General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: gases, Dalton, kinetic theory and real gases (202-SN1-RE)

Gases are the most calculation-heavy chapter of 202-SN1-RE (formerly 202-NYA), and the most regular: four laws, two constants, and a discipline of units. Almost every question comes down to choosing between the combined gas law, which compares two states, and the ideal gas law, which describes a single state.

This sheet fixes the units, separates the two families of laws, and gathers the traps that come back from one exam to the next: water vapour, mole fractions, the molar mass in kilograms and the direction of Graham's law.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Every law in the chapter involves a RATIO of temperatures, so it demands kelvins. A degree Celsius left in a gas calculation does not give a small error, it gives a result with no relation to reality.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

The constants and the units, before anything else

• $R=8.314$ kPa·L/(mol·K) for everything involving $PV=nRT$ ; $R=8.314$ J/(mol·K) for everything involving energy and speeds. It is the same number, not the same unit.
• $101.3$ kPa $=1$ atm $=760$ mm Hg. $T(\text{K})=T(°\text{C})+273$ (a rounding of $273.15$ , which the worked example keeps).
• Molar volume: $22.4$ L/mol at STP ( $0$ °C, $101.3$ kPa) and $24.5$ L/mol at $25$ °C and $101.3$ kPa.
• Combined gas law, for a CLOSED amount of gas: $\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$ . Boyle's, Charles's and Gay-Lussac's laws are only special cases of it.
• Ideal gas law, for a single state: $PV=nRT$ , hence $M=\dfrac{mRT}{PV}$ and $\rho=\dfrac{PM}{RT}$ .

On the left, the SAME amount of gas goes from one state to another: the combined gas law links the two states without ever asking for

n

. On the right, a single state:

PV=nRT

requires

n

The combined gas law compares TWO states and does not need $n$ ; the ideal gas law describes ONE state and requires $n$ . Choosing the wrong one sends you looking for data the problem does not supply.

Gas mixtures: everything goes through the mole fraction

• Dalton's law: $P_{\text{total}}=\sum P_{i}$ , and $P_{i}=x_{i}P_{\text{total}}$ where $x_{i}=\dfrac{n_{i}}{n_{\text{total}}}$ is the MOLE FRACTION.
• In a rigid container, the pressure depends only on the TOTAL NUMBER of moles of gas, never on the nature of the gases.
• Gas collected over water: the gas is saturated with vapour, so $P_{\text{gas}}=P_{atm}-P_{vap}(T)$ . The vapour pressure depends ONLY on the temperature.
• At fixed $T$ and $P$ , gas volumes are proportional to amounts of substance: the coefficients of the equation can then be read directly as VOLUMES.
• Mole fraction and mass fraction differ as soon as the molar masses differ, and the problem often gives the second one to set a trap.

Each gas exerts the pressure it would have ALONE in the whole volume: six molecules out of nine make two thirds of the pressure, whatever their nature.

Any question that mentions “collected over water”, “over water” or “by displacement of water” requires subtracting the vapour pressure. It is the most profitable keyword of the chapter.

Kinetic theory: what depends on $T$ and what depends on $M$

• The average kinetic ENERGY depends ONLY on the temperature: $\overline{E_{k}}=\tfrac{3}{2}RT$ per mole. At $25$ °C, all gases have the same one.
• The SPEED, on the other hand, depends on the mass: $v_{rms}=\sqrt{\dfrac{3RT}{M}}$ , with $R$ in J/(mol·K) and $M$ IN KILOGRAMS PER MOLE.
• Graham's law: $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{M_{2}}{M_{1}}}$ . The LIGHT gas effuses FASTEST.
• Maxwell-Boltzmann distribution: $v_{p}<v_{avg}<v_{rms}$ , and the area under the curve is always $1$ .
• Real gases: $P=\dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}}$ . The $b$ term corrects for the molecular volume, the $a$ term corrects for the attractions.

Heating shifts the maximum to the right AND flattens the curve, since the area stays equal to

1

. It is the right-hand tail, that of the very fast molecules, that swells the most.

Heating shifts the maximum of the distribution to the right AND flattens it, since the area is constant. It is this flattening, more than the shift, that explains the explosion in the number of very fast molecules.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Leaving a temperature in degrees Celsius in a gas law

the whole question, 5 marks

What not to write

“The gas goes from $20$ °C to $40$ °C at constant pressure: its volume doubles.”

