- C—Using Dalton’s law (PTotal=PA+PB+.. .),
the partial pressure may be found by:
756 mm Hg −41 mm Hg =715 mm Hg. - A—The answer may be found using the com-
bined gas law. Removing the constant pressure
leaves Charles’s law: V 1 /T 1 = V 2 /T 2. This is
rearranged to: T 2 =V 2 T 1 /V 1 =(20.00 L ×400. K)/
(8.00 L) = 1000 K (= 727 °C). The other
answers result from common errors in this
problem. - A—The average kinetic energy of the molecules
depends on the temperature. The correct answer
involves a temperature difference (333 K – 303 K).
Do not forget that ALL gas law calculations
require Kelvin temperatures. - A—You can begin by removing the volume
(constant) from the combined gas law to pro-
duce Gay-Lussac’s law = P 1 /T 1 = P 2 /T 2. This
equation may be rearranged to: P 2 =P 1 T 2 /T 1 =
(0.300 atm × 290. K)/(300. K) = 0.290 atm.
The moles are not important since they do not
change. Some of the other answers result from
common errors. - B—The molar mass may be obtained by dividing
the grams by the number of moles (calculated
from the ideal gas equation). Do not forget to
convert the temperature to kelvin. - C—Choice I requires an increase in volume.
Choice II requires an increase in temperature.
Choice III requires a change in the composition
of the gas. - 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: - 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 (molecules) of the gas and the
pressure (forcing the molecules closer together).
A constant temperature means there will be
a constant average speed. - B—The pressure is calculated using the ideal gas
equation. A common mistake is forgetting that
5 mol of gas are produced for every 2 mol of
solid reactant. The ideal gas equation is
rearranged to P = nRT/V = (^5 ⁄ 2 )(0.2 mol)
(0.082 L atm mol−^1 K−^1 ). (300. K)/(2 L) =6 atm. - E—The lighter the gas, the faster it effuses
(escapes). Equal moles of gases in the same con-
tainer would give equal initial partial pressures.
The partial pressures would be reduced relative
to the masses of the molecules, with the lightest
gas being reduced the most. - C—Deviations from ideal behavior depend on
the size and the intermolecular 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 mol-
ecule listed. - D—The molar mass of gas must be determined.
The simplest method to find the molar mass is:
(5.47 g/L) ×(22.4 L/mol) =123 g/mol (simple
factor label). The molar mass may also be deter-
mined by dividing the mass of the gas by the
moles (using 22.4 L/mol for a gas at STP and
using 1 L). If you did not recognize the condi-
tions 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/mol. - D—The hot-air balloon rises because it has a
lower density than air. Less dense objects will
float on more dense objects. In other words
“lighter” objects will float on “heavy” objects.
MMAB= ()^122
118 Step 4. Review the Knowledge You Need to Score High