1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A sample of argon gas in a rigid container is cooled from 25 °C to a temperature just above the one at which it begins to condense. Which of the numbered boxes in the diagram best represents the sample at the low temperature?
Answer and reasoning
ABox 1 A student who thinks gas particles never attract one another picks this evenly spread, non-interacting picture, which is the ideal-gas model. Just above condensation the atoms move slowly enough that attractions hold some of them close together for a time.
BBox 2 A student who thinks the particles join by covalent bonds as a gas nears condensation picks this picture of Ar₂ molecules. The attractions between Ar atoms are intermolecular; no electrons are shared and no new molecules form.
CBox 3Correct Just above condensation the Ar atoms move slowly, so the attractions between them (dashed lines) hold some atoms close together for part of the time. The atoms keep their size, no bonds form, and all eight atoms are still present. These attractions are why the gas deviates from ideal behavior near condensation.
DBox 4 A student who thinks particles shrink when a gas is cooled picks this. The size of each Ar atom does not depend on temperature; cooling only slows the atoms down.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.6.A.1 Ideal gas Fix
Ideal gas
A model of a gas in which the particles have negligible volume and no attractions for one another. The pressure, volume, temperature and amount of an ideal gas are related exactly by PV = nRT.
Real gas
An actual gas, whose particles have a volume of their own and attract one another. A real gas follows the ideal gas law closely at low pressures and at temperatures well above those at which it condenses, and deviates from it measurably near condensation and at extremely high pressures.
Deviation from ideal behavior
A difference between a measured property of a real gas (for example its pressure or volume) and the value calculated from the ideal gas law for the same n, T and the other variable. It can be expressed as the ratio PV/nRT, which equals 1 for an ideal gas, is less than 1 when attractions dominate and is greater than 1 when particle volume dominates.
Effect of interparticle attractions
Attractions between gas particles pull them toward one another, so they strike the walls less often and less forcefully than particles of an ideal gas would. The measured pressure (at fixed volume) or volume (at fixed pressure) is then smaller than the ideal-gas value. The effect is largest at conditions close to those that cause condensation, such as low temperature and high pressure.
Effect of particle volume
The particles of a real gas occupy part of the container, so the space in which they move is smaller than the container volume, and the gas cannot be compressed into the space the particles themselves fill. At extremely high pressures this makes the measured volume (at fixed pressure) or pressure (at fixed volume) larger than the ideal-gas value.
Conditions near condensation
Conditions, such as a low temperature or a high pressure, at which the particles' kinetic energies are not large compared with the attractions between them and the particles spend more time close together. A gas deviates most from ideal behavior because of attractions under these conditions.
Students often think Every gas obeys the ideal gas law exactly under all conditions, because gas particles have no attractions for one another and take up no space. In fact No. PV = nRT describes an ideal gas, whose particles have no volume and no attractions. A real gas follows it closely at low pressure and at temperatures well above condensation, but deviates measurably near condensation and at extremely high pressures.
Students often think Deviations from ideal behavior are always caused by attractions between the particles, so a real gas always has a lower pressure or smaller volume than the ideal gas law predicts. In fact No. Attractions make the pressure or volume of a real gas smaller than ideal and dominate near condensation, but the volume of the particles makes the pressure or volume larger than ideal and dominates at extremely high pressures.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
A sample of N₂ gas at 25 °C is compressed to an extremely high pressure. The measured volume of the sample is compared with the volume calculated from the ideal gas law for the same amount of gas at the same temperature and pressure. Which claim, with its reasoning, is correct?
Answer and reasoning
ALarger than calculated, because the N₂ molecules themselves fill a significant part of the spaceCorrect At extremely high pressure the molecules are packed so closely that their own volume is a significant fraction of the sample's volume. The gas cannot be compressed into the space the molecules already fill, so the measured volume is larger than the ideal-gas volume, which treats the molecules as points.
BSmaller than calculated, because the N₂ molecules take up space and leave less room for the gas A student who thinks particle volume makes a real gas occupy less space picks this. The molecules' own volume stops the sample from being compressed as far as an ideal gas, so the measured volume is larger, not smaller.
CSmaller than calculated, because attractions pull the closely packed N₂ molecules together A student who thinks attractions always cause the deviation picks this. Attractions do act, but at extremely high pressure the volume of the molecules has the larger effect, and the measured volume is larger than the ideal-gas volume.
DLarger than calculated, because collisions among crowded N₂ molecules add to the gas pressure A student who thinks collisions between particles cause deviations from ideal behavior picks this. The claim is right but the reasoning is not: pressure comes from collisions with the walls, and collisions between particles are already part of the ideal-gas model. The volume is larger because the molecules themselves occupy a significant part of the space.