What to write

“ $293$ K becomes $313$ K: the volume only increases by $6.8\%$ , not by $100\%$ .”

The volume is proportional to the temperature in KELVINS: the green line starts at

-273

°C. The red one, starting from

0

°C, is the one you assume by keeping degrees Celsius.

Why: The gas laws involve RATIOS of temperatures, and a ratio only makes sense on a scale whose zero is absolute. On the Celsius scale, $0$ °C would give a division by zero.

2. Mixing the two units of the gas constant

3 marks, and a factor of

1000

What not to write

“ $P=101.3$ kPa, $V=2.00$ L, $T=298$ K and $R=8.314$ J/(mol·K): $n=\dfrac{PV}{RT}$ with $P$ in pascals... or in kilopascals?”

What to write

“With $P$ in kPa and $V$ in L, take $R=8.314$ kPa·L/(mol·K) and $n=0.0818$ mol. With $R$ in J/(mol·K), $P$ would have to be in Pa and $V$ in m $^{3}$ .”

Why: $1$ J $=1$ Pa·m $^{3}$ $=1$ kPa·L: the two forms are numerically identical provided you are consistent. Mixing the two is what produces the factors of a thousand.

3. Forgetting the water vapour pressure on a gas collected by displacement of water

3 marks, and an amount of substance overestimated by

2.7\%

What not to write

“The gas is collected over water at $22$ °C under $101.0$ kPa: so $P_{\text{gas}}=101.0$ kPa.”

What to write

“The gas is SATURATED with water vapour: $P_{\text{gas}}=101.0-2.64=98.4$ kPa, the vapour pressure at $22$ °C being $2.64$ kPa.”

With the levels equalized, the pressure in the jar equals the atmospheric pressure, but it is shared between TWO gases: subtract the water vapour before any calculation.

Why: The collecting jar contains two gases, the one under study and water vapour. Dalton's law requires removing the contribution of the second, and its value depends only on the temperature.

4. Using a mass fraction as a mole fraction in Dalton's law

the whole question, 6 marks

What not to write

“The mixture contains $50\%$ by mass of oxygen gas and $50\%$ of hydrogen gas: each gas exerts half of the total pressure.”

What to write

“For $100$ g, $n(O_{2})=1.56$ mol and $n(H_{2})=24.8$ mol: the mole fractions are $0.059$ and $0.941$ . Hydrogen gas exerts $94\%$ of the pressure.”

Fifty grams of hydrogen gas make sixteen times more molecules than fifty grams of oxygen gas: it is this number, not the mass, that fixes the mole fraction.

Why: Pressure is produced by COLLISIONS, hence by a number of particles, not by a mass. Two gases with very different molar masses never have the same mass fraction and mole fraction.

5. Leaving the molar mass in grams per mole in the root-mean-square speed

3 marks, and a factor of

31.6

What not to write

“ $v_{rms}=\sqrt{\dfrac{3(8.314)(295)}{2.016}}=60.4$ m/s for hydrogen gas.”

What to write

“ $M$ must be in KILOGRAMS per mole: $\sqrt{\dfrac{3(8.314)(295)}{0.002016}}=1.91\times 10^{3}$ m/s.”

Why: The joule is a kilogram metre squared per second squared: the mass must therefore enter in kilograms. The error factor is $\sqrt{1000}$ , about $32$ , which jumps out at you on the order of magnitude.

6. Inverting the ratio in Graham's law

3 marks

What not to write

“Oxygen gas is heavier than hydrogen gas, so it effuses four times faster.”

What to write

“ $\dfrac{v_{H_{2}}}{v_{O_{2}}}=\sqrt{\dfrac{M_{O_{2}}}{M_{H_{2}}}}=\sqrt{\dfrac{32.00}{2.016}}=3.98$ : it is the LIGHT one that goes almost four times faster.”

Same temperature, hence same average kinetic energy: the molecule sixteen times lighter goes

\sqrt{16}

, that is almost four times faster, and leaves first through the hole.

Why: At the same temperature, the two gases have the same average kinetic energy. A larger mass therefore forces a smaller speed, since the energy is $\tfrac{1}{2}mv^{2}$ .

