Hello dear student! Welcome to Unit 3. In this unit, we will study the three physical states of matter — solid, liquid, and gas — and understand why substances behave differently in each state. We will learn about the kinetic theory, gas laws (with many calculations!), and properties of liquids and solids. This unit has a lot of problem-solving, so let’s build a strong foundation step by step.
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3.1 Introduction
Matter exists in three common physical states: solid, liquid, and gas. Have you ever asked yourself: why is a rock hard, water flows, and air is invisible? The answer lies in how the particles are arranged and how they move.
Let’s compare the three states:
| Property | Solid | Liquid | Gas |
|---|---|---|---|
| Shape | Fixed shape | Takes shape of container | Takes shape of container |
| Volume | Fixed volume | Fixed volume | Fills entire container |
| Particle arrangement | Ordered, closely packed | Disordered, close | Very far apart, random |
| Particle motion | Vibrate in fixed positions | Slide past each other | Move freely at high speed |
| Compressibility | Nearly incompressible | Slightly compressible | Highly compressible |
| Intermolecular forces | Very strong | Moderate | Very weak |
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3.2 Kinetic Theory and Properties of Matter
3.2.1 The Kinetic Theory of Matter
The kinetic theory of matter states that all matter is made of tiny particles (atoms or molecules) that are in constant motion. The amount of motion (kinetic energy) depends on temperature.
Main postulates of the kinetic theory:
- All matter consists of tiny particles (atoms, molecules, or ions).
- These particles are in continuous random motion.
- The kinetic energy of particles increases with temperature.
- There are forces of attraction between particles. In solids these are very strong; in liquids, moderate; in gases, very weak.
- When particles collide with each other or with the walls of the container, the collisions are perfectly elastic (no net loss of kinetic energy).
The kinetic theory beautifully explains why heating a solid turns it into liquid, and further heating turns the liquid into gas: as temperature increases, particles gain more kinetic energy and can overcome the intermolecular forces holding them together.
3.2.2 Properties of Matter
Key measurable properties include:
- Mass (m): Amount of matter in a substance, measured in kg or g.
- Volume (V): Space occupied, measured in m³, L, or mL.
- Density (ρ): Mass per unit volume: $$\rho = \frac{m}{V}$$
- Temperature (T): Measure of average kinetic energy, measured in °C, K (Kelvin).
- Pressure (P): Force per unit area: $$P = \frac{F}{A}$$
Important: Kelvin and Celsius are related by: $$T(\text{K}) = T(°\text{C}) + 273.15$$
Never use Celsius in gas law calculations — always convert to Kelvin!
Key Exam Notes — Kinetic Theory:
• All matter has particles in continuous motion — even in solids (vibration).
• Higher temperature = greater kinetic energy = faster particle movement.
• State changes occur when particles gain/lose enough energy to overcome/establish intermolecular forces.
• Always convert °C to K for gas law calculations: K = °C + 273.15.
• Standard Temperature and Pressure (STP): T = 273 K (0 °C), P = 1 atm (101.325 kPa).
• All matter has particles in continuous motion — even in solids (vibration).
• Higher temperature = greater kinetic energy = faster particle movement.
• State changes occur when particles gain/lose enough energy to overcome/establish intermolecular forces.
• Always convert °C to K for gas law calculations: K = °C + 273.15.
• Standard Temperature and Pressure (STP): T = 273 K (0 °C), P = 1 atm (101.325 kPa).
Practice Questions:
1. Convert 37 °C and −40 °C to Kelvin.
2. A block of metal has a mass of 150 g and a volume of 20 cm³. Calculate its density.
1. Convert 37 °C and −40 °C to Kelvin.
2. A block of metal has a mass of 150 g and a volume of 20 cm³. Calculate its density.
Answer 1:
37 °C = 37 + 273.15 = 310.15 K
−40 °C = −40 + 273.15 = 233.15 K
Answer 2:
$$\rho = \frac{m}{V} = \frac{150 \text{ g}}{20 \text{ cm}^3} = 7.5 \text{ g/cm}^3$$
37 °C = 37 + 273.15 = 310.15 K
−40 °C = −40 + 273.15 = 233.15 K
Answer 2:
$$\rho = \frac{m}{V} = \frac{150 \text{ g}}{20 \text{ cm}^3} = 7.5 \text{ g/cm}^3$$
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3.3 The Gaseous State
Gases are the simplest state to study mathematically because gas particles are far apart and intermolecular forces are negligible. Let’s start with the theory behind gases, then learn each gas law one by one with worked examples.
3.3.1 The Kinetic Molecular Theory of Gases
This is a more detailed version of the kinetic theory, specifically for gases. It has five postulates:
- Gases consist of many tiny particles (atoms or molecules) that are far apart relative to their size. The volume of the particles themselves is negligible compared to the volume of the container.
- Gas particles are in constant random, straight-line motion. They change direction only when they collide with each other or with the container walls.
- Collisions are perfectly elastic. No kinetic energy is lost during collisions — total kinetic energy remains constant.
- There are no intermolecular forces between gas particles (except during collisions). Particles do not attract or repel each other.
- The average kinetic energy of gas particles is proportional to the absolute temperature (in Kelvin). KE ∝ T. At a given temperature, all gases have the same average kinetic energy regardless of the gas identity.
From postulate 5, since KE = ½mv² and KE ∝ T:
$$\text{KE} = \frac{3}{2}kT \quad \text{(per molecule)}$$
$$\text{KE}_{\text{avg}} = \frac{3}{2}RT \quad \text{(per mole)}$$
where k = Boltzmann constant (1.38 × 10⁻²³ J/K) and R = gas constant (8.314 J/(mol·K)).
An important consequence: at the same temperature, lighter gas molecules move faster than heavier ones (since KE is the same but KE = ½mv²).
Key Exam Notes — KMT of Gases:
• “Ideal gas” behavior follows all five postulates exactly. Real gases deviate slightly.
• Real gases behave most like ideal gases at HIGH temperature and LOW pressure.
• At high P or low T, gas particles are closer together → intermolecular forces matter → deviation from ideal behavior.
• Heavier gases move more slowly at the same temperature (same KE, but larger m).
• “Ideal gas” behavior follows all five postulates exactly. Real gases deviate slightly.
• Real gases behave most like ideal gases at HIGH temperature and LOW pressure.
• At high P or low T, gas particles are closer together → intermolecular forces matter → deviation from ideal behavior.
• Heavier gases move more slowly at the same temperature (same KE, but larger m).
Practice Questions:
1. At the same temperature, which moves faster: O₂ molecules or H₂ molecules? Explain.
2. Why do real gases deviate from ideal behavior at high pressure?
1. At the same temperature, which moves faster: O₂ molecules or H₂ molecules? Explain.
2. Why do real gases deviate from ideal behavior at high pressure?
Answer 1:
H₂ molecules move faster. At the same temperature, both gases have the same average kinetic energy (KE = 3/2 kT). Since KE = ½mv², if KE is the same and m is smaller for H₂ (2 g/mol vs 32 g/mol for O₂), then v must be larger for H₂.
Answer 2:
At high pressure, gas particles are forced very close together. This means: (a) the volume of the particles themselves is no longer negligible compared to the container volume (violating postulate 1), and (b) intermolecular forces become significant because particles are close together (violating postulate 4). Both effects cause real gases to deviate from ideal behavior.
H₂ molecules move faster. At the same temperature, both gases have the same average kinetic energy (KE = 3/2 kT). Since KE = ½mv², if KE is the same and m is smaller for H₂ (2 g/mol vs 32 g/mol for O₂), then v must be larger for H₂.
Answer 2:
At high pressure, gas particles are forced very close together. This means: (a) the volume of the particles themselves is no longer negligible compared to the container volume (violating postulate 1), and (b) intermolecular forces become significant because particles are close together (violating postulate 4). Both effects cause real gases to deviate from ideal behavior.
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3.3.2 The Gas Laws
Now we come to the most important part of this unit. The gas laws describe how the volume of a gas changes when pressure, temperature, or amount of gas changes. Let’s learn each one carefully.
There is a very useful tool called the gas law variables triangle — just remember that for a fixed amount of gas, changing one variable affects another:
P and V: inversely proportional (Boyle’s Law)
V and T: directly proportional (Charles’s Law)
P and T: directly proportional (Gay-Lussac’s Law)
V and n: directly proportional (Avogadro’s Law)
A. Boyle’s Law (Pressure-Volume Relationship)
In 1662, Robert Boyle discovered that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
Boyle’s Law: At constant T and n: $$P \propto \frac{1}{V} \quad \text{or} \quad P_1 V_1 = P_2 V_2$$
This means if you double the pressure, the volume halves. If you triple the volume, the pressure becomes one-third.
Worked Example 3.1: A gas occupies a volume of 250 mL at a pressure of 1.5 atm. If the pressure is increased to 3.0 atm at constant temperature, what is the new volume?
Solution:
$$P_1 V_1 = P_2 V_2$$
$$(1.5 \text{ atm})(250 \text{ mL}) = (3.0 \text{ atm})(V_2)$$
$$V_2 = \frac{(1.5)(250)}{3.0} = 125 \text{ mL}$$
The volume halved because the pressure doubled. This makes sense!
Solution:
$$P_1 V_1 = P_2 V_2$$
$$(1.5 \text{ atm})(250 \text{ mL}) = (3.0 \text{ atm})(V_2)$$
$$V_2 = \frac{(1.5)(250)}{3.0} = 125 \text{ mL}$$
The volume halved because the pressure doubled. This makes sense!
Worked Example 3.2: A gas in a syringe has a volume of 60 mL at 95 kPa. What pressure is needed to compress it to 40 mL?
