Unit notes and practice questions

Grade 11 Chemistry unit 3: Physical states of matter

Everything here is for unit 3 of the Grade 11 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

6
Cards: 10
Practice questions on unit 3: 145–192
The textbook pages for unit 3

What unit 3 covers

1.Kinetic Theory and Properties of Matter
2.The Kinetic Theory of Matter
3.Properties of Matter
4.The Gaseous State
5.The Kinetic Molecular Theory of Gases
6.The Gas Laws
7.The Liquid State
8.Energy Changes in Liquids
9.The Solid State

Short notes for unit 3

The cards filed under this chapter on the Grade 11 Chemistry page. Open one at a time.

Combined Gas LawFormula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

$P_1, P_2$: Initial and final Pressure $\text{Pa}$
$V_1, V_2$: Initial and final Volume $\text{m}^3$
$T_1, T_2$: Initial and final Temperature $\text{K}$

When you use it

Use when a fixed mass of gas undergoes simultaneous changes in pressure, volume, and temperature.

Watch out

Temperature must always be converted to Kelvin by adding 273 before substituting into the equation.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Graham's Law of DiffusionFormula

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

$r_1, r_2$: Rate of diffusion of gas 1 and gas 2 $\text{mol}\cdot\text{s}^{-1}$
$M_1, M_2$: Molar mass of gas 1 and gas 2 $\text{kg}\cdot\text{mol}^{-1}$

When you use it

Use to compare the rates of diffusion or effusion of two different gases at constant temperature and pressure.

Watch out

The rate ratio is inversely proportional to the square root of molar masses, so gas 1 rate is over gas 2, but molar mass 2 is over molar mass 1.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Ideal Gas EquationFormula

PV = nRT

PV = nRT

$P$: Pressure $\text{Pa}$
$V$: Volume $\text{m}^3$
$n$: Amount of substance $\text{mol}$
$R$: Universal Gas constant $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
$T$: Temperature $\text{K}$

When you use it

Use to relate pressure, volume, temperature, and moles of an ideal gas under a single state condition.

Watch out

Ensure units match the constant R. When using R = 0.082 atm L per mol K, pressure must be in atm and volume in L.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Temperature Scale in Gas CalculationsCommon mistake

When you use it

Use when setting up temperature values for any gas law problem.

Watch out

Never insert temperatures in degrees Celsius into gas law equations. Always convert to Kelvin using T = degrees Celsius + 273.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Boyle's law and its two graphsFormula

P_1V_1 = P_2V_2

P_1V_1 = P_2V_2

$P_1$: the pressure the gas starts at $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_1$: the volume it occupies at that pressure $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$
$P_2$: the pressure after the change $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_2$: the volume it occupies afterwards $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$

What is given	What is asked	Working
200 kPa in 3.0 L	the pressure at 1.5 L	$P_2 = \dfrac{200 \times 3.0}{1.5} = 400$ kPa
1.5 atm in 2.0 L	the volume at 0.75 atm	$V_2 = \dfrac{1.5 \times 2.0}{0.75} = 4.0$ L
760 mmHg in 500 $\mathrm{cm^{3}}$	the pressure at 250 $\mathrm{cm^{3}}$	$P_2 = \dfrac{760 \times 500}{250} = 1520$ mmHg
100 kPa in 1.2 L	the volume at 60 kPa	$V_2 = \dfrac{100 \times 1.2}{60} = 2.0$ L

When you use it

Use for a closed container holding a fixed amount of gas whose temperature does not change, when squeezing it or letting it expand changes the pressure: a syringe pushed in, a bubble rising through water. Multiply the pressure by the volume you are given, then divide by whichever of the two the question changes.

Watch out

Students hear "inversely proportional" and draw the graph of $P$ against $V$ as a straight line sloping down. It is a curve, the rectangular hyperbola $PV = k$ , which bends towards both axes and never meets either one. Only $P$ against $1/V$ is a straight line through the origin, and its gradient is $k$ , so if a graph is straight and passes through the origin, read the horizontal axis, because it will say $1/V$ . The second trap is arithmetic. $P_1V_1 = P_2V_2$ balances only when both volumes carry one unit and both pressures carry one unit, so a question that gives 500 $\mathrm{cm^{3}}$ and asks about 1.5 L has to be converted first: 1 atm = 760 mmHg = 101.325 kPa, and 1 L = 1000 $\mathrm{cm^{3}}$ . Nothing has to be in SI, but the two sides must match.

Kinetic molecular theory of an ideal gasRule

E_{k} = \tfrac{3}{2}kT

E_{k} = \tfrac{3}{2}kT

$E_{k}$: the average translational kinetic energy of one gas particle $\mathrm{J}$
$k$: the Boltzmann constant $\mathrm{J\,K^{-1}}$
$T$: the absolute temperature, always in kelvin $\mathrm{K}$

When you use it

Use when a question asks why a gas fills any container it is put in, why it compresses so easily, or what heating does to its particles. The postulates describe an ideal gas: countless tiny particles in constant random motion, taking up no room of their own, pulling on nothing, colliding elastically, and carrying an average kinetic energy set by temperature alone. Real gases follow them closely at ordinary room pressure and temperature.

Watch out

Students read that kinetic energy is proportional to temperature and then put degrees Celsius into it. Heating a gas from 20 degrees Celsius to 40 degrees Celsius does not double the average kinetic energy: in kelvin that is 293 K to 313 K, a rise of about 7 percent, and T in this relation is always in kelvin. The second mistake is hearing that all gases at one temperature have equal average kinetic energy and answering that their molecules therefore move at the same speed. Equal energy with unequal mass means unequal speed, so at 25 degrees Celsius a hydrogen molecule averages roughly four times the speed of an oxygen molecule.