7. Believing that the average kinetic energy depends on the nature of the gas

4 marks on the reasoning question

What not to write

“At $25$ °C, oxygen molecules have more kinetic energy than helium molecules, since they are heavier.”

What to write

“ $\overline{E_{k}}=\tfrac{3}{2}RT$ depends ONLY on the temperature: both gases have exactly the same average kinetic energy. It is the SPEED that differs.”

25

°C, helium and oxygen gas have the SAME average kinetic energy: only the speed differs, and it is the lightest gas that goes fastest.

Why: It is the very definition of temperature in kinetic theory. If two gases at the same temperature did not have the same average energy per molecule, they would exchange heat on contact, which would contradict thermal equilibrium.

8. Making the pressure in a rigid container depend on the nature of the gases

the whole question, 5 marks

What not to write

“After the reaction, carbon dioxide is left instead of oxygen gas, which is heavier, so the pressure changes.”

What to write

“At constant volume and temperature, $P=\dfrac{n_{\text{total}}RT}{V}$ : only a change in the TOTAL NUMBER of moles of gas changes the pressure.”

The two gauges read the same pressure: at fixed volume and temperature, only the NUMBER of molecules counts, not their mass.

Why: A heavy molecule hits harder but less often: the two effects compensate exactly. It is the very content of Avogadro's law.

9. Getting the signs wrong in the two van der Waals corrections

3 marks

What not to write

“ $P=\dfrac{nRT}{V+nb}+\dfrac{an^{2}}{V^{2}}$ : we add the volume of the molecules and the effect of the attractions.”

What to write

“ $P=\dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}}$ : the AVAILABLE volume is reduced by the molecular volume, and the attractions LOWER the pressure exerted on the walls.”

Real methane goes below

Z=1

while the attractions dominate, then above it when the molecular volume wins: the two corrections act in opposite directions.

Why: Molecules take up room, so the free space is smaller than $V$ . And a molecule pulled inward hits the wall less hard, which lowers the measured pressure.

Which method to choose

Which gas law to use

The number of states described by the problem and the quantities supplied

If the problem describes TWO states of the same amount of gas → combined gas law $\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$ , crossing out the constant quantities

Example: a balloon compressed and cooled
If the problem describes a SINGLE state and talks about moles, mass or molar mass → $PV=nRT$ , with $n=\dfrac{m}{M}$ if needed

Example: identifying a gas from its density
If the problem talks about a MIXTURE of gases → Dalton, $P_{i}=x_{i}P_{\text{total}}$ with MOLE fractions

Example: air, a diving mixture, gas collected over water
If the problem mentions “collected over water” → subtract the vapour pressure first, THEN apply $PV=nRT$

Example: $P_{\text{gas}}=101.0-2.64=98.4$ kPa
If the problem asks for a speed, an energy or an effusion → kinetic theory, with $R$ in J/(mol·K) and $M$ in kg/mol

Example: $v_{rms}$ , Graham's law
If the problem gives the parameters $a$ and $b$ , or talks about high pressure → van der Waals, and compare with the ideal result to comment on the difference

at low pressure and high temperature, the two results coincide

If a piece of data supplied is useless in the law you chose, it is almost always the law that is wrong. The parameters $a$ and $b$ , in particular, are never given for decoration.

What depends on temperature, on mass, or on both

The quantity asked for

If average kinetic energy → depends ONLY on $T$ : $\overline{E_{k}}=\tfrac{3}{2}RT$ per mole

Example: all gases at $25$ °C have $3.72$ kJ/mol
If root-mean-square speed → depends on $T$ AND on $M$ : $v_{rms}=\sqrt{\dfrac{3RT}{M}}$

Example: $1911$ m/s for $H_{2}$ at $22$ °C, $480$ m/s for $O_{2}$
If ratio of two effusion rates at the same temperature → Graham, $\sqrt{\dfrac{M_{2}}{M_{1}}}$ , the temperature drops out

Example: $H_{2}$ effuses $3.98$ times faster than $O_{2}$
If vapour pressure of a liquid → depends ONLY on $T$ , never on the volume or the amount of liquid

Example: $2.64$ kPa for water at $22$ °C, whatever the size of the beaker
If position of the maximum of the Maxwell-Boltzmann distribution → $v_{p}=\sqrt{\dfrac{2RT}{M}}$ , so shifted to the right if $T$ rises or if $M$ falls

the curve flattens as it shifts, because the area under the curve stays equal to 1

The distinction between what depends on the mass and what does not is the heart of the reasoning questions in this chapter: the energy depends only on the temperature, the speed depends on both.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Working with a gas collected over water

When to use it: The problem mentions a gas collected by displacement of water, over a water trough or in an inverted test tube.