Solution:
$$P_1 V_1 = P_2 V_2$$
$$(95 \text{ kPa})(60 \text{ mL}) = P_2(40 \text{ mL})$$
$$P_2 = \frac{95 \times 60}{40} = 142.5 \text{ kPa}$$
Solution:
$$P_1 V_1 = P_2 V_2$$
$$(95 \text{ kPa})(60 \text{ mL}) = P_2(40 \text{ mL})$$
$$P_2 = \frac{95 \times 60}{40} = 142.5 \text{ kPa}$$
Key Exam Notes — Boyle’s Law:
• P₁V₁ = P₂V₂ (T and n must be constant).
• Inverse relationship: P goes up → V goes down.
• Any pressure units work as long as they are the SAME on both sides.
• Graphically: P vs 1/V is a straight line through the origin; P vs V is a hyperbola.
• P₁V₁ = P₂V₂ (T and n must be constant).
• Inverse relationship: P goes up → V goes down.
• Any pressure units work as long as they are the SAME on both sides.
• Graphically: P vs 1/V is a straight line through the origin; P vs V is a hyperbola.
Practice Questions:
1. A gas at 2.5 atm occupies 400 mL. What volume will it occupy at 1.0 atm (constant T)?
2. A sample of gas is compressed from 800 mL to 200 mL. If the original pressure was 1.2 atm, what is the final pressure?
1. A gas at 2.5 atm occupies 400 mL. What volume will it occupy at 1.0 atm (constant T)?
2. A sample of gas is compressed from 800 mL to 200 mL. If the original pressure was 1.2 atm, what is the final pressure?
Answer 1:
$$P_1 V_1 = P_2 V_2$$
$$(2.5)(400) = (1.0)(V_2)$$
$$V_2 = 1000 \text{ mL}$$
Answer 2:
$$(1.2)(800) = P_2(200)$$
$$P_2 = \frac{1.2 \times 800}{200} = 4.8 \text{ atm}$$
$$P_1 V_1 = P_2 V_2$$
$$(2.5)(400) = (1.0)(V_2)$$
$$V_2 = 1000 \text{ mL}$$
Answer 2:
$$(1.2)(800) = P_2(200)$$
$$P_2 = \frac{1.2 \times 800}{200} = 4.8 \text{ atm}$$
B. Charles’s Law (Volume-Temperature Relationship)
In 1787, Jacques Charles discovered that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (in Kelvin).
Charles’s Law: At constant P and n: $$V \propto T \quad \text{or} \quad \frac{V_1}{T_1} = \frac{V_2}{T_2}$$
Warning: Temperature MUST be in Kelvin! If you use Celsius, you will get wrong answers.
Worked Example 3.3: A gas occupies 300 mL at 27 °C. What volume will it occupy at 127 °C if pressure is constant?
Solution:
Convert to Kelvin first!
$$T_1 = 27 + 273 = 300 \text{ K}$$
$$T_2 = 127 + 273 = 400 \text{ K}$$
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
$$\frac{300}{300} = \frac{V_2}{400}$$
$$V_2 = 400 \text{ mL}$$
Solution:
Convert to Kelvin first!
$$T_1 = 27 + 273 = 300 \text{ K}$$
$$T_2 = 127 + 273 = 400 \text{ K}$$
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
$$\frac{300}{300} = \frac{V_2}{400}$$
$$V_2 = 400 \text{ mL}$$
Worked Example 3.4: A balloon has a volume of 2.0 L at 25 °C. It is placed in a freezer at −15 °C. What is its new volume (P constant)?
Solution:
$$T_1 = 25 + 273 = 298 \text{ K}$$
$$T_2 = -15 + 273 = 258 \text{ K}$$
$$\frac{2.0}{298} = \frac{V_2}{258}$$
$$V_2 = \frac{2.0 \times 258}{298} = 1.73 \text{ L}$$
Solution:
$$T_1 = 25 + 273 = 298 \text{ K}$$
$$T_2 = -15 + 273 = 258 \text{ K}$$
$$\frac{2.0}{298} = \frac{V_2}{258}$$
$$V_2 = \frac{2.0 \times 258}{298} = 1.73 \text{ L}$$
Key Exam Notes — Charles’s Law:
• V₁/T₁ = V₂/T₂ (P and n must be constant).
• Direct relationship: T goes up → V goes up.
• Temperature MUST be in Kelvin. Always convert first!
• Graphically: V vs T (in K) is a straight line through the origin. Extrapolating to V = 0 gives T = −273.15 °C (absolute zero).
• V₁/T₁ = V₂/T₂ (P and n must be constant).
• Direct relationship: T goes up → V goes up.
• Temperature MUST be in Kelvin. Always convert first!
• Graphically: V vs T (in K) is a straight line through the origin. Extrapolating to V = 0 gives T = −273.15 °C (absolute zero).
Practice Questions:
1. At 300 K, a gas has a volume of 500 mL. At what temperature (in K and °C) will the volume be 600 mL?
2. Why can’t you use Celsius directly in Charles’s Law?
1. At 300 K, a gas has a volume of 500 mL. At what temperature (in K and °C) will the volume be 600 mL?
2. Why can’t you use Celsius directly in Charles’s Law?
Answer 1:
$$\frac{500}{300} = \frac{600}{T_2}$$
$$T_2 = \frac{600 \times 300}{500} = 360 \text{ K}$$
In °C: 360 − 273 = 87 °C
Answer 2:
If you use Celsius, you would get V/T not being a constant. For example, at 0 °C a gas has some volume V, but V/0 is undefined. The relationship V ∝ T only works with absolute temperature (Kelvin), which starts at absolute zero where particle motion theoretically stops. The Kelvin scale shifts the zero point correctly.
$$\frac{500}{300} = \frac{600}{T_2}$$
$$T_2 = \frac{600 \times 300}{500} = 360 \text{ K}$$
In °C: 360 − 273 = 87 °C
Answer 2:
If you use Celsius, you would get V/T not being a constant. For example, at 0 °C a gas has some volume V, but V/0 is undefined. The relationship V ∝ T only works with absolute temperature (Kelvin), which starts at absolute zero where particle motion theoretically stops. The Kelvin scale shifts the zero point correctly.
C. Gay-Lussac’s Law (Pressure-Temperature Relationship)
At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.
Gay-Lussac’s Law: At constant V and n: $$P \propto T \quad \text{or} \quad \frac{P_1}{T_1} = \frac{P_2}{T_2}$$
This explains why aerosol cans warn you not to incinerate them — heating increases pressure in a fixed volume, which can cause explosion!
Worked Example 3.5: A sealed container of gas has a pressure of 1.0 atm at 27 °C. If heated to 127 °C, what is the new pressure?
Solution:
$$T_1 = 27 + 273 = 300 \text{ K}; \quad T_2 = 127 + 273 = 400 \text{ K}$$
$$\frac{1.0}{300} = \frac{P_2}{400}$$
$$P_2 = \frac{1.0 \times 400}{300} = 1.33 \text{ atm}$$
Solution:
$$T_1 = 27 + 273 = 300 \text{ K}; \quad T_2 = 127 + 273 = 400 \text{ K}$$
$$\frac{1.0}{300} = \frac{P_2}{400}$$
$$P_2 = \frac{1.0 \times 400}{300} = 1.33 \text{ atm}$$
Key Exam Notes — Gay-Lussac’s Law:
• P₁/T₁ = P₂/T₂ (V and n must be constant).
• Direct relationship: T goes up → P goes up.
• Must use Kelvin temperature.
• This law applies to rigid containers (fixed volume).
• P₁/T₁ = P₂/T₂ (V and n must be constant).
• Direct relationship: T goes up → P goes up.
• Must use Kelvin temperature.
• This law applies to rigid containers (fixed volume).
Practice Questions:
1. A gas in a rigid container has a pressure of 2.5 atm at 30 °C. At what temperature will the pressure be 3.0 atm?
1. A gas in a rigid container has a pressure of 2.5 atm at 30 °C. At what temperature will the pressure be 3.0 atm?
Answer:
$$T_1 = 30 + 273 = 303 \text{ K}$$
$$\frac{2.5}{303} = \frac{3.0}{T_2}$$
$$T_2 = \frac{3.0 \times 303}{2.5} = 363.6 \text{ K} = 90.6 °C$$
$$T_1 = 30 + 273 = 303 \text{ K}$$
$$\frac{2.5}{303} = \frac{3.0}{T_2}$$
$$T_2 = \frac{3.0 \times 303}{2.5} = 363.6 \text{ K} = 90.6 °C$$
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D. The Combined Gas Law
What if temperature AND pressure both change? We combine Boyle’s and Charles’s laws:
Combined Gas Law: For a fixed amount of gas: $$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
This is very powerful — it works even if only one variable changes (just keep the others the same on both sides).
Worked Example 3.6: A gas occupies 500 mL at 27 °C and 1.0 atm. What volume does it occupy at 77 °C and 0.8 atm?
Solution:
$$T_1 = 27 + 273 = 300 \text{ K}; \quad T_2 = 77 + 273 = 350 \text{ K}$$
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$\frac{(1.0)(500)}{300} = \frac{(0.8)(V_2)}{350}$$
$$V_2 = \frac{(1.0)(500)(350)}{(300)(0.8)} = 729.2 \text{ mL}$$
Solution:
$$T_1 = 27 + 273 = 300 \text{ K}; \quad T_2 = 77 + 273 = 350 \text{ K}$$
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$\frac{(1.0)(500)}{300} = \frac{(0.8)(V_2)}{350}$$
$$V_2 = \frac{(1.0)(500)(350)}{(300)(0.8)} = 729.2 \text{ mL}$$
Worked Example 3.7: A sample of gas at 45 °C and 1.2 atm occupies a volume of 350 mL. If the volume is compressed to 250 mL and the temperature drops to 25 °C, what is the new pressure?