A student plans to measure the pressure of 1.00 mol of a gas in a rigid 10.0 L container as the gas is cooled from 500 K to a temperature just above the one at which it begins to condense. At each temperature the student will calculate the ratio Pmeasured/Pideal, where Pideal = nRT/V. Which result should the student predict?
Answer and reasoning
AThe ratio remains equal to 1 at every temperature from 500 K down to condensation. A student who thinks every gas obeys the ideal gas law exactly picks this. Near condensation the attractions between the particles make the measured pressure lower than nRT/V.
BThe ratio is lowest at 500 K and gets closer to 1 as the gas is cooled toward condensation. A student who thinks faster-moving, more often colliding particles deviate more picks this. Higher temperatures make a gas behave more ideally, because attractions matter less when the particles move fast; the deviation grows on cooling.
CThe ratio is close to 1 at 500 K and falls below 1 as the gas nears condensation.Correct At 500 K the particles move fast and the attractions between them have little effect, so the gas behaves almost ideally. As the gas is cooled toward condensation the particles move more slowly, attractions pull them toward one another and reduce the force and frequency of their collisions with the walls, so the measured pressure falls below nRT/V.
DThe ratio is close to 1 at 500 K and climbs above 1 as the gas approaches condensation. A student who pairs the two causes of deviation with the wrong directions, taking attractions to raise the ratio above 1, picks this. Attractions pull the particles toward one another and reduce the force and frequency of their collisions with the walls, so near condensation the measured pressure is lower than nRT/V and the ratio falls below 1.
The graph shows PV/nRT as a function of pressure for a sample of a hypothetical real gas at constant temperature; the dashed line shows the value for an ideal gas. Over which range of pressure do attractions between the gas particles have a greater effect on the gas's behavior than the volume of the particles?
Answer and reasoning
AFrom 0 to 90 atm A student who reads the falling part of a curve as the region of low values picks this. Between 90 and 500 atm the curve is rising, but it is still below 1, so attractions still have the greater effect there.
BFrom 0 to 500 atmCorrect Attractions make PV/nRT smaller than 1 and particle volume makes it larger than 1. The curve lies below the ideal-gas line from 0 atm until it crosses the line at 500 atm, so over that whole range attractions have the greater effect, even where the curve is already rising.
CFrom 500 to 1000 atm A student who thinks attractions raise PV/nRT above 1 picks this. Above 500 atm PV/nRT is greater than 1, which is the effect of the particles' own volume, not of attractions.
DFrom 0 to 1000 atm A student who thinks attractions are always the cause of deviations picks this. Above 500 atm PV/nRT is greater than 1, so there the volume of the particles has the greater effect.
Working PV/nRT < 1 means the gas's PV is smaller than ideal, which is the effect of attractions; PV/nRT > 1 is the effect of particle volume. Read where the curve lies below the dashed line: from 0 atm up to the crossing point at 500 atm. Above 500 atm the curve is above the line, so particle volume dominates there.
A 40.0 g sample of a hypothetical gas, X₂, is held in a rigid 1.00 L container at 327 °C. The atomic mass of X is 20.0 g/mol. The measured pressure of the gas is 45.0 atm. What is the value of PV/nRT for the sample? (R = 0.08206 L·atm/(mol·K))
Answer and reasoning
A1.68 A student who substitutes the Celsius temperature into the ideal gas law picks this: 45.0 ÷ (1.00 × 0.08206 × 327) = 1.68. The gas law needs the absolute temperature, 600 K.
B0.46 A student who uses the atomic mass of X, 20.0 g/mol, as the molar mass of X₂ picks this: n = 2.00 mol and PV/nRT = 0.46. Each X₂ molecule has two X atoms, so its molar mass is 40.0 g/mol and n = 1.00 mol.
C1.00 A student who thinks every gas obeys the ideal gas law exactly takes the ratio to be 1 without using the data. The measured pressure, 45.0 atm, is lower than the 49.2 atm the ideal gas law gives, so PV/nRT = 0.91.
D0.91Correct n = 40.0 g ÷ 40.0 g/mol = 1.00 mol and T = 600 K, so nRT/V = 49.2 atm. PV/nRT = 45.0 ÷ 49.2 = 0.91. The measured pressure is lower than the ideal-gas pressure, so attractions between the particles have the greater effect.
Working Molar mass of X₂ = 2 × 20.0 = 40.0 g/mol, so n = 40.0 g ÷ 40.0 g/mol = 1.00 mol. T = 327 + 273.15 = 600.15 K. PV/nRT = (45.0 atm)(1.00 L) ÷ [(1.00 mol)(0.08206 L·atm/(mol·K))(600.15 K)] = 45.0 ÷ 49.25 = 0.914, which is 0.91 to two decimal places. A value below 1 shows that attractions between the particles have the greater effect at these conditions.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