1 Write Dalton's law applied to the jar: $P_{atm}=P_{\text{gas}}+P_{vap}$ , justifying that the gas is saturated with vapour.
2 Read the vapour pressure in the table at the TEMPERATURE of the experiment, and subtract it.
3 Convert: volume in litres, temperature in kelvins, pressure in kilopascals.
4 Apply $PV=nRT$ with $R=8.314$ kPa·L/(mol·K) to obtain the amount of gas.
5 Carry on with stoichiometry if the problem asks for it, starting from this amount.

Concluding sentence

“Since the gas is collected over water, it is saturated with vapour: $P_{H_{2}}=101.0-2.64=98.4$ kPa. The ideal gas law then gives $n=\dfrac{98.4\times 0.245}{8.314\times 295.15}=9.82\times 10^{-3}$ mol.”

The trap: Subtracting the vapour pressure AFTER calculating $n$ . The correction applies to the pressure, so it must be made before any other operation.

Marking: Typically 1 mark for citing Dalton's law, 1 for the subtraction, 2 for the conversions, 2 for the amount of substance, 2 for the stoichiometry that follows.

Comparing two gases with kinetic theory

When to use it: The problem asks for a speed, an effusion time, or the comparison of two gases at the same temperature.

1 First check whether the question is about ENERGY, which depends only on $T$ , or about SPEED, which also depends on $M$ .
2 Convert each molar mass to kilograms per mole, writing it out explicitly.
3 Apply $v_{rms}=\sqrt{\dfrac{3RT}{M}}$ with $R=8.314$ J/(mol·K), or Graham's law if only the ratio is asked for.
4 Conclude by naming the faster gas and checking that it is indeed the lighter one.

Concluding sentence

“At $22$ °C, $v_{rms}(H_{2})=\sqrt{\dfrac{3(8.314)(295.15)}{0.002016}}=1.91\times 10^{3}$ m/s against $4.80\times 10^{2}$ m/s for $O_{2}$ , in the ratio $\sqrt{\dfrac{32.00}{2.016}}=3.98$ predicted by Graham's law.”

The trap: Comparing two gases at different temperatures with Graham's law. It assumes the same temperature, otherwise you must go back through $v_{rms}$ for each one.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

All temperatures in kelvins

Reread each line of the calculation and point at the temperatures: none should be below $200$ in a room-temperature problem.

A $T=22$ in a ratio of temperatures: the kelvins were forgotten, the result is off by a factor of $13$ .
The order of magnitude of a molecular speed

A root-mean-square speed at room temperature is a few hundred metres per second for a common gas, and up to two kilometres per second for hydrogen gas.

A $v_{rms}=60$ m/s: the molar mass was left in grams per mole.
The sum of the partial pressures

Add up all the partial pressures you found: the total must equal exactly the total pressure given in the problem.

A sum of $95$ kPa when the problem announces $101$ kPa: a mole fraction is wrong or the water vapour was forgotten.
The direction of change before the calculation

Before applying the combined gas law, say whether the volume should increase or decrease. The result must go in that direction.

A compressed gas whose calculated volume increases: the two states have been swapped.
The deviation from the ideal gas

The pressure given by van der Waals should stay close to that of the ideal gas at moderate pressure, and depart clearly from it only at high pressure.

A difference of $40\%$ at $1$ atm: one of the two correction terms has the wrong sign.

The typical problem, taken apart

A hydrogen evolution collected over water

Zinc reacts with an EXCESS of hydrochloric acid according to $Zn+2\,HCl\rightarrow ZnCl_{2}+H_{2}$ . The hydrogen gas is collected by displacement of water: the volume is $245$ mL at $22.0$ °C, the atmospheric pressure is $101.0$ kPa and the vapour pressure of water at this temperature is $2.64$ kPa.