Solution:
$$T_1 = 45 + 273 = 318 \text{ K}; \quad T_2 = 25 + 273 = 298 \text{ K}$$
$$\frac{(1.2)(350)}{318} = \frac{P_2(250)}{298}$$
$$P_2 = \frac{(1.2)(350)(298)}{(318)(250)} = 1.57 \text{ atm}$$
Solution:
$$T_1 = 45 + 273 = 318 \text{ K}; \quad T_2 = 25 + 273 = 298 \text{ K}$$
$$\frac{(1.2)(350)}{318} = \frac{P_2(250)}{298}$$
$$P_2 = \frac{(1.2)(350)(298)}{(318)(250)} = 1.57 \text{ atm}$$
Key Exam Notes — Combined Gas Law:
• P₁V₁/T₁ = P₂V₂/T₂ (n must be constant).
• Must use Kelvin for temperature.
• Pressure and volume units must match on both sides.
• This equation reduces to Boyle’s, Charles’s, or Gay-Lussac’s law when one variable is held constant.
• P₁V₁/T₁ = P₂V₂/T₂ (n must be constant).
• Must use Kelvin for temperature.
• Pressure and volume units must match on both sides.
• This equation reduces to Boyle’s, Charles’s, or Gay-Lussac’s law when one variable is held constant.
Practice Questions:
1. A gas at 100 °C and 760 mmHg occupies 1.50 L. What volume at 0 °C and 760 mmHg?
2. A 600 mL sample of gas at 25 °C and 1.5 atm is compressed to 400 mL and heated to 50 °C. Find the new pressure.
1. A gas at 100 °C and 760 mmHg occupies 1.50 L. What volume at 0 °C and 760 mmHg?
2. A 600 mL sample of gas at 25 °C and 1.5 atm is compressed to 400 mL and heated to 50 °C. Find the new pressure.
Answer 1:
Since P₁ = P₂ = 760 mmHg, this reduces to Charles’s Law.
$$T_1 = 100 + 273 = 373 \text{ K}; \quad T_2 = 0 + 273 = 273 \text{ K}$$
$$\frac{1.50}{373} = \frac{V_2}{273}$$
$$V_2 = \frac{1.50 \times 273}{373} = 1.10 \text{ L}$$
Answer 2:
$$T_1 = 298 \text{ K}; \quad T_2 = 323 \text{ K}$$
$$\frac{(1.5)(600)}{298} = \frac{P_2(400)}{323}$$
$$P_2 = \frac{(1.5)(600)(323)}{(298)(400)} = 2.44 \text{ atm}$$
Since P₁ = P₂ = 760 mmHg, this reduces to Charles’s Law.
$$T_1 = 100 + 273 = 373 \text{ K}; \quad T_2 = 0 + 273 = 273 \text{ K}$$
$$\frac{1.50}{373} = \frac{V_2}{273}$$
$$V_2 = \frac{1.50 \times 273}{373} = 1.10 \text{ L}$$
Answer 2:
$$T_1 = 298 \text{ K}; \quad T_2 = 323 \text{ K}$$
$$\frac{(1.5)(600)}{298} = \frac{P_2(400)}{323}$$
$$P_2 = \frac{(1.5)(600)(323)}{(298)(400)} = 2.44 \text{ atm}$$
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E. Avogadro’s Law
In 1811, Amedeo Avogadro proposed that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles (molecules).
Avogadro’s Law: At constant T and P: $$V \propto n \quad \text{or} \quad \frac{V_1}{n_1} = \frac{V_2}{n_2}$$
One mole of any ideal gas at STP occupies 22.4 L. This is called the molar volume of a gas.
Worked Example 3.8: What volume does 0.5 mol of gas occupy at STP?
Solution:
At STP, 1 mol = 22.4 L
$$V = 0.5 \times 22.4 = 11.2 \text{ L}$$
Solution:
At STP, 1 mol = 22.4 L
$$V = 0.5 \times 22.4 = 11.2 \text{ L}$$
Worked Example 3.9: If 2.0 mol of a gas occupies 44.8 L at a certain T and P, what volume will 5.0 mol occupy at the same T and P?
Solution:
$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$
$$\frac{44.8}{2.0} = \frac{V_2}{5.0}$$
$$V_2 = \frac{44.8 \times 5.0}{2.0} = 112 \text{ L}$$
Solution:
$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$
$$\frac{44.8}{2.0} = \frac{V_2}{5.0}$$
$$V_2 = \frac{44.8 \times 5.0}{2.0} = 112 \text{ L}$$
Key Exam Notes — Avogadro’s Law:
• V₁/n₁ = V₂/n₂ (T and P must be constant).
• At STP: 1 mol of any gas = 22.4 L.
• STP = 273 K (0 °C) and 1 atm (101.325 kPa).
• This law explains why the molar volume is the same for all gases — it depends only on the number of particles, not their identity.
• V₁/n₁ = V₂/n₂ (T and P must be constant).
• At STP: 1 mol of any gas = 22.4 L.
• STP = 273 K (0 °C) and 1 atm (101.325 kPa).
• This law explains why the molar volume is the same for all gases — it depends only on the number of particles, not their identity.
Practice Questions:
1. How many moles of gas are in 67.2 L at STP?
2. A gas sample occupies 15.0 L and contains 0.75 mol. If 0.50 more mol is added at the same T and P, what is the new volume?
1. How many moles of gas are in 67.2 L at STP?
2. A gas sample occupies 15.0 L and contains 0.75 mol. If 0.50 more mol is added at the same T and P, what is the new volume?
Answer 1:
$$n = \frac{V}{22.4} = \frac{67.2}{22.4} = 3.0 \text{ mol}$$
Answer 2:
New amount = 0.75 + 0.50 = 1.25 mol
$$\frac{15.0}{0.75} = \frac{V_2}{1.25}$$
$$V_2 = \frac{15.0 \times 1.25}{0.75} = 25.0 \text{ L}$$
$$n = \frac{V}{22.4} = \frac{67.2}{22.4} = 3.0 \text{ mol}$$
Answer 2:
New amount = 0.75 + 0.50 = 1.25 mol
$$\frac{15.0}{0.75} = \frac{V_2}{1.25}$$
$$V_2 = \frac{15.0 \times 1.25}{0.75} = 25.0 \text{ L}$$
F. The Ideal Gas Equation
We can combine all four gas laws into one powerful equation — the Ideal Gas Law:
Ideal Gas Equation: $$PV = nRT$$
where:
P = pressure
V = volume
n = number of moles
R = universal gas constant
T = absolute temperature (Kelvin)
Values of R (choose based on your units):
| R Value | P Units | V Units | T Units |
|---|---|---|---|
| 0.0821 L·atm/(mol·K) | atm | L | K |
| 8.314 J/(mol·K) | Pa | m³ | K |
| 8.314 kPa·L/(mol·K) | kPa | L | K |
| 62.36 L·mmHg/(mol·K) | mmHg | L | K |
The most commonly used value in problems is R = 0.0821 L·atm/(mol·K).
Worked Example 3.10: Calculate the volume occupied by 2.5 mol of gas at 3.0 atm and 300 K.
Solution:
$$PV = nRT$$
$$V = \frac{nRT}{P} = \frac{(2.5)(0.0821)(300)}{3.0} = 20.5 \text{ L}$$
Solution:
$$PV = nRT$$
$$V = \frac{nRT}{P} = \frac{(2.5)(0.0821)(300)}{3.0} = 20.5 \text{ L}$$
Worked Example 3.11: What pressure is exerted by 0.10 mol of gas in a 5.0 L container at 25 °C?
Solution:
$$T = 25 + 273 = 298 \text{ K}$$
$$P = \frac{nRT}{V} = \frac{(0.10)(0.0821)(298)}{5.0} = 0.489 \text{ atm}$$
Solution:
$$T = 25 + 273 = 298 \text{ K}$$
$$P = \frac{nRT}{V} = \frac{(0.10)(0.0821)(298)}{5.0} = 0.489 \text{ atm}$$
Worked Example 3.12: A 250 mL container holds a gas at 27 °C and 750 mmHg. How many moles of gas are present?
Solution:
Convert: V = 250 mL = 0.250 L; T = 300 K; Use R = 62.36 L·mmHg/(mol·K)
$$n = \frac{PV}{RT} = \frac{(750)(0.250)}{(62.36)(300)} = 0.0100 \text{ mol}$$
Solution:
Convert: V = 250 mL = 0.250 L; T = 300 K; Use R = 62.36 L·mmHg/(mol·K)
$$n = \frac{PV}{RT} = \frac{(750)(0.250)}{(62.36)(300)} = 0.0100 \text{ mol}$$
Key Exam Notes — Ideal Gas Law:
• PV = nRT — choose R to match your pressure and volume units!
• Most common: R = 0.0821 when P is in atm and V is in L.
• Temperature MUST be in Kelvin.
• If mass (m) is given instead of moles: n = m/M (M = molar mass).
• This equation contains ALL four gas laws within it.
• PV = nRT — choose R to match your pressure and volume units!
• Most common: R = 0.0821 when P is in atm and V is in L.
• Temperature MUST be in Kelvin.
• If mass (m) is given instead of moles: n = m/M (M = molar mass).
• This equation contains ALL four gas laws within it.
