Gases Test – typical questions
STP = 101.3 kPa, 0oCSATP = 100 kPa, 25oC
0oC = 273KPV = nRTR = 8.31 kPa∙L/(mol∙K)
Boyle’s Law: P α 1/V, so … P1V1 = P2V2
Charle’s Law: V α T, so …. V1/T1 = V2/T2
Guy-Lussac’s Law: P α T … P1/T1 = P2/T2
1 atm= 101.3 kPa
= 14.7 lb∙in-2
= 760 mm Hg
8.1
What is absolute pressure?
The sum of relative (gauge) pressure and atmospheric pressure, Pabs = Prel + Patm.
A bike tire is filled to 60.0 psi (pounds per square inch, lb∙in-2. Convert to SI units. Then, calculate the absolute pressure on a day with SATP conditions?
P (kPa) = 60.0 psi x 101.3 kPa .= 413 kPa
14.7 psi
Pabs = 413 + 100 kPa = 513 kPa.
Compare a manometer and a barometer (similarities and differences).
Both devices measure gas pressure, and in their simplest forms, both utilize the downwards pressure on a liquid in a U-shaped tube, which pushes the liquid higher on the other side of the U. Barometers measure the external, atmospheric pressure, whereas manometers measure the internal (gauge, relative) pressure of a gas inside a container.
8.2
What should you get if you extrapolate a graph of V vs. T (oC) backwards?
The graph would be linear for the most part, until very cold temperatures. Extrapolating it back to the horizontal, Temperature axis, it would intercept at -273oC, known as absolute zero, a theoretical state at which all motion ceases, and matter is compressed into perfectly ordered crystals.
Jason lit a fire without realizing the chimney had been bricked in years ago by his father. The living room was well sealed with closed windows and doors. The initial conditions in the room were SATP, but just before he passed out, Jason (ever the scientist) noticed that the temperature had risen to 50oC. Calculate the air pressure. (P.S. You need to assume that there is no increase in moles of gas, which is clearly wrong, but do it anyway.) Whose law is needed to answer this question?
P1/T1 = P2/T2, so P2 = P1T2/T1. Substituting in the variables, with temperatures in Kelvins, and evaluating leads to P2 = 108.4 kPa.
This follows Guy-Lussac’s Law.
A cylinder with a movable piston is held in a large water bath which keeps the internal temperature of the cylinder relatively constant. If the piston compresses the gas to 1/3 of its initial volume what happens to the pressure? If the piston is raised to expand the gas to 2x the initial volume, what happens to the pressure? Use the kinetic-molecular theory to explain why these pressure changes happen. Whose law is needed?
If V2 = 1/3 V1, then P2 = 3x the initial pressure.
If V2 = 2V1, then P2 = ½ P1
The pressure increases when the volume decreases because the particles are now closer together, and collide more frequently with the walls of the container, increasing the pressure they exert on the container.
The pressure-volume relationship is Boyle’s law.
If the pressure of a gas remains constant, what is the new volume when the gas is cooled to half its original temperature? Whose law is needed to answer this question?
When T2 = ½ T1, the new volume, V2 = ½ V1. This follows Charles’ Law.
8.3
What is meant by an ideal gas? Why do the ideal gas laws become less and less accurate as the pressure on an amount of gas increase, squeezing it into a smaller and smaller volume?
An ideal gas has particles that are not attracted to each other, that take up a negligible volume compared to the volume of the container, move in straight-lines, collide with one another and with the container without losing any energy, and have the same average kinetic energy at a particular temperature no matter the identity of the gas.
As a gas is compressed, the particles begin to experience an increasing force of attraction for one another, and since the volume is decreasing, the volume taken up by the particles of gas becomes a greater and greater fraction of the total volume.
How many times faster is a sample of helium than a sample of radon? (They are both at the same temperature.) Why is it faster?
This is because they have the same average kinetic energy, but Ra is much heavier. Kinetic energy is ½ mv2. The heavier the particle, the slower it travels if it has the same energy.
8.4
Winnie the Pooh was walking home on a warm, 30.0oC day, under those clear blue skies you only get on high pressure days. (It was 104 kPa.) Pooh was carrying a helium balloon filled with 5.00 L of gas. What amount of helium was in the balloon? What was the mass?