Determine the mass of zinc consumed, then calculate the root-mean-square speed of the hydrogen molecules at this temperature and compare it with that of oxygen gas.

Data: $M(Zn)=65.38$ g/mol, $M(H_{2})=2.016$ g/mol, $M(O_{2})=32.00$ g/mol, $R=8.314$ kPa·L/(mol·K) $=8.314$ J/(mol·K).

The hydrogen gas pushes the water out of the inverted test tube; the gas collected there is mixed with water vapour, at

22.0

°C under

101.0

kPa.

Step 1

Since the gas is collected over water, it is saturated with vapour: $P_{H_{2}}=P_{atm}-P_{vap}=101.0-2.64=98.4$ kPa.

Why

This is the first line, not the last: the whole rest of the calculation uses this corrected pressure. Justifying it in one sentence earns a mark and keeps you from having to come back to it.

Step 2

Conversions: $V=245$ mL $=0.245$ L and $T=22.0+273.15=295.15$ K.

Why

Both conversions are done at the same time, before any calculation, because they are exactly the two places where this chapter costs marks.

Step 3

$n(H_{2})=\dfrac{PV}{RT}=\dfrac{98.4\times 0.245}{8.314\times 295.15}=9.82\times 10^{-3}$ mol.

Why

The constant is taken in kPa·L/(mol·K) because the pressure is in kilopascals and the volume in litres. No other conversion is then needed.

Step 4

The equation gives $1$ mole of zinc for $1$ mole of hydrogen gas, so $n(Zn)=9.82\times 10^{-3}$ mol and $m=9.82\times 10^{-3}\times 65.38=0.642$ g.

Why

With the acid in excess, zinc is the limiting reactant and the measured amount of gas tells you directly. No limiting reactant calculation is needed here.

Step 5

Speed: $M(H_{2})=2.016$ g/mol $=2.016\times 10^{-3}$ kg/mol, so $v_{rms}=\sqrt{\dfrac{3(8.314)(295.15)}{2.016\times 10^{-3}}}=1.91\times 10^{3}$ m/s.

Why

The conversion to kilograms per mole is written on its own line: it is the only way not to forget it, and forgetting it costs a factor of $32$ on the result.

Step 6

For oxygen gas: $v_{rms}=\sqrt{\dfrac{3(8.314)(295.15)}{32.00\times 10^{-3}}}=4.80\times 10^{2}$ m/s.

Why

The same calculation, apart from the molar mass: this is what makes the comparison meaningful, the temperature being identical.

Step 7

Ratio: $\dfrac{1911}{480}=3.98$ , to be compared with $\sqrt{\dfrac{32.00}{2.016}}=3.98$ given by Graham's law.

Why

Two independent routes lead to the same number: it is the strongest check of the speed calculation, and it requires no extra data.

Step 8

Energy check: $\overline{E_{k}}=\tfrac{3}{2}(8.314)(295.15)=3.68\times 10^{3}$ J/mol for BOTH gases, despite speeds in a ratio of $4$ .

Why

This check recalls the distinction that structures the whole chapter: the energy depends only on the temperature, the speed also depends on the mass.

The conclusion, written out

“After correcting for the vapour pressure, the hydrogen gas collected amounts to $9.82\times 10^{-3}$ mol, which corresponds to $0.642$ g of zinc consumed; its molecules move at $1.91$ km/s root-mean-square, which is $3.98$ times faster than those of oxygen gas, exactly the ratio predicted by Graham's law.”

The classic mistake on this problem: Using $101.0$ kPa instead of $98.4$ kPa. You get $0.659$ g of zinc instead of $0.642$ g: the difference of $2.7\%$ is too small to shock and too large to get past the marking scheme, which makes it the most profitable trap of the exam.