Practice Questions:
1. What volume does 4.0 g of O₂ (M = 32 g/mol) occupy at 1.0 atm and 27 °C?
2. A 10.0 L tank contains 0.50 mol of N₂ at 25 °C. What is the pressure in kPa?
1. What volume does 4.0 g of O₂ (M = 32 g/mol) occupy at 1.0 atm and 27 °C?
2. A 10.0 L tank contains 0.50 mol of N₂ at 25 °C. What is the pressure in kPa?
Answer 1:
$$n = \frac{4.0}{32} = 0.125 \text{ mol}; \quad T = 300 \text{ K}$$
$$V = \frac{nRT}{P} = \frac{(0.125)(0.0821)(300)}{1.0} = 3.08 \text{ L}$$
Answer 2:
$$T = 298 \text{ K}; \quad \text{Use } R = 8.314 \text{ kPa·L/(mol·K)}$$
$$P = \frac{nRT}{V} = \frac{(0.50)(8.314)(298)}{10.0} = 123.9 \text{ kPa}$$
$$n = \frac{4.0}{32} = 0.125 \text{ mol}; \quad T = 300 \text{ K}$$
$$V = \frac{nRT}{P} = \frac{(0.125)(0.0821)(300)}{1.0} = 3.08 \text{ L}$$
Answer 2:
$$T = 298 \text{ K}; \quad \text{Use } R = 8.314 \text{ kPa·L/(mol·K)}$$
$$P = \frac{nRT}{V} = \frac{(0.50)(8.314)(298)}{10.0} = 123.9 \text{ kPa}$$
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G. Gas Density and Molar Mass from the Ideal Gas Law
We can rearrange the ideal gas law to find the density of a gas or its molar mass.
Since n = m/M, substituting into PV = nRT:
$$PV = \frac{m}{M}RT \quad \Rightarrow \quad PM = \frac{m}{V}RT = \rho RT$$
So we get two very useful equations:
Density of a gas: $$\rho = \frac{PM}{RT}$$ Molar mass of a gas: $$M = \frac{\rho RT}{P} \quad \text{or} \quad M = \frac{mRT}{PV}$$
Worked Example 3.13: Calculate the density of O₂ (M = 32 g/mol) at STP.
Solution:
At STP: P = 1 atm, T = 273 K, R = 0.0821
$$\rho = \frac{PM}{RT} = \frac{(1)(32)}{(0.0821)(273)} = 1.43 \text{ g/L}$$
Solution:
At STP: P = 1 atm, T = 273 K, R = 0.0821
$$\rho = \frac{PM}{RT} = \frac{(1)(32)}{(0.0821)(273)} = 1.43 \text{ g/L}$$
Worked Example 3.14: A gas has a density of 1.96 g/L at 27 °C and 1.0 atm. Calculate its molar mass and identify the gas.
Solution:
$$T = 300 \text{ K}$$
$$M = \frac{\rho RT}{P} = \frac{(1.96)(0.0821)(300)}{1.0} = 48.3 \text{ g/mol}$$
This molar mass (≈ 48) is close to O₃ (ozone, 48 g/mol).
Solution:
$$T = 300 \text{ K}$$
$$M = \frac{\rho RT}{P} = \frac{(1.96)(0.0821)(300)}{1.0} = 48.3 \text{ g/mol}$$
This molar mass (≈ 48) is close to O₃ (ozone, 48 g/mol).
Worked Example 3.15: 0.582 g of a gas occupies 250 mL at 100 °C and 740 mmHg. Find the molar mass.
Solution:
V = 0.250 L; T = 373 K; R = 62.36 L·mmHg/(mol·K)
$$M = \frac{mRT}{PV} = \frac{(0.582)(62.36)(373)}{(740)(0.250)} = 73.8 \text{ g/mol}$$
Solution:
V = 0.250 L; T = 373 K; R = 62.36 L·mmHg/(mol·K)
$$M = \frac{mRT}{PV} = \frac{(0.582)(62.36)(373)}{(740)(0.250)} = 73.8 \text{ g/mol}$$
Key Exam Notes — Gas Density and Molar Mass:
• ρ = PM/(RT) — heavier gases are denser at the same T and P.
• M = mRT/(PV) — very common exam question type!
• Gas density is usually in g/L (not g/mL or g/cm³).
• At STP, all gases have the same molar volume (22.4 L/mol) but different densities (heavier gases are denser).
• ρ = PM/(RT) — heavier gases are denser at the same T and P.
• M = mRT/(PV) — very common exam question type!
• Gas density is usually in g/L (not g/mL or g/cm³).
• At STP, all gases have the same molar volume (22.4 L/mol) but different densities (heavier gases are denser).
Practice Questions:
1. Which is denser at STP: CO₂ (44 g/mol) or N₂ (28 g/mol)? Calculate both densities.
2. 1.20 g of an unknown gas occupies 500 mL at 27 °C and 1.5 atm. Find the molar mass.
1. Which is denser at STP: CO₂ (44 g/mol) or N₂ (28 g/mol)? Calculate both densities.
2. 1.20 g of an unknown gas occupies 500 mL at 27 °C and 1.5 atm. Find the molar mass.
Answer 1:
CO₂: ρ = (1)(44)/((0.0821)(273)) = 1.96 g/L
N₂: ρ = (1)(28)/((0.0821)(273)) = 1.25 g/L
CO₂ is denser because it has a higher molar mass.
Answer 2:
$$M = \frac{(1.20)(0.0821)(300)}{(1.5)(0.500)} = \frac{29.56}{0.75} = 39.4 \text{ g/mol}$$
CO₂: ρ = (1)(44)/((0.0821)(273)) = 1.96 g/L
N₂: ρ = (1)(28)/((0.0821)(273)) = 1.25 g/L
CO₂ is denser because it has a higher molar mass.
Answer 2:
$$M = \frac{(1.20)(0.0821)(300)}{(1.5)(0.500)} = \frac{29.56}{0.75} = 39.4 \text{ g/mol}$$
H. Dalton’s Law of Partial Pressures
In a mixture of non-reacting gases, each gas exerts its own pressure independently. The partial pressure of a gas in a mixture is the pressure that gas would exert if it alone occupied the entire container.
Dalton’s Law: $$P_{\text{total}} = P_1 + P_2 + P_3 + \cdots$$
The partial pressure of each gas is related to its mole fraction (X):
$$P_i = X_i \times P_{\text{total}} \quad \text{where} \quad X_i = \frac{n_i}{n_{\text{total}}}$$
Worked Example 3.16: A mixture of gases contains 0.5 mol N₂, 0.3 mol O₂, and 0.2 mol He. If the total pressure is 2.0 atm, find the partial pressure of each gas.
Solution:
Total moles = 0.5 + 0.3 + 0.2 = 1.0 mol
$$P_{N_2} = \frac{0.5}{1.0} \times 2.0 = 1.0 \text{ atm}$$
$$P_{O_2} = \frac{0.3}{1.0} \times 2.0 = 0.6 \text{ atm}$$
$$P_{He} = \frac{0.2}{1.0} \times 2.0 = 0.4 \text{ atm}$$
Check: 1.0 + 0.6 + 0.4 = 2.0 atm ✓
Solution:
Total moles = 0.5 + 0.3 + 0.2 = 1.0 mol
$$P_{N_2} = \frac{0.5}{1.0} \times 2.0 = 1.0 \text{ atm}$$
$$P_{O_2} = \frac{0.3}{1.0} \times 2.0 = 0.6 \text{ atm}$$
$$P_{He} = \frac{0.2}{1.0} \times 2.0 = 0.4 \text{ atm}$$
Check: 1.0 + 0.6 + 0.4 = 2.0 atm ✓
Collecting gases over water: When a gas is collected over water, it is mixed with water vapor. The total pressure = pressure of dry gas + vapor pressure of water.
$$P_{\text{dry gas}} = P_{\text{total}} – P_{\text{water vapor}}$$
Worked Example 3.17: Oxygen is collected over water at 25 °C. The total pressure is 755 mmHg. The vapor pressure of water at 25 °C is 23.8 mmHg. What is the partial pressure of the dry O₂?
Solution:
$$P_{O_2} = P_{\text{total}} – P_{H_2O} = 755 – 23.8 = 731.2 \text{ mmHg}$$
Solution:
$$P_{O_2} = P_{\text{total}} – P_{H_2O} = 755 – 23.8 = 731.2 \text{ mmHg}$$
Key Exam Notes — Dalton’s Law:
• P_total = P₁ + P₂ + P₃ + … (gases must not react with each other).
• Partial pressure = mole fraction × total pressure.
• Gas collected over water: P_dry = P_total − P_water vapor.
• Vapor pressure of water increases with temperature (given in tables).
• P_total = P₁ + P₂ + P₃ + … (gases must not react with each other).
• Partial pressure = mole fraction × total pressure.
• Gas collected over water: P_dry = P_total − P_water vapor.
• Vapor pressure of water increases with temperature (given in tables).
Practice Questions:
1. A mixture has 2.0 mol H₂ and 3.0 mol Ar at total pressure 5.0 atm. Find P_H₂ and P_Ar.
2. H₂ gas is collected over water at 30 °C. Total pressure = 780 mmHg. Vapor pressure of water at 30 °C = 31.8 mmHg. Find P_H₂ (dry).
1. A mixture has 2.0 mol H₂ and 3.0 mol Ar at total pressure 5.0 atm. Find P_H₂ and P_Ar.
2. H₂ gas is collected over water at 30 °C. Total pressure = 780 mmHg. Vapor pressure of water at 30 °C = 31.8 mmHg. Find P_H₂ (dry).
Answer 1:
Total moles = 5.0 mol
$$P_{H_2} = \frac{2.0}{5.0} \times 5.0 = 2.0 \text{ atm}$$
$$P_{Ar} = \frac{3.0}{5.0} \times 5.0 = 3.0 \text{ atm}$$
Answer 2:
$$P_{H_2} = 780 – 31.8 = 748.2 \text{ mmHg}$$
Total moles = 5.0 mol
$$P_{H_2} = \frac{2.0}{5.0} \times 5.0 = 2.0 \text{ atm}$$
$$P_{Ar} = \frac{3.0}{5.0} \times 5.0 = 3.0 \text{ atm}$$
Answer 2:
$$P_{H_2} = 780 – 31.8 = 748.2 \text{ mmHg}$$
I. Graham’s Law of Diffusion and Effusion
Diffusion is the gradual mixing of gases. Effusion is the escape of gas molecules through a tiny hole.
