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Gases EXAM #3
(a) The van der Waals equation of state for one mole
of a real gas is as follows:
(P + a/V2)(V - b) = RT
EXAM #1 For any given gas, the values of the constants a and b
2 HCOONa + H2SO4 2 CO + 2 H2O + Na2SO4 can be experimentally calculated. Indicate which
A 0.964 gram sample of a mixture of sodium chloride physical properties of a molecule determine the values
and sodium formate was analyzed by adding acid. The of the constants a and b.
equation is shown above. The carbon monoxide (b) Which of the two molecules, H2 or H2S, has the
formed measured 242 mL when collected over water at higher value for a and which has the higher value for
752 torr and 22.0C. At 22.0C, the vapor pressure of b? Explain.
water is 19.8 torr. Calculate the percentage of sodium (c) One of the van der Waals constants can be
formate in the original mixture. correlated with the boiling point of a substance. Which
constant and how it is related to boiling point?
Answer
PCO = Patm - PH2O = (752 - 19.8) torr = 732.2 torr Answer:
PV=nRT (a) a indicates intermolecular forces in real gases.
(.963 atm)(.242L) = n(.0821Latm/molK)(295K) b indicates actual volume of real molecules.
-3
n = 9.62x10 mol CO (b) H2S would have a larger a, because it is a dipole
2 mol HCOONa = 2 mol CO (see eqn.) and has stronger IMF. It would have a larger b,
-3
So we have 9.62x10 mol HCOONa also because it is a molecule with a larger volume.
9.62x10-3 mol HCOONa x 68.0g/mol (c) a is correlated with the boiling point. The larger the
= 0.654 g HCOONa value, the stronger the IMF and the higher the boiling
0.654 g/0.964 g x 100 = 67.8% point.
EXAM #2 EXAM #4
(a) From the standpoint of the kinetic-molecular Give a scientific explanation for the following
theory, discuss the properties of gas molecules observation. A hot-air balloon must be larger than a
that cause deviations from ideal behavior. helium-filled balloon in order to lift the same mass.
(b) At 298K and 1 atmosphere pressure, which of the
following gases shows the greatest deviation from Answer:
ideal behavior? Give two reasons. Hot air is denser than helium. Hot air, therefore, has
CH4 SO2 O2 H2 much less lifting power per unit volume than helium.
(c) Real gases approach ideality at high temperature (This is a scientific explanation of the volume/lift
and low pressure. Explain these observations. relationship).
Answer:
(a) Real molecules have actual volumes.
Real molecules have attractive forces.
(b) SO2 is the least ideal gas.
It has the largest volume.
It has the strongest attractive forces (dipole-dipole)
(c) High temperature results in high kinetic energies.
This energy overcomes the attractive forces.
Low pressure increases the distance between
molecules.
EXAM #5 EXAM #6
A mixture of H2(g), O2(g), and 2 mL of H2O(l) is present Observations about real gases can be explained at the
in a 0.500 L rigid container at 25C. The number of molecular level according to the kinetic molecular
moles of H2 and the number of moles of O2 are equal. theory of gases and ideas about intermolecular forces.
The total pressure was initially 1,146 torr Clarify how each of the following observations can be
(The vapor pressure of pure water at 25C is 24 torr.) interpreted according to these ideas, including how the
The mixture is sparked, and H2 and O2 react until one observation supports the correctness of these theories.
reactant is completely consumed. (a) When a gas-filled balloon is cooled, it shrinks in
(a) Identify the reactant remaining and calculate the volume; this occurs no matter what the gas is.
number of moles of the reactant remaining. (b) When the balloon described in (a) is cooled
(b) Calculate the total pressure in the container at the further, the volume does not become zero; rather,
end of the reaction if the final temperature is the gas becomes a liquid or a solid.
90C. (The vapor pressure of water at 90C is 526 torr.) (c) When ammonia gas is introduced at one end of a
(c) Calculate the moles of water present as vapor in long tube while HCl gas is introduced
the container at 90C. simultaneously at the other end, a ring of white
ammonium chloride is observed to form in the
Answer: tube after a few minutes. This ring is closer to the
HCl end of the tube than the NH3 end.
