Chem 150 Answers
Problem Set 4
Figure A
Gases
Figure B
Figure C
1. The three gas mixtures depicted in the figures above (helium and nitrogen) are all at the same
temperature and in containers of the same volume. Rank the samples A, B and C according to:
a) total pressure A = C > B
b) partial pressure of helium A > B > C
c) density C > A > B (since D = m/V and V = const. simply rank after mass: He: 4 u, N2: 28 u)
d) average kinetic energy A = B = C (since they are all at the same temperature, the average
kinetic energy of the particles must all be the same)
Use <, > or = for ranking (e.g. A>B=C)
2. A mixture containing 3.00 g helium, 3.00 g neon and 3.00 g argon is confined to a 50.0 L vessel
at 20.0 °C. Calculate the partial pressure of each of the gases and the total pressure of the mixture.
We need to find the number of moles of each gas: 3g He x
= 0.750 mol He
Similar we find 0.148 mol Ne and 0.0750 mol of Ar (or a total of 0.973 mol gases). We can
calculate the individual partial pressures of each gas and sum them up to get p total. For example:
pHe =
=
= 36.5 kPa
Similar we get pNe = 7.21 kPa and pAr = 3.65 kPa and ptotal = pHe + pNe + pAr = 47.4 kPa
Alternatively and perhaps more elegantly we could calculate the total pressure first:
Ptotal =
=
= 47.4 kPa
and derive the partial pressure from the mol fraction:
pHe =
x ptotal =
x 47.4 kPa = 36.5 kPa (do pNe and pAr the same way)
Does the result make sense? We have close to 1 mole of gases slightly above standard
temperature where they should occupy a volume close to 22.4 L ( the molar volume). The gas is
however confined to approximately twice the molar volum (2 x 22.4 L = 44.4 L approx. 50 L)
which means it would exert approximately half the standard pressure. 47.4 kPa is roughly half of
101.3 kPa. The result makes sense.
3. 1.00 g dry ice is added to a 1.00 L flask containing only air (approx 80 % N 2 and 20 % O2 by
volume) at 25 C and 112 kPa. Calculate the partial pressure of N2, O2 and CO2 as well as the total
pressure in the flask after all the dry ice has sublimed and the temperature in the flask has
equilibrated with the surrounding.
We need to find the number of moles of each gas in the flask: there are 0.8L N2 and 0.2 L O2 in
the flask at 112 kPa and 298 K: using pV = nRT we calculate the numbers of moles for nitrogen
0.03616 mol and oxygen: 0.00904 mol.
We can get to the numbers of moles of CO2 more easily because we have the mass: 1 g/ 44.01
gmol-1 = 0.02227 mol.
The total pressure depends on the total number of moles of gases: 0.0679 mol, which yields
(using pV = nRT) 168.3 kPa. The partial pressures of the gases are fractions of the total pressure
given by the relative ratio of the numbers of moles of that gas to the total number of mole
pCO2 = (0.02227 mol/0.0679 mol) x 168.3 kPa = 55.2 kPa
pN2 = (0.03616 mol/0.0679 mol) x 168.3 kPa = 89.6 kPa
pO2 = (0.00904 mol/0.0679 mol) x 168.3 kPa = 22.4 kPa
4. Zinc reacts with dilute hydrochloric acid to produce zinc chloride and hydrogen gas. Write a
balanced equation for this reaction. Hint: In compounds Zinc always carries a charge of +2
Zn
+ 2HCl → ZnCl2 + H2
If 55.00 g zinc are dissolved in excess hydrochloric acid, how much hydrogen gas is produced
in L. Assume 25 °C and 1 atm.
nZn = 55 g/65.39 gmol-1 = nH2 = 0.841 mol H2 Since molar ratio is 1:1)
Using pV = nRT we get a volume of 20.6 L which makes sense since we have just under one
mole (e.g.22.4 L/mol)
5. The decomposition of ammonium nitrate at elevated temperatures yields dinitrogen monoxide
and water. Write a balanced equation for this reaction. Using the ideal gas law calculate the total
volume of gases produced if 1.22 mol reactant is completely consumed. Assume both the
dinitrogen monoxide and the water to be in the gaseous state at the reaction temperature of 350
°C and standard pressure.
NH4NO3
→
N2O + 2H2O
The total volume of gas produced depends on the number of moles of gaseous reactants formed.
Under the reaction conditions (350 °C and standard pressure e.g 1 atm = 101.3 kPa) both
reactants are in the gaseous state as indicated in the question. Thus 1.22 mol ammonium nitrate
will form 1.22 mol of nitrous oxide and 2 x 1.22 mol of water. The total amount of gases formed
is 3.66 mol. Using pV = nRT and solving for V yields:
V=
=
= 187 L
Does the result make sense? The molar volume is 22.4 L/mol so 3.66 mol gases should occupy a
volume of 82 L at 273 K and 1 atm. Doubling the thermodynamic temperature (in K) should
double the volume so raising the temperature from 273 K to 546 K would increase the volume
from 82 L to 164 L. The reaction temperature is yet a bit higher (623 K) so we would expect the
volume to be a bit above 164 L which it is. The result makes sense.