Thomas Graham found that the rate of diffusion/effusion of a gas is inversely proportional to the square root of its molar mass:
Graham’s Law: $$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$$
Lighter gases diffuse faster. For example, NH₃ (17 g/mol) diffuses faster than HCl (36.5 g/mol).
Worked Example 3.18: Compare the rates of diffusion of H₂ (M = 2 g/mol) and O₂ (M = 32 g/mol).
Solution:
$$\frac{\text{Rate}_{H_2}}{\text{Rate}_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$
H₂ diffuses 4 times faster than O₂.
Solution:
$$\frac{\text{Rate}_{H_2}}{\text{Rate}_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$
H₂ diffuses 4 times faster than O₂.
Worked Example 3.19: An unknown gas diffuses 1.5 times slower than O₂. What is its molar mass?
Solution:
$$\frac{\text{Rate}_{\text{unknown}}}{\text{Rate}_{O_2}} = \frac{1}{1.5}$$
$$\frac{1}{1.5} = \sqrt{\frac{32}{M_{\text{unknown}}}}$$
$$\left(\frac{1}{1.5}\right)^2 = \frac{32}{M_{\text{unknown}}}$$
$$\frac{1}{2.25} = \frac{32}{M_{\text{unknown}}}$$
$$M_{\text{unknown}} = 32 \times 2.25 = 72 \text{ g/mol}$$
Solution:
$$\frac{\text{Rate}_{\text{unknown}}}{\text{Rate}_{O_2}} = \frac{1}{1.5}$$
$$\frac{1}{1.5} = \sqrt{\frac{32}{M_{\text{unknown}}}}$$
$$\left(\frac{1}{1.5}\right)^2 = \frac{32}{M_{\text{unknown}}}$$
$$\frac{1}{2.25} = \frac{32}{M_{\text{unknown}}}$$
$$M_{\text{unknown}} = 32 \times 2.25 = 72 \text{ g/mol}$$
Key Exam Notes — Graham’s Law:
• Rate₁/Rate₂ = √(M₂/M₁) — lighter gas in denominator → faster rate.
• “Gas A diffuses twice as fast as Gas B” means Rate_A/Rate_B = 2.
• Time and rate are inversely related: if diffusion takes longer, rate is slower.
• Application: separating isotopes (e.g., ²³⁵UF₆ vs ²³⁸UF₆).
• Rate₁/Rate₂ = √(M₂/M₁) — lighter gas in denominator → faster rate.
• “Gas A diffuses twice as fast as Gas B” means Rate_A/Rate_B = 2.
• Time and rate are inversely related: if diffusion takes longer, rate is slower.
• Application: separating isotopes (e.g., ²³⁵UF₆ vs ²³⁸UF₆).
Practice Questions:
1. Which diffuses faster: CO₂ (44 g/mol) or CH₄ (16 g/mol)? By what factor?
2. Gas X takes 60 seconds to effuse. Under the same conditions, O₂ takes 30 seconds. Find the molar mass of X.
1. Which diffuses faster: CO₂ (44 g/mol) or CH₄ (16 g/mol)? By what factor?
2. Gas X takes 60 seconds to effuse. Under the same conditions, O₂ takes 30 seconds. Find the molar mass of X.
Answer 1:
$$\frac{\text{Rate}_{CH_4}}{\text{Rate}_{CO_2}} = \sqrt{\frac{44}{16}} = \sqrt{2.75} = 1.66$$
CH₄ diffuses 1.66 times faster than CO₂.
Answer 2:
Since time is inversely proportional to rate: Rate_X/Rate_O₂ = 30/60 = 0.5
$$0.5 = \sqrt{\frac{32}{M_X}}$$
$$0.25 = \frac{32}{M_X}$$
$$M_X = 128 \text{ g/mol}$$
$$\frac{\text{Rate}_{CH_4}}{\text{Rate}_{CO_2}} = \sqrt{\frac{44}{16}} = \sqrt{2.75} = 1.66$$
CH₄ diffuses 1.66 times faster than CO₂.
Answer 2:
Since time is inversely proportional to rate: Rate_X/Rate_O₂ = 30/60 = 0.5
$$0.5 = \sqrt{\frac{32}{M_X}}$$
$$0.25 = \frac{32}{M_X}$$
$$M_X = 128 \text{ g/mol}$$
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3.4 The Liquid State
Liquids are intermediate between gases and solids. The particles are close together (like solids) but can move past each other (like gases). Let’s study the key properties of liquids.
Properties of Liquids
- Definite volume but no definite shape (takes the shape of the container).
- Slightly compressible — much less than gases but more than solids.
- Fluidity — particles can flow past each other.
- Diffusion — much slower than in gases because particles are closer together.
- Surface tension — the tendency of liquids to minimize their surface area (cohesion between surface molecules).
- Viscosity — resistance to flow. Honey has high viscosity; water has low viscosity. Viscosity decreases with increasing temperature.
- Capillary action — the ability of a liquid to rise in a narrow tube (adhesion to the tube wall vs cohesion within the liquid).
3.4.1 Energy Changes in Liquids
Vapor Pressure
In a closed container, some liquid molecules at the surface gain enough kinetic energy to escape into the gas phase. This process is called evaporation (or vaporization). Some gas molecules also return to the liquid phase (condensation). When the rate of evaporation equals the rate of condensation, a dynamic equilibrium is established. The pressure exerted by the vapor at equilibrium is called the vapor pressure.
Factors affecting vapor pressure:
- Temperature: Higher temperature → more molecules have enough energy to escape → higher vapor pressure.
- Intermolecular forces: Stronger IMFs → fewer molecules can escape → lower vapor pressure. Water (H-bonding) has lower vapor pressure than ether (weak LDF) at the same temperature.
- Type of molecules: Lighter molecules tend to have higher vapor pressure.
Boiling Point
The boiling point of a liquid is the temperature at which its vapor pressure equals the external (atmospheric) pressure. At this point, bubbles of vapor form throughout the liquid, not just at the surface.
The normal boiling point is the boiling point at standard pressure (1 atm = 101.325 kPa = 760 mmHg). For water, the normal boiling point is 100 °C.
At higher altitudes, atmospheric pressure is lower, so liquids boil at lower temperatures. For example, in Addis Ababa (altitude ~2355 m), water boils at about 88–90 °C instead of 100 °C.
Evaporation vs Boiling
| Property | Evaporation | Boiling |
|---|---|---|
| Where it occurs | Only at the surface | Throughout the liquid |
| Temperature | Occurs at any temperature | Occurs at one specific temperature |
| Condition | Some molecules have enough energy | Vapor pressure = external pressure |
| Rate | Slow | Rapid (vigorous) |
| Bubbles | No bubbles | Bubbles form throughout |
Specific Heat Capacity
The specific heat capacity (c) of a substance is the amount of heat energy required to raise the temperature of 1 g of the substance by 1 °C (or 1 K).
$$q = mc\Delta T$$
where q = heat energy (J), m = mass (g), c = specific heat capacity (J/(g·°C)), ΔT = temperature change.
Water has a very high specific heat capacity (4.184 J/(g·°C)), which is why it heats up slowly and cools down slowly — very important for climate regulation!
Heat of Vaporization
The heat of vaporization (ΔH_vap) is the amount of heat energy required to convert 1 mol (or 1 g) of liquid to gas at its boiling point, at constant pressure.
$$q = n \Delta H_{\text{vap}} \quad \text{or} \quad q = m \Delta H_{\text{vap}}$$
Water has a high heat of vaporization (40.7 kJ/mol) because of hydrogen bonding.
Worked Example 3.20: How much heat is required to convert 100 g of water at 25 °C to steam at 100 °C?
(c_water = 4.184 J/(g·°C), ΔH_vap = 2260 J/g for water)
Solution:
Step 1: Heat water from 25 °C to 100 °C:
$$q_1 = mc\Delta T = (100)(4.184)(100 – 25) = 31,380 \text{ J}$$
Step 2: Vaporize water at 100 °C:
$$q_2 = m\Delta H_{\text{vap}} = (100)(2260) = 226,000 \text{ J}$$
Total: $$q = q_1 + q_2 = 31,380 + 226,000 = 257,380 \text{ J} = 257.4 \text{ kJ}$$
(c_water = 4.184 J/(g·°C), ΔH_vap = 2260 J/g for water)
Solution:
Step 1: Heat water from 25 °C to 100 °C:
$$q_1 = mc\Delta T = (100)(4.184)(100 – 25) = 31,380 \text{ J}$$
Step 2: Vaporize water at 100 °C:
$$q_2 = m\Delta H_{\text{vap}} = (100)(2260) = 226,000 \text{ J}$$
Total: $$q = q_1 + q_2 = 31,380 + 226,000 = 257,380 \text{ J} = 257.4 \text{ kJ}$$
Key Exam Notes — Liquid State:
• Vapor pressure increases with temperature and decreases with stronger IMFs.
• Boiling point = temperature where vapor pressure = external pressure.
• Normal boiling point = boiling point at 1 atm.
• High altitude → lower atmospheric pressure → lower boiling point.
• Evaporation: surface only, any temperature. Boiling: throughout, specific temperature.
• q = mcΔT for heating; q = nΔH_vap for vaporization.
• Water has high specific heat and high heat of vaporization (due to H-bonding).
• Vapor pressure increases with temperature and decreases with stronger IMFs.
• Boiling point = temperature where vapor pressure = external pressure.
• Normal boiling point = boiling point at 1 atm.
