gas. The pres-
sure in each container will be the same because
the number of moles, the temperature, and the
volume are the same. Ideal gases at the same
temperature will have the same average kinetic
energy. However, the heavier molecules do not
need to travel as fast as the lighter molecules to