6. Magnesium can be uses as a “getter” in tungsten light bulbs to remove residual amounts of
oxygen and nitrogen which would shorten the lifetime of the tungsten wire because it forms
magnesium oxide and magnesium nitride respectively. Write a balanced equation for this
reaction. Assume a residual nitrogen partial pressure p N2 of 3.45 x 10 -4 kPa in a light bulb of
0.320 L volume at 20°C (and no residual oxygen). What mass of magnesium would react?
3Mg + N2 → Mg3N2
In order to solve this stoichiometry problem we need to find the number of moles of nitrogen
available for reacting with magnesium. Use pV = nRT and solve for n
nN₂ =
= 4.53 x 10-8 mol N2
=
Now we have to convert moles of N2 to moles of Mg using the reaction equation:
4.53 x 10-8 mol N2 x
₂
= 1.35 x 10-7 mol Mg
Almost finished, just convert moles of Mg to grams of Mg:
1.35 x 10-7 mol Mg x
= 3.30 x 10-6 g Mg.
Does the result make sense? This is not as easy to approximate as in the examples above.
However. The partial pressure of nitrogen is very small. So we would expect the mass of Mg that
would react to be very small as well. So if your result states 106 g that cannot be right (e.g. 106 g
= 103 kg = 1 t of Mg which would not be able to fit in an enclosure of 0.32 L.)
7. Use the Van der Waals equation to calculate the pressure of 1mol of CCl 4 confined to a volume of
28 L at 40 °C. The van der Waals constants for CCl4 are: a = 2070 L2kPa/mol2 and b = 0.1383
L/mol.
This involves solving the van der Waals equation for pressure
[p + n2a/V2][V – nb] = nRT solve for p: make sure you end up with
p = [nRT/(V-nb)] –[n2a/V2] = 93.4 kPa – 2.6 kPa = 90.8 kPa
8. Here is another data set from our in class demonstration on the molar volume of a hydrogen
where we reacted magnesium with hydrochloric acid to produce hydrogen gas (which we
collected in a gas burette) and aqueous magnesium chloride. Calculate the molar volume of H2.
mass of Mg:
54 mg
ambient air pressure
102.8 kPa
temperature
20 °C
Volume of Hydrogen gas
48.1 ml
vapour pressure of water at 20 °C
2.3 kPa
Mg + 2HCl → MgCl2 + H2
Not all the pressure in the gas burette is caused by hydrogen molecules. Because we are
collecting the gas over water some water vapour is also present. To find the partial pressure
of hydrogen we use Daltons law of partial pressures.
Ptotal =p(H2Ovapour) + p(H2)
p(H2) = 102.8 kPa – 2.3 kPa = 100.5 kPa
Now we need to compare the volume the hydrogen occupied in our gas burette under the
prevailing conditions to the volume it would have occupied at STP (1 atm and 273 K)
piVi(Ti)-1 = pfVf(Tf)-1
(100.5 kPa)(0.0481 L)/(293 K) = (101.3 Kpa)(V f)/(273 K)
Vf = 44.46 mL
Last we relate the volume of hydrogen to moles of hydrogen which we easily find from the mass
of magnesium which was completely consumed (balanced chemical equation!).
0.054 g Mg x
Vmolar =
=
x
₂
= 0.00222 mol H2
= 20 L (Our experiment in class was better!!!)
9. A power failure on board a submarine disables the vessel to manoeuvre, surface or regenerate air
which puts the crew of 26 under severe distress. To sustain vital body functions until help arrives
rations are in place. Assume a diet of 300 g of carbohydrates per sailor per day which is
metabolized according to the following equation:
C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
Calculate the partial pressure of oxygen and carbon dioxide in the submarine after two days. Also
express the concentration of CO2 in % vol. (e.g. mL of CO2 in 100 mL “air”) and comment on the
crew’s time given that drowsiness in people starts to occur at CO 2 concentrations of 1 % vol
(10,000 ppm) and unconsciousness follows after a few minutes of exposures to 7-10 % vol CO2.
The initial conditions in the submarine that has an air volume of 360 m3 are:
T = constant = 18°C, pressure 119 kPa, air composition approximately 20 % vol O2 and 80 % vol
N2. The initial CO2 concentration present in air is 365 ppm = 365 ml/m3.
p(O2)initial = 0.2 x 119 kPa = 23.8 kPa
n(O2)initial =
=
= 3541 mol
n(O2)consumed will depend on the amount of glucose metabolized:
x 26 sailors x 2 days = 15,600 g glucose which gives 15,600 g x
86.6 mol glucose x
= 86.6 mol
= 519.5 mol O2 consumed = 519.5 mol CO2 produced
n(O2)after 2 days = n(O2)initial - n(O2)consumed = 3541 mol - 519.5 mol = 3021.5 mol
p(O2) after 2 days =
=
=20.3 kPa
The initial concentration of CO2 is so small that we could ignore it but I will include it here:
360 ppm CO2 = 365 mL/m3 = 0.365 L/m3 and since we have 360 m3 of air we have 129.6 L of
CO2 in the entire submarine to start with.
Using the ideal gas equation we find n(CO2)initial = 6.46 mol
n(CO2)after 2 days = n(CO2)initial + n(CO2)produced = 6.46 mol + 519.5 mol = 526.0 mol
p(CO2) after 2 days =
=
=
= 3.53 kPa
and ptotal = constant since the mol of CO2 produced equal the mol O2 consumed.
x 100 % = 2.96 % CO2 vol. The crew is drowsy and might survive another day.
Formulas and constants:
pV = nRT
piVi(Ti)-1 = pfVf(Tf)-1
Ptotal = p1 + p2 + p3 + .... + pn
[p + n2a/V2][V – nb] = nRT
Vmol = 22.4 L mol-1 at STP (1 atm = 101.325 kPa and 273.15 K)
R = 8.3145 L kPa mol-1 K-1
Avogadro’s number
Na = 6.022 x 1023 mol-1
Unified atomic mass unit
1 u = 1.660 x 10-27 kg
Π = MRT
PA = xAP°A