• High altitude → lower atmospheric pressure → lower boiling point.
• Evaporation: surface only, any temperature. Boiling: throughout, specific temperature.
• q = mcΔT for heating; q = nΔH_vap for vaporization.
• Water has high specific heat and high heat of vaporization (due to H-bonding).
Practice Questions:
1. Why does water have a lower vapor pressure than ethanol (CH₃CH₂OH) at the same temperature?
2. How much heat is needed to raise the temperature of 200 g of water from 20 °C to 80 °C? (c = 4.184 J/(g·°C))
1. Why does water have a lower vapor pressure than ethanol (CH₃CH₂OH) at the same temperature?
2. How much heat is needed to raise the temperature of 200 g of water from 20 °C to 80 °C? (c = 4.184 J/(g·°C))
Answer 1:
Water molecules are connected by hydrogen bonding (H bonded to O), which is a very strong intermolecular force. Ethanol has hydrogen bonding too, but also has a larger nonpolar group (CH₃CH₂—) that weakens the overall intermolecular attraction compared to water. Additionally, water can form up to 4 H-bonds per molecule (2 as donor, 2 as acceptor), while ethanol can form fewer. The stronger IMFs in water mean fewer molecules have enough energy to escape to the vapor phase, resulting in lower vapor pressure.
Answer 2:
$$q = mc\Delta T = (200)(4.184)(80 – 20) = (200)(4.184)(60) = 50,208 \text{ J} = 50.2 \text{ kJ}$$
Water molecules are connected by hydrogen bonding (H bonded to O), which is a very strong intermolecular force. Ethanol has hydrogen bonding too, but also has a larger nonpolar group (CH₃CH₂—) that weakens the overall intermolecular attraction compared to water. Additionally, water can form up to 4 H-bonds per molecule (2 as donor, 2 as acceptor), while ethanol can form fewer. The stronger IMFs in water mean fewer molecules have enough energy to escape to the vapor phase, resulting in lower vapor pressure.
Answer 2:
$$q = mc\Delta T = (200)(4.184)(80 – 20) = (200)(4.184)(60) = 50,208 \text{ J} = 50.2 \text{ kJ}$$
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3.5 The Solid State
In a solid, particles are tightly packed in fixed positions and can only vibrate. Solids have definite shape and definite volume.
Types of Solids
| Type | Particles | Bonding | Melting Point | Examples |
|---|---|---|---|---|
| Ionic | Ions | Electrostatic | High | NaCl, MgO |
| Covalent network | Atoms | Covalent bonds | Very high | Diamond, SiO₂ |
| Molecular | Molecules | IMFs | Low | Ice, dry ice, I₂ |
| Metallic | Metal cations + e⁻ | Metallic bonds | Variable | Fe, Cu, Na |
| Amorphous | Various | Various | No sharp m.p. | Glass, rubber, plastic |
Crystalline vs Amorphous Solids
Crystalline solids have a regular, repeating arrangement of particles (crystal lattice). They have sharp melting points. Examples: NaCl, diamond, ice.
Amorphous solids lack an ordered arrangement. They do not have a sharp melting point — they gradually soften over a temperature range. Examples: glass, rubber, plastics.
Unit Cells and Crystal Lattices
A crystal lattice is a 3D arrangement of particles. The smallest repeating unit that shows the full symmetry is called a unit cell.
Types of unit cells (cubic systems):
- Simple cubic (SC): Particles only at the 8 corners. 1 atom per unit cell (8 × ⅛).
- Body-centered cubic (BCC): 8 corners + 1 in the center. 2 atoms per unit cell (8 × ⅛ + 1).
- Face-centered cubic (FCC): 8 corners + 6 face centers. 4 atoms per unit cell (8 × ⅛ + 6 × ½).
Worked Example 3.21: Calculate the number of atoms per unit cell in FCC and BCC.
Solution:
FCC: Corner atoms: 8 × ⅛ = 1 atom. Face atoms: 6 × ½ = 3 atoms. Total = 4 atoms.
BCC: Corner atoms: 8 × ⅛ = 1 atom. Body center: 1 × 1 = 1 atom. Total = 2 atoms.
Solution:
FCC: Corner atoms: 8 × ⅛ = 1 atom. Face atoms: 6 × ½ = 3 atoms. Total = 4 atoms.
BCC: Corner atoms: 8 × ⅛ = 1 atom. Body center: 1 × 1 = 1 atom. Total = 2 atoms.
Key Exam Notes — Solid State:
• Corner atom contributes ⅛; face atom contributes ½; edge atom contributes ¼; body atom contributes 1.
• SC = 1 atom; BCC = 2 atoms; FCC = 4 atoms per unit cell.
• Crystalline solids have sharp melting points; amorphous solids soften gradually.
• Melting point order: covalent network > ionic > metallic > molecular (generally).
• The type of bonding determines the properties of the solid.
• Corner atom contributes ⅛; face atom contributes ½; edge atom contributes ¼; body atom contributes 1.
• SC = 1 atom; BCC = 2 atoms; FCC = 4 atoms per unit cell.
• Crystalline solids have sharp melting points; amorphous solids soften gradually.
• Melting point order: covalent network > ionic > metallic > molecular (generally).
• The type of bonding determines the properties of the solid.
Practice Questions:
1. How many atoms per unit cell in a simple cubic structure?
2. Why does diamond have a much higher melting point than sodium chloride?
1. How many atoms per unit cell in a simple cubic structure?
2. Why does diamond have a much higher melting point than sodium chloride?
Answer 1:
Simple cubic: 8 corners × ⅛ = 1 atom per unit cell.
Answer 2:
Diamond is a covalent network solid — every C atom is bonded to 4 others by strong covalent bonds in a continuous 3D network. Melting requires breaking ALL these covalent bonds. NaCl is ionic — to melt it, you only need to overcome the electrostatic attraction between Na⁺ and Cl⁻ ions. Covalent bonds are much stronger than ionic attractions, so diamond has a much higher melting point (~3550 °C vs 801 °C for NaCl).
Simple cubic: 8 corners × ⅛ = 1 atom per unit cell.
Answer 2:
Diamond is a covalent network solid — every C atom is bonded to 4 others by strong covalent bonds in a continuous 3D network. Melting requires breaking ALL these covalent bonds. NaCl is ionic — to melt it, you only need to overcome the electrostatic attraction between Na⁺ and Cl⁻ ions. Covalent bonds are much stronger than ionic attractions, so diamond has a much higher melting point (~3550 °C vs 801 °C for NaCl).
Quick Revision Notes — Exam Focus
1. Important Definitions
• Kinetic theory: All matter consists of particles in continuous random motion.
• Ideal gas: A gas that perfectly follows all five postulates of KMT (no intermolecular forces, negligible particle volume, elastic collisions).
• STP: Standard Temperature and Pressure = 273 K (0 °C) and 1 atm (101.325 kPa).
• Molar volume: Volume of 1 mol of gas at STP = 22.4 L.
• Vapor pressure: Pressure exerted by vapor in equilibrium with its liquid.
• Boiling point: Temperature at which vapor pressure equals external pressure.
• Normal boiling point: Boiling point at 1 atm.
• Specific heat capacity: Heat needed to raise 1 g of substance by 1 °C.
• Heat of vaporization: Heat needed to convert liquid to gas at constant T and P.
• Partial pressure: Pressure a gas in a mixture would exert if alone.
• Effusion: Escape of gas through a tiny hole.
• Diffusion: Gradual mixing of gases.
• Ideal gas: A gas that perfectly follows all five postulates of KMT (no intermolecular forces, negligible particle volume, elastic collisions).
• STP: Standard Temperature and Pressure = 273 K (0 °C) and 1 atm (101.325 kPa).
• Molar volume: Volume of 1 mol of gas at STP = 22.4 L.
• Vapor pressure: Pressure exerted by vapor in equilibrium with its liquid.
• Boiling point: Temperature at which vapor pressure equals external pressure.
• Normal boiling point: Boiling point at 1 atm.
• Specific heat capacity: Heat needed to raise 1 g of substance by 1 °C.
• Heat of vaporization: Heat needed to convert liquid to gas at constant T and P.
• Partial pressure: Pressure a gas in a mixture would exert if alone.
• Effusion: Escape of gas through a tiny hole.
• Diffusion: Gradual mixing of gases.
2. All Gas Law Formulas
$$P_1 V_1 = P_2 V_2 \quad \text{(Boyle’s, constant T and n)}$$ $$\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(Charles’s, constant P and n)}$$ $$\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \text{(Gay-Lussac’s, constant V and n)}$$ $$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \quad \text{(Combined, constant n)}$$ $$\frac{V_1}{n_1} = \frac{V_2}{n_2} \quad \text{(Avogadro’s, constant T and P)}$$ $$PV = nRT \quad \text{(Ideal Gas Law)}$$ $$\rho = \frac{PM}{RT} \quad \text{(Gas density)}$$ $$M = \frac{mRT}{PV} \quad \text{(Molar mass)}$$ $$P_{\text{total}} = P_1 + P_2 + P_3 + \cdots \quad \text{(Dalton’s Law)}$$ $$P_i = X_i \times P_{\text{total}} \quad \text{(Partial pressure)}$$ $$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \quad \text{(Graham’s Law)}$$ $$q = mc\Delta T \quad \text{(Heating)}$$ $$q = n\Delta H_{\text{vap}} \quad \text{(Vaporization)}$$
3. Values of R
R = 0.0821 L·atm/(mol·K) [P in atm, V in L]
R = 8.314 J/(mol·K) [P in Pa, V in m³]
R = 8.314 kPa·L/(mol·K) [P in kPa, V in L]
R = 62.36 L·mmHg/(mol·K) [P in mmHg, V in L]
4. Unit Conversions
1 atm = 101.325 kPa = 760 mmHg = 760 torr
T(K) = T(°C) + 273.15
1 L = 1000 mL = 0.001 m³
1 mol at STP = 22.4 L
5. Unit Cell Summary
Type Corners Faces Body Atoms/Cell
SC 8×⅛ — — 1
BCC 8×⅛ — 1 2
FCC 8×⅛ 6×½ — 4
6. State Comparison Quick Guide
| Solid | Liquid | Gas | |
|---|---|---|---|
| Shape | Fixed | Variable | Variable |
| Volume | Fixed | Fixed | Variable |
| Compressibility | Nearly 0 | Slight | High |
| Particle arrangement | Ordered | Disordered | Random, far apart |
| Particle motion | Vibrate | Slide past | Free movement |
| IMF strength | Very strong | Moderate | Very weak |
7. Common Mistakes to Avoid
❌ Using Celsius instead of Kelvin in gas law calculations.