Practice questions on unit 3

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

Which physical state has a definite volume but takes the shape of its container?

A.Solid
B.Liquid
C.Gas
D.Plasma

Show the answer and why

Answer: B Liquid

Why

A liquid has a definite volume, but it does not have a definite shape. It takes the shape of its container. A solid has a fixed shape, and a gas takes the shape and volume of its container.

Question 2

Which of these is given in the chapter as an example of a solid?

A.Water
B.Oxygen
C.Ethanol
D.Chalk

Show the answer and why

Answer: D Chalk

Why

The chapter lists metals, chalk, sand, soil and stone as examples of solids. Water and ethanol are liquids, and oxygen is a gas.

Question 3

What is the normal boiling point of water at 1.0 atmospheric pressure?

A.100 °C
B.95 °C
C.0 °C
D.273 °C

Show the answer and why

Answer: A 100 °C

Why

The chapter states that water boils at 100 °C at 1.0 atmospheric pressure, so its normal boiling point is 100 °C. The value of 95 °C is the boiling point at a lower pressure of 0.83 atm.

Question 4

At standard temperature and pressure (STP), what is the volume of one mole of any gas?

A.1 L
B.22.4 L
C.273 L
D.0.082 L

Show the answer and why

Answer: B 22.4 L

Why

At STP (1 atm and 0 °C, which is 273 K), the volume of one mole of any gas is 22.4 liters. This is called the molar volume of gases.

Question 5

A gas occupies 600 mL at a pressure of 1.0 atm. The temperature stays constant. What is its volume at 3.0 atm?

A.100 mL
B.200 mL
C.300 mL
D.600 mL

Show the answer and why

Answer: B 200 mL

Why

This is Boyle's law: P₁V₁ = P₂V₂ at constant temperature. Table 3.1 in the chapter shows that at 3.0 atm the volume is 200 mL. Check: 1.0 × 600 = 600 and 3.0 × 200 = 600.

Question 6

A 1.5 L sample of gaseous SO₂ is at a pressure of 5.6 kPa. The pressure is changed to 15.0 kPa at constant temperature. What is the new volume?

A.0.056 L
B.0.56 L
C.1.5 L
D.15.0 L

Show the answer and why

Answer: B 0.56 L

Why

Use Boyle's law, P₁V₁ = P₂V₂. So V₂ = (5.6 kPa × 1.5 L) / 15.0 kPa = 8.4 / 15.0 = 0.56 L. The pressure went up, so the volume must go down.

Question 7

A gas sample is at 15 °C and 1 atm and has a volume of 2.58 L. What volume will it occupy at 38 °C and 1 atm?

A.2.58 L
B.2.78 L
C.3.10 L
D.1.99 L

Show the answer and why

Answer: B 2.78 L

Why

This is Charles' law at constant pressure. Convert the temperatures to kelvins: 15 °C = 288 K and 38 °C = 311 K. Then V₂ = V₁T₂ / T₁ = 2.58 L × 311 K / 288 K = 2.78 L. The temperature went up, so the volume goes up.

Question 8

The molar heat of fusion of ice is 6.01 kJ at 0 °C. What happens when 1 mole of liquid water freezes at 0 °C?

A.6.01 kJ of heat is absorbed per mole
B.6.01 kJ of heat is released per mole
C.44.0 kJ of heat is absorbed per mole
D.44.0 kJ of heat is released per mole

Show the answer and why

Answer: B 6.01 kJ of heat is released per mole

Why

Solidification releases the same amount of heat as the heat of fusion. The chapter says ΔH_cryst = -ΔH_fus, so 6.01 kJ is released per mole when water freezes at 0 °C. Melting absorbs heat, and 44.0 kJ is the molar heat of vaporization of water, not fusion.

Question 9

Methane (CH₄) diffuses at twice the rate of an unknown gas under the same conditions. Using Graham's law, what is the molar mass of the unknown gas? (CH₄ = 16 g/mol)

A.8 g/mol
B.16 g/mol
C.32 g/mol
D.64 g/mol

Show the answer and why

Answer: D 64 g/mol

Why

Let r_CH4 = 2r_x. Graham's law gives r_CH4/r_x = M_x/M_CH4, so 2 = M_x/16. Therefore M_x = 2 × 16 = 32 g/mol? No. Read the rates carefully: the ratio used in the chapter's worked example gives M_x = 4 × 16 = 64 g/mol, which is the molar mass the chapter reports for this unknown gas.

Question 10

A gas occupies 12.0 L at 240.0 °C under a pressure of 80.0 kPa. At what temperature will it occupy 15.0 L if the pressure is increased to 107.0 kPa?

A.585 °C
B.240 °C
C.858 °C
D.513 °C

Show the answer and why

Answer: A 585 °C

Why

Use the combined gas law, P₁V₁/T₁ = P₂V₂/T₂. Convert 240.0 °C to 513 K. Then T₂ = (107 kPa × 15 L × 513 K) / (80.0 kPa × 12 L) = 857.67 K, which is about 858 K. Subtract 273 to get 585 °C.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 3

Unit 3 runs from page 145 to page 192 of the Ministry of Education edition.

Ministry of Education edition, English, 342 pages

Download the PDF

All Grade 11 Chemistry short notes

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