Learn by heart

• $T$ ALWAYS in kelvins: $T(\text{K})=T(°\text{C})+273$ (rounded from $273.15$ ).
• $R=8.314$ kPa·L/(mol·K) with $P$ in kPa and $V$ in L; $R=8.314$ J/(mol·K) for energies and speeds.
• Two states of the same amount: combined gas law. A single state with moles: $PV=nRT$ .
• Dalton: $P_{i}=x_{i}P_{\text{total}}$ with MOLE fractions, never mass fractions.
• Collected over water: $P_{\text{gas}}=P_{atm}-P_{vap}(T)$ , and $P_{vap}$ depends only on $T$ .
• $\overline{E_{k}}=\tfrac{3}{2}RT$ depends only on $T$ ; $v_{rms}=\sqrt{\dfrac{3RT}{M}}$ with $M$ in kg/mol.
• Graham: the LIGHT gas effuses fastest, in the ratio $\sqrt{\dfrac{M_{\text{heavy}}}{M_{\text{light}}}}$ .
• van der Waals: $V-nb$ in the denominator, $-\dfrac{an^{2}}{V^{2}}$ subtracted. $Z<1$ : attractions; $Z>1$ : molecular volume.

Frequently asked questions

Why do temperatures have to be converted to kelvins for gases?

Because all the gas laws involve a ratio of temperatures, and a ratio only makes sense on a scale whose zero corresponds to a real absence of agitation. On the Celsius scale, zero degrees is not a physical zero, and a negative temperature would even make the calculation absurd.

Why subtract the vapour pressure from a gas collected over water?

Because the jar contains two gases: the one under study and water vapour, which saturates the space above the liquid. The pressure measured is their sum, according to Dalton's law. You must therefore subtract the vapour pressure, which is read from a table at the temperature of the experiment and depends only on it.

Which gas effuses faster, the light one or the heavy one?

The light one. At the same temperature, all gases have the same average kinetic energy per molecule. Since this energy equals half the mass times the square of the speed, a smaller mass forces a larger speed. Hydrogen gas thus effuses almost four times faster than oxygen gas.

Does the average kinetic energy depend on the gas chosen?

No, it depends only on the temperature. It is even the kinetic definition of temperature. Two different gases at 25 degrees have exactly the same average kinetic energy per molecule, about 3.7 kilojoules per mole. Only their speeds differ, because their masses differ.

Which unit should the molar mass have in the root-mean-square speed?

The kilogram per mole, never the gram per mole. The gas constant is then 8.314 joules per mole per kelvin, and the joule contains a kilogram. Using grams gives a result about 32 times too small, which is spotted immediately from the order of magnitude.

What does a compressibility factor below one mean?

That the attractive forces between molecules dominate: they pull the molecules inward, so they hit the walls less hard than in an ideal gas. Conversely, a factor above one signals that the volume of the molecules themselves becomes dominant, which happens at very high pressures.

Practise it

Corrected exercises: Gases: Dalton, kinetic theory and real gases, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn1-gases. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: gases, Dalton, kinetic theory and real gases (202-SN1-RE)

The essentials

The constants and the units, before anything else

Gas mixtures: everything goes through the mole fraction

Kinetic theory: what depends on TTT and what depends on MMM

The mistakes that cost marks

1. Leaving a temperature in degrees Celsius in a gas law

2. Mixing the two units of the gas constant

3. Forgetting the water vapour pressure on a gas collected by displacement of water

4. Using a mass fraction as a mole fraction in Dalton's law

5. Leaving the molar mass in grams per mole in the root-mean-square speed

6. Inverting the ratio in Graham's law

7. Believing that the average kinetic energy depends on the nature of the gas

8. Making the pressure in a rigid container depend on the nature of the gases

9. Getting the signs wrong in the two van der Waals corrections

Which method to choose

Which gas law to use

What depends on temperature, on mass, or on both

How the answer is expected to be written

Working with a gas collected over water

Comparing two gases with kinetic theory

Check before you hand in

The typical problem, taken apart

A hydrogen evolution collected over water

Learn by heart

Frequently asked questions

Why do temperatures have to be converted to kelvins for gases?

Why subtract the vapour pressure from a gas collected over water?

Which gas effuses faster, the light one or the heavy one?

Does the average kinetic energy depend on the gas chosen?

Which unit should the molar mass have in the root-mean-square speed?

What does a compressibility factor below one mean?

Corrected exercises: Gases: Dalton, kinetic theory and real gases, 202-SN1-RE

See also

Looking for a 202-SN1-RE chemistry tutor in Montreal?

Kinetic theory: what depends on $T$ and what depends on $M$