❌ Forgetting to convert mL to L (divide by 1000) when using R = 0.0821.
❌ Using the wrong value of R for the given pressure units.
❌ Confusing diffusion rate with diffusion time (they are inversely related).
❌ Thinking boiling and evaporation are the same process.
❌ Confusing vapor pressure with total pressure in gas-over-water problems.
❌ Forgetting that all gas laws (except ideal gas law) require constant amount of gas (n).
❌ Putting the heavier molar mass in the numerator in Graham’s Law (it should be in the denominator under the square root for the faster gas).
❌ Forgetting to add/subtract 273.15 when converting between °C and K.
❌ Confusing boiling point (equilibrium property) with evaporation (surface process).
❌ Forgetting to convert mL to L (divide by 1000) when using R = 0.0821.
❌ Using the wrong value of R for the given pressure units.
❌ Confusing diffusion rate with diffusion time (they are inversely related).
❌ Thinking boiling and evaporation are the same process.
❌ Confusing vapor pressure with total pressure in gas-over-water problems.
❌ Forgetting that all gas laws (except ideal gas law) require constant amount of gas (n).
❌ Putting the heavier molar mass in the numerator in Graham’s Law (it should be in the denominator under the square root for the faster gas).
❌ Forgetting to add/subtract 273.15 when converting between °C and K.
❌ Confusing boiling point (equilibrium property) with evaporation (surface process).
Challenge Exam Questions
Test yourself thoroughly! Try each question before checking the answer.
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Section A: Multiple Choice Questions
Q1. A gas at 27 °C is heated to 127 °C at constant pressure. Its volume will become:
A) 4 times the original B) 2 times the original C) 1.33 times the original D) 0.75 times the original
A) 4 times the original B) 2 times the original C) 1.33 times the original D) 0.75 times the original
Answer: C
$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow \frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{400}{300} = 1.33$$
The volume becomes 1.33 times the original.
$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow \frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{400}{300} = 1.33$$
The volume becomes 1.33 times the original.
Q2. At STP, 11.2 L of a gas has a mass of 14 g. The molar mass of the gas is:
A) 14 g/mol B) 28 g/mol C) 32 g/mol D) 44 g/mol
A) 14 g/mol B) 28 g/mol C) 32 g/mol D) 44 g/mol
Answer: B
At STP, 22.4 L = 1 mol. So 11.2 L = 0.5 mol.
M = mass/n = 14/0.5 = 28 g/mol (N₂).
At STP, 22.4 L = 1 mol. So 11.2 L = 0.5 mol.
M = mass/n = 14/0.5 = 28 g/mol (N₂).
Q3. Which gas will diffuse fastest under identical conditions?
A) O₂ (32 g/mol) B) N₂ (28 g/mol) C) CO₂ (44 g/mol) D) NH₃ (17 g/mol)
A) O₂ (32 g/mol) B) N₂ (28 g/mol) C) CO₂ (44 g/mol) D) NH₃ (17 g/mol)
Answer: D
According to Graham’s Law, rate ∝ 1/√M. The gas with the smallest molar mass diffuses fastest. NH₃ (17 g/mol) has the smallest molar mass, so it diffuses fastest.
According to Graham’s Law, rate ∝ 1/√M. The gas with the smallest molar mass diffuses fastest. NH₃ (17 g/mol) has the smallest molar mass, so it diffuses fastest.
Q4. The vapor pressure of a liquid increases when:
A) Intermolecular forces increase B) Surface area decreases C) Temperature increases D) Atmospheric pressure increases
A) Intermolecular forces increase B) Surface area decreases C) Temperature increases D) Atmospheric pressure increases
Answer: C
Vapor pressure increases with temperature because more molecules have sufficient kinetic energy to escape the liquid surface. Stronger IMFs would decrease vapor pressure (option A is wrong). Surface area and atmospheric pressure do not affect the equilibrium vapor pressure.
Vapor pressure increases with temperature because more molecules have sufficient kinetic energy to escape the liquid surface. Stronger IMFs would decrease vapor pressure (option A is wrong). Surface area and atmospheric pressure do not affect the equilibrium vapor pressure.
Q5. An FCC unit cell contains:
A) 1 atom B) 2 atoms C) 4 atoms D) 6 atoms
A) 1 atom B) 2 atoms C) 4 atoms D) 6 atoms
Answer: C
FCC: 8 corners × ⅛ = 1 atom + 6 faces × ½ = 3 atoms. Total = 4 atoms per unit cell.
FCC: 8 corners × ⅛ = 1 atom + 6 faces × ½ = 3 atoms. Total = 4 atoms per unit cell.
Section B: Fill in the Blanks
Q6. The volume occupied by 0.25 mol of any gas at STP is ________ L.
Answer: 5.6 L. At STP, 1 mol = 22.4 L. So 0.25 × 22.4 = 5.6 L.
Q7. Real gases deviate most from ideal behavior at ________ temperature and ________ pressure.
Answer: Low temperature and high pressure. At low T, particles move slowly and intermolecular forces become significant. At high P, particles are forced close together, making their volume significant and intermolecular forces noticeable. Both conditions violate KMT postulates.
Q8. The normal boiling point of water is ________ °C, which corresponds to a vapor pressure of ________ atm.
Answer: 100 °C and 1 atm. The normal boiling point is defined as the temperature at which vapor pressure equals standard atmospheric pressure (1 atm).
Q9. According to the kinetic molecular theory, the average kinetic energy of gas molecules is proportional to the ________ temperature.
Answer: Absolute (Kelvin). KE_avg = 3/2 kT, where T must be in Kelvin. At 0 K, kinetic energy is zero.
Q10. A gas collected over water at 20 °C has a total pressure of 755 mmHg. If the vapor pressure of water at 20 °C is 17.5 mmHg, the partial pressure of the dry gas is ________ mmHg.
Answer: 737.5 mmHg. P_dry = P_total − P_water = 755 − 17.5 = 737.5 mmHg.
Section C: Short Answer Questions
Q11. Explain why a pressure cooker cooks food faster than an open pot.
Answer:
A pressure cooker is a sealed container that traps steam, increasing the pressure inside. Since boiling point is the temperature at which vapor pressure equals external pressure, higher pressure inside the cooker means water boils at a temperature above 100 °C (typically 110–120 °C). At this higher temperature, food cooks faster because chemical reactions (including those that break down food) proceed faster at higher temperatures.
A pressure cooker is a sealed container that traps steam, increasing the pressure inside. Since boiling point is the temperature at which vapor pressure equals external pressure, higher pressure inside the cooker means water boils at a temperature above 100 °C (typically 110–120 °C). At this higher temperature, food cooks faster because chemical reactions (including those that break down food) proceed faster at higher temperatures.
Q12. Distinguish between evaporation and boiling. Why does evaporation cause cooling?
Answer:
Evaporation occurs only at the surface, at any temperature. Boiling occurs throughout the liquid, at a specific temperature (where vapor pressure = external pressure).
Why evaporation causes cooling: Only the most energetic (fastest) molecules at the surface have enough energy to escape the liquid. When these high-energy molecules leave, the average kinetic energy of the remaining molecules decreases. Since temperature is proportional to average kinetic energy, the temperature of the remaining liquid drops — this is evaporative cooling.
Evaporation occurs only at the surface, at any temperature. Boiling occurs throughout the liquid, at a specific temperature (where vapor pressure = external pressure).
Why evaporation causes cooling: Only the most energetic (fastest) molecules at the surface have enough energy to escape the liquid. When these high-energy molecules leave, the average kinetic energy of the remaining molecules decreases. Since temperature is proportional to average kinetic energy, the temperature of the remaining liquid drops — this is evaporative cooling.
Q13. Why does ethyl ether (C₂H₅OC₂H₅) have a higher vapor pressure than water at the same temperature, even though ether has a higher molar mass?
Answer:
Vapor pressure depends primarily on the strength of intermolecular forces, not molar mass. Water molecules are connected by strong hydrogen bonding (H—O bonds), which holds molecules tightly in the liquid phase. Ether cannot form hydrogen bonds — its intermolecular forces are only dipole-dipole and London dispersion forces, which are much weaker than H-bonds. Weaker IMFs mean more molecules can escape to the vapor phase, resulting in higher vapor pressure for ether despite its larger mass.
Vapor pressure depends primarily on the strength of intermolecular forces, not molar mass. Water molecules are connected by strong hydrogen bonding (H—O bonds), which holds molecules tightly in the liquid phase. Ether cannot form hydrogen bonds — its intermolecular forces are only dipole-dipole and London dispersion forces, which are much weaker than H-bonds. Weaker IMFs mean more molecules can escape to the vapor phase, resulting in higher vapor pressure for ether despite its larger mass.
Section D: Calculation Questions
Q14. A sample of gas at 25 °C occupies 400 mL at 1.2 atm. It is heated to 80 °C and compressed to 300 mL. Calculate the new pressure.