(a) 2 H2 + O2 2 H2O
(d) A flag waves in the wind.
I x x
C -2y -y
Answer:
E x-2y x-y
(a) Reducing the temperature of a gas reduces the
mol H2 = mol O2 initially, but 2 mole H2 react average kinetic energy of the gas molecules. This
for every 1 mol of O2, therefore, O2 is left. would reduce the number of collisions of gas
PT = PH2 + PO2 + PH2O molecules with the surface of the balloon. In
1146 torr = PH2 + PO2 + 24 torr order to maintain a constant pressure, the volume
PH2 + PO2 = 1122 torr must decrease.
At the end of the rxn, H2 is used up, so ½ of the (b) The molecules of the gas do have volume, when
pressure is gone they are cooled sufficiently, the attractive forces
1122/2 = 561 torr (just O2), but also ½ of the O2 gets cause them to liquefy or solidify.
uedup as well... (c) The molecules of gas are in constant motion so the
561/2 = 280.5 torr PO2 = 280.5 torr HCl and NH3 diffuse along the tube. Where they
PV=nRT meet, NH4Cl(s) is formed. Since HCl has a higher
(.369 atm)(.500L) = n(.0821Latm/molK)(298K) molar mass, its velocity is lower, therefore, it
n = 7.54 x10-3 mol O2 doesn’t diffuse as fast as the NH3.
(d) The wind is moving molecules of air that are
(b) PV=nRT
(P O2)(.500L) = (7.54 x10-3 mol O2) (.0821Latm/molK)(363K) going mostly in one direction. Upon encountering
P O2 = 0.449 atm a flag, they transfer some of their energy to it and
The vapor pressure of water at 90C is 526 torr = .692 atm cause it to flap..
PT = PO2 + PH2O = (.449 +.692) = 1.14 atm
(c) PV=nRT
(.692 atm)(.500L) = n (.0821Latm/molK)(363K)
n = 0.0116 moles
EXAM #7 EXAM #8
Above are five identical balloons, each filled to the
A student collected a sample of hydrogen gas by water same volume at 298K and 1.00 atm with a pure gas.
displacement as shown above. The relevant data are (a) Which balloon contains the greatest mass of gas?
given as follows: Explain.
GAS SAMPLE DATA (b) Compare the average kinetic energies of the gas
molecules in all of the balloons. Explain.
Volume of sample 90.0 mL
(c) Which balloon has the gas that would deviate
Temperature 25C most from the ideal behavior? Explain.
Atmospheric Pressure 745 mm Hg (d) Twelve hours after being filled, all the balloons
Vapor Pressure of H2O at 25C 23.8 mm Hg have decreased in volume. Predict which balloon
will be the smallest. Explain.
(a) Calculate the number of moles of H2 collected.
(b) Calculate the number of molecules of water vapor Answer:
in the sample of gas.
(a) CO2; according to Avogadro’s Hypothesis, they
(c) Calculate the ratio of the average speed of the all contain the same number of particles,
hydrogen molecules to that of the water vapor therefore, the heaviest molecule, CO2, will have
molecules in the sample. the greatest mass.
(d) Which of the two gases, H2 or H2O, deviates more (b) All the same; at the same temperature all gases
from ideal behavior? Explain. have the same kinetic energy.
(c) CO2; since they are all non-polar, the largest
Answer: intermolecular (London) force would be greatest
(a) PH2 = Patm - PH2O = (745 - 23.8) mm Hg in the molecule/atom with the most electrons.
= 721.2 mm Hg (d) He has the smallest size and has the greatest
PV=nRT (721.2 mm Hg/760 mmHg = 0.949 atm) particle speed (Graham’s Law); therefore it’s the
(0.949 atm) (.0900 L) =n(0.0821Latm/molK)(298K) easiest to effuse through the balloon walls.