(Amount usually means moles). n = PV/RT, so in this case:
n = (104 kPa)(5.00 L) / (8.31 x 303 K)
= 0.2065188… mol
= approx. 0.207 mol (3 sig. dig’s)
m = n x M
= 0.2065188 mol x 4.00 g/mol
= 0.826075 g
= approx. 0.826 g
He then tied the balloon to his wrist and lay down for a nap. As he slumbered, a low pressure system moved in, dropping down to 97.0 kPa. The temperature fell to 28.0oC. What was the new volume?
P1V1 = P2V2 so V2 = P1V1T2 After substituting (using Kelvins), V2 = 5.33 L (appox.)
T1 T2 P2T1
8.5
A typical high pressure reading at the Dead Sea is 106.5 kPa. Meanwhile, Lake Louise is at high altitude, and may often reach an air pressure reading of 92.0 kPa. What can be said about the boiling point of water at each location?
In the Dead Sea region, the boiling point will be much higher, since the vapour pressure of the heated water must increase until it equals the atmospheric pressure, which is higher than normal. Conversely, the low air pressure at Lake Louise means that at a lower than normal temperature, the water’s vapour pressure will be high enough for boiling.
While running a marathon at high altitude, a contestant passed out, and was placed in an oxygen tent to revive. The total air pressure in the tent was maintained at 105 kPa. It consisted of 65% nitrogen, 25% oxygen, 9% water vapour, and 1% argon. Calculate the partial pressure of each gas. If there were 2000 L of gas in the tent, what volume of each gas was there?
PN2 = 0.65 x 105 kPaPO2 = 0.25 x 105 kPa
= 68.25 kPaetc.
= 68 kPa (2 sig.digs)
VN2 = 0.65 x 2 000 L
= 1 300 L
9.1
Write a balanced reaction for the complete* combustion of liquid propanol, C3H7OH.
If 5.0 g was ignited inside a rigid 1.0-L container (with an excess of oxygen gas to ensure complete combustion), what mass of gas is produced? What combined pressure do these gas products exert if the final temperature is 60oC?
*complete combustion reactions produce only carbon dioxide and water.
2C3H7OH(l) + 9O2(g)6CO2(g) + 8H2O(g)
n (propanol) = 5.0 g / 60.11g/mol
= 0.0831808… mol
n (gases) = 0.0831808… x 14/2
= 0.582265… mol
n (CO2) = 0.0831808… x 6/2n (H2O) = 0.0831808… x 8/2
= 0.24954… mol= 0.16636167 mol
m (CO2) = 0.24954… mol x 44.01 g/molm (H2O) =0.16636167 mol x 18.02 g/mol
= 10.9823 g= 2.997837 g
= approx.. 11 g= 3.0 g
P = nRT/V
= 0.582265… mol x 8.31 kPa∙L/(mol∙K) x 333 K / 1.0 L
= 1 611.2635… kPa
= 1 600 kPa
= 1.6 MPa
Note – I have ignored the contribution of any leftover oxygen. If the question was worded to say that there had been just the right amount of oxygen to ensure complete combustion, then the assumption is valid. In this case, since it says there was an excess of oxygen, the final pressure would be higher. Hindsight is 20-20?
9.2
At STP, a certain container holds 0.25 L of gas. How many moles is this? Calculate this in two ways to double check your answer – using the ideal gas law, and using the STP relationship between a gas volume and a molar quantity.
Using the STP relationship, we know that 1 L at STP occupies 22.4 L. Therefore,
n = 0.25 L x 1 mol/22.4 L (in other words, we divide 0.25 by 22.4 mol/L)
= 0.0111607 mol
= approx..0.011 mol
Or, since PV = nRT, we can rearrange the equation to get: n = PV/RT
n = (101.3 x 0.25) / (8.31 x 273)
= 0.011 mol
9.3
At 15oC and 80kPa, 625 mL of oxygen gas are produced over water. The vapour pressure of water at that temperature is 1.7 kPa. Calculate the moles of oxygen.
The atmospheric pressure of 80 kPa is due to the two partial pressures of water vapour and the oxygen gas. Therefore:
80 kPa = 1.7 kPa + PO2 . By subtraction, PO2 = 69.3 kPa
Then use this partial pressure in the ideal gas law, PV = nRT, to calculate the moles.
Answer: n = 0.018 mol
9.4
Sodium metal will react violently with hydrochloric acid, producing hydrogen gas and aqueous sodium chloride. If 24.8 L of hydrogen gas were produced at SATP, and the concentration of the acid was 0.30 mol/L, what volume of the acid was needed?