= 3.4910-3 mol H2
(b) PV=nRT (23.8 mm Hg/760 mmHg = .0313 atm)
(0.0313 atm) (.0900 L) =n(0.0821Latm/molK)(298K)
= 1.1510-4 mol H2O
1.1510-4 mol H2O x 6.021023 molecules/mol
= 6.931019 molecules of water
(c) KE per molcule = ½ (massH2)(velocityH2)2 =
½ (massH2O)(velocityH2O)2
(Note KE is the same because the temp. is the same)
(½) 2(vH )2 = (½) 18(vH O)2
2 2
v2H / v2H O = 9; vH / vH O = 3
2 2 2 2
(d) H2O deviates more from ideal behavior:
(i) Water has more electrons = greater attractive forces
(ii) It is a polar molecule with strong polar attraction
(iii) It has hydrogen bonding to other water molecules
EXAM #9
A rigid 5.00 L cylinder contains 24.5 g of N2(g) and
28.0 g of O2(g)
(a) Calculate the total pressure of the gas mixture in
the cylinder at 25 oC.
(b) The temperature of the gas mixture in the cylinder
is decreased to 7oC. Calculate each of the
following.
(i) The mole fraction of N2(g) in the cylinder.
(ii) The partial pressureof N2(g) in the cylinder.
(c) If the cylinder develops a pinhole-sized leak and
some of the gaseous mixture escapes, would the
N2 ( g )
ratio in the cylinder increase, decrease, or
O2 ( g )
stay the same? Justify your answer.
A different rigid 5.00 L cylinder contains 0.176 mol of
NO(g) at 25oC. A 0.176 mol sample of O2(g) is added
to the cylinder. A reaction occurs to produce NO2(g).
(d) Write the balanced equation for this reaction.
(e) Calculate the total pressure in the cylinder at
25oC after the reaction is complete.
Answer:
(a) 24.5 g N2 1mol/28.0 g = 0.875 mol N2
28.0 g O2 1mol/32,0 g = 0.875 mol O2
PV=nRT nTotal = 0.875 + 0.875 = 1.75moles
(P) (5.00 L) =(1.75 moles)(0.0821Latm/molK)(298K)
= 8.56 atm
0.875 mol N 2
(b) (i) = 0.500 mole fraction N2
1.75 mol mix
P P P T (8.56atm)(280K)
(ii) 1 2 ; P2 1 2
T1 T2 T1 298K
= 8.05 atm mole fraction = 8.05 atm 0.500
= 4.02 atm N2
(c) decrease; N2 molecules are smaller than O2 they
have a higher speed and will escape more rapidly
(Graham’s Law), decreasing the N2/O2 ratio
(d) 2 NO + O2 2 NO2
(e) I 0.0176 0.0176 0
C -0.0176 -0.0088 +0.0176
E 0 0.0088 mol 0.0176 mol
0.176 + 0.088 = 0.264 mol of gas is in the container.
(P) (5.00 L) =(0.264 mol)(0.0821Latm/molK)(298K)
= 1.29 atm
EXAM #10 (iii) less; Avogadro’s Hypothesis, equal volumes of
Answer the following questions about carbon gas at the same temperature and pressure contain
monoxide and carbon dioxide. Assume that both gases equal number of molecules. Since the pressure of
exhibit ideal behavior. CO2 is half the pressure of the CO, it must contain
(a) A 1.0 mol sample of CO(g) is heated at constant half as many molecules.
pressure. On the graph below, sketch the expected
plot of volume vs. temperature.
(b) Samples of CO(g) and CO2(g) are placed in 1.00 L
containers at the conditions below:
(i) Indicate whether the average kinetic energy
of the CO2 is greater than, equal to, or less
than that of CO molecules. Justify answer.
(ii) Indicate whether the root-mean-square speed
of the CO2(g) molecules is greater than, equal
to or less than that of the CO(g) molecules.
Justify answer.
(iii) Indicate whether the number of CO2
molecules is greater than, equal, or less than
that of CO molecules. Justify answer.
Answer:
(a)
(b) (i) equal to; at the same temperature, all gas
molecules have the same kinetic energy.
(ii) less, since CO2 has a molar mass of 44 g/mol and
CO has a mass of 28 g/mol, the lighter molecule is
faster at the same temperature.(Graham’s Law)