2Na(s) + 2HCl(aq) 2NaCl(aq) + H2(g)
nH2 = (100 kPa x 24.8 L) / (8.31 x 298 K)
= 1.001461811 mol
~ 1.0 mol/L
From this we can get the moles of acid, which is twice the moles of hydrogen:
nHCl = 2.00292… mol
VHCl = n / L
= 2.00292… mol / 0.30 mol/L
= 6.6764 L
~ 6.7 L
(That’s a lot of acid, and seems unrealistic. So then I started thinking about the volume of gas produced in the question, 24.8 L. That’s a lot of gas! That’s why so much acid is needed. We could calculate how much Na(s) must have been reacted, i.e.,
2.00292 mol x 22.99 g/mol = 46 g, which is quite a bit.)
9.5
Draw a diagram of the 5 layers of our atmosphere (TSMThE). Which layer contains a gas that helps absorb UV radiation and thus protects us from it? What gas is that? Why/how is it being depleted? What international attempts have been made to reduce the damage?
Exosphere
Thermosphere
Mesosphere
Stratosphere – contains ozone, O3(g), to absorb UV rays. It participates in this cycle:
uv uv
(1)O2 2O (unstable)(2) O2 + O O3 (3) O3O2 + O
Troposphere
Ozone is being depleted by chemicals such as chlorofluorocarbons, which react with the ozone. Unfortunately those chemicals are quite stable, and so last for a long time in the stratosphere, destroying many ozone molecules. International efforts have focused on reducing (or eliminating) the use of CFCs. Alternatives, such as HCFCs, have been used in air conditioners and refrigerators.
On a separate diagram, focusing on the lowest level, show the Sun’s incoming visible wavelengths, and the various ways it is reflected, absorbed, and reradiated. Explain the greenhouse effect. Name some gases that increase the greenhouse effect. How/how is this effect being exacerbated? What international efforts have been made to deal with this? What can an individual do to reduce the harm?
To save time, I will “cheat” by pasting diagrams: The first is from Wikipedia:
We can ignore the power per area figures (Watts per m2, but if you are curious, this is called “irradiance”). What is important for our purposes is that out of the Sun’s radiation that arrives at our planet, about 1/3 is reflected by the atmosphere. Of the roughly 2/3 that passes through, half is reflected by the Earth’s surface (esp. light coloured areas such as the polar ice caps). The other half of the 2/3 that reaches the surface is absorbed. Much of the absorbed radiation is converted to infrared radiation, and reradiated into the atmosphere. Of that, some passes back out into space, but some is absorbed by greenhouse gases (GHGs).
As this shows, the energy absorbed by GHGs in the troposphere and stratosphere is reradiated in all directions, warming our surface and the troposphere.
The most significant greenhouse gas is water vapour. It is more potent than some, though not all, gases, and is the most plentiful by far. However its levels are not increasing over time. Other GHGs are increasing as the result of human activities. Of these, carbon dioxide is being produced in relatively large quantities, and so is a major concern. In addition, methane is a very potent GHG. Others include nitrogen oxides, hydrofluorocarbons (HFCs), perfluorocarbons (PFCs), and sulfur hexafluoride (SF6).
The Kyoto Protocol was signed in 1998 to limit emissions of GHGs. Unfortunately not all of the countries that signed this have achieved their targets, and regrettably, Canada has officially withdrawn (as of Dec., 2011) from the legally binding plan. The United States, one of the world’s two biggest contributors of GHGs, signed the protocol in 1998, but their government never ratified it at home. China, the other of the biggest contributors, ratified in 2002.
Worse, the limits of the Kyoto Protocol expired on Jan.1, 2013. A weaker version was extended until 2015 (I need to check that date - can anyone confirm?), but it still omits the US, and also omits China. A new international agreement is to be negotiated for 2015 or 2020.
In the meantime, quite recently, U.S. President Obama has indicated he will use his executive power (which bypasses the House of Congress and the Senate), to require 30% reductions in emissions by fossil fuel power generating stations, largely coal, which produces a large portion of the U.S.’s GHGs, by 2030. It remains to be seen whether his plan will succeed, given the legal challenges that will be launched by those opposed.
Individually, one can try to reduce one’s own family’s production of GHGs, e.g., by switching to more fuel-efficient vehicles, reusing and reducing, and by searching out alternative energy sources. One can also lobby governments and businesses, and seek to do business with companies that are reducing their own carbon footprint.