Answer:
$$T_1 = 25 + 273 = 298 \text{ K}; \quad T_2 = 80 + 273 = 353 \text{ K}$$
Using the combined gas law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$\frac{(1.2)(400)}{298} = \frac{P_2(300)}{353}$$
$$P_2 = \frac{(1.2)(400)(353)}{(298)(300)} = \frac{169,440}{89,400} = 1.90 \text{ atm}$$
$$T_1 = 25 + 273 = 298 \text{ K}; \quad T_2 = 80 + 273 = 353 \text{ K}$$
Using the combined gas law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$\frac{(1.2)(400)}{298} = \frac{P_2(300)}{353}$$
$$P_2 = \frac{(1.2)(400)(353)}{(298)(300)} = \frac{169,440}{89,400} = 1.90 \text{ atm}$$
Q15. 0.350 g of a volatile liquid is vaporized and found to occupy 125 mL at 99 °C and 748 mmHg. Calculate the molar mass of the liquid.
Answer:
V = 0.125 L; T = 99 + 273 = 372 K; R = 62.36 L·mmHg/(mol·K)
$$M = \frac{mRT}{PV} = \frac{(0.350)(62.36)(372)}{(748)(0.125)}$$
$$= \frac{8,123.2}{93.5} = 86.9 \text{ g/mol}$$
V = 0.125 L; T = 99 + 273 = 372 K; R = 62.36 L·mmHg/(mol·K)
$$M = \frac{mRT}{PV} = \frac{(0.350)(62.36)(372)}{(748)(0.125)}$$
$$= \frac{8,123.2}{93.5} = 86.9 \text{ g/mol}$$
Q16. A mixture contains 0.40 mol N₂, 0.30 mol O₂, and 0.10 mol CO₂ in a 10.0 L container at 27 °C. Calculate:
(a) Total moles of gas
(b) Total pressure
(c) Partial pressure of each gas
(d) Mole fraction of O₂
(a) Total moles of gas
(b) Total pressure
(c) Partial pressure of each gas
(d) Mole fraction of O₂
Answer:
(a) n_total = 0.40 + 0.30 + 0.10 = 0.80 mol
(b) T = 300 K; PV = nRT
$$P_{\text{total}} = \frac{(0.80)(0.0821)(300)}{10.0} = \frac{19.704}{10.0} = 1.97 \text{ atm}$$
(c) $$P_{N_2} = \frac{0.40}{0.80} \times 1.97 = 0.985 \text{ atm}$$
$$P_{O_2} = \frac{0.30}{0.80} \times 1.97 = 0.739 \text{ atm}$$
$$P_{CO_2} = \frac{0.10}{0.80} \times 1.97 = 0.246 \text{ atm}$$
Check: 0.985 + 0.739 + 0.246 = 1.97 ✓
(d) $$X_{O_2} = \frac{0.30}{0.80} = 0.375$$
(a) n_total = 0.40 + 0.30 + 0.10 = 0.80 mol
(b) T = 300 K; PV = nRT
$$P_{\text{total}} = \frac{(0.80)(0.0821)(300)}{10.0} = \frac{19.704}{10.0} = 1.97 \text{ atm}$$
(c) $$P_{N_2} = \frac{0.40}{0.80} \times 1.97 = 0.985 \text{ atm}$$
$$P_{O_2} = \frac{0.30}{0.80} \times 1.97 = 0.739 \text{ atm}$$
$$P_{CO_2} = \frac{0.10}{0.80} \times 1.97 = 0.246 \text{ atm}$$
Check: 0.985 + 0.739 + 0.246 = 1.97 ✓
(d) $$X_{O_2} = \frac{0.30}{0.80} = 0.375$$
Q17. Gas A (molar mass 64 g/mol) and Gas B (molar mass 16 g/mol) are allowed to diffuse from opposite ends of a tube. Where will the gases meet? (The tube is 100 cm long.)
Answer:
$$\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{16}{64}} = \sqrt{0.25} = 0.5$$
Gas A diffuses at half the rate of Gas B. In the same time, Gas B travels twice as far as Gas A.
If Gas A travels distance d, Gas B travels 2d.
Since the tube is 100 cm: d + 2d = 100 → 3d = 100 → d = 33.3 cm
The gases meet 33.3 cm from the end where Gas A started (or 66.7 cm from where Gas B started).
$$\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{16}{64}} = \sqrt{0.25} = 0.5$$
Gas A diffuses at half the rate of Gas B. In the same time, Gas B travels twice as far as Gas A.
If Gas A travels distance d, Gas B travels 2d.
Since the tube is 100 cm: d + 2d = 100 → 3d = 100 → d = 33.3 cm
The gases meet 33.3 cm from the end where Gas A started (or 66.7 cm from where Gas B started).
Q18. How much energy is required to convert 50.0 g of ice at −10 °C to steam at 110 °C?
Given: c_ice = 2.09 J/(g·°C), ΔH_fus = 334 J/g, c_water = 4.184 J/(g·°C), ΔH_vap = 2260 J/g, c_steam = 2.01 J/(g·°C).
Given: c_ice = 2.09 J/(g·°C), ΔH_fus = 334 J/g, c_water = 4.184 J/(g·°C), ΔH_vap = 2260 J/g, c_steam = 2.01 J/(g·°C).
Answer:
This requires 5 steps:
Step 1: Heat ice from −10 °C to 0 °C:
$$q_1 = mc\Delta T = (50.0)(2.09)(0 – (-10)) = (50.0)(2.09)(10) = 1,045 \text{ J}$$
Step 2: Melt ice at 0 °C:
$$q_2 = m\Delta H_{\text{fus}} = (50.0)(334) = 16,700 \text{ J}$$
Step 3: Heat water from 0 °C to 100 °C:
$$q_3 = mc\Delta T = (50.0)(4.184)(100) = 20,920 \text{ J}$$
Step 4: Vaporize water at 100 °C:
$$q_4 = m\Delta H_{\text{vap}} = (50.0)(2260) = 113,000 \text{ J}$$
Step 5: Heat steam from 100 °C to 110 °C:
$$q_5 = mc\Delta T = (50.0)(2.01)(10) = 1,005 \text{ J}$$
$$q_{\text{total}} = 1,045 + 16,700 + 20,920 + 113,000 + 1,005 = 152,670 \text{ J} = 152.7 \text{ kJ}$$
This requires 5 steps:
Step 1: Heat ice from −10 °C to 0 °C:
$$q_1 = mc\Delta T = (50.0)(2.09)(0 – (-10)) = (50.0)(2.09)(10) = 1,045 \text{ J}$$
Step 2: Melt ice at 0 °C:
$$q_2 = m\Delta H_{\text{fus}} = (50.0)(334) = 16,700 \text{ J}$$
Step 3: Heat water from 0 °C to 100 °C:
$$q_3 = mc\Delta T = (50.0)(4.184)(100) = 20,920 \text{ J}$$
Step 4: Vaporize water at 100 °C:
$$q_4 = m\Delta H_{\text{vap}} = (50.0)(2260) = 113,000 \text{ J}$$
Step 5: Heat steam from 100 °C to 110 °C:
$$q_5 = mc\Delta T = (50.0)(2.01)(10) = 1,005 \text{ J}$$
$$q_{\text{total}} = 1,045 + 16,700 + 20,920 + 113,000 + 1,005 = 152,670 \text{ J} = 152.7 \text{ kJ}$$
Q19. A 5.0 L container holds 3.0 mol of gas at 300 K. If 2.0 mol of gas are removed and the temperature is increased to 400 K, what is the new volume if the pressure remains constant?
Answer:
Method 1: Using Avogadro’s Law then Charles’s Law:
Step 1 (Avogadro’s, constant T and P):
$$\frac{V_1}{n_1} = \frac{V_{\text{temp}}}{n_2} \Rightarrow \frac{5.0}{3.0} = \frac{V_{\text{temp}}}{1.0} \Rightarrow V_{\text{temp}} = 1.67 \text{ L}$$
Step 2 (Charles’s, constant n and P):
$$\frac{1.67}{300} = \frac{V_2}{400} \Rightarrow V_2 = \frac{1.67 \times 400}{300} = 2.22 \text{ L}$$
Method 2: Combined approach using PV = nRT at constant P:
$$\frac{V_1}{n_1 T_1} = \frac{V_2}{n_2 T_2}$$
$$\frac{5.0}{(3.0)(300)} = \frac{V_2}{(1.0)(400)}$$
$$V_2 = \frac{5.0 \times 400}{3.0 \times 300} = \frac{2000}{900} = 2.22 \text{ L}$$
Method 1: Using Avogadro’s Law then Charles’s Law:
Step 1 (Avogadro’s, constant T and P):
$$\frac{V_1}{n_1} = \frac{V_{\text{temp}}}{n_2} \Rightarrow \frac{5.0}{3.0} = \frac{V_{\text{temp}}}{1.0} \Rightarrow V_{\text{temp}} = 1.67 \text{ L}$$
Step 2 (Charles’s, constant n and P):
$$\frac{1.67}{300} = \frac{V_2}{400} \Rightarrow V_2 = \frac{1.67 \times 400}{300} = 2.22 \text{ L}$$
Method 2: Combined approach using PV = nRT at constant P:
$$\frac{V_1}{n_1 T_1} = \frac{V_2}{n_2 T_2}$$
$$\frac{5.0}{(3.0)(300)} = \frac{V_2}{(1.0)(400)}$$
$$V_2 = \frac{5.0 \times 400}{3.0 \times 300} = \frac{2000}{900} = 2.22 \text{ L}$$
Q20. Two flasks are connected by a valve. Flask A contains He gas at 2.0 atm in 3.0 L. Flask B contains Ne gas at 4.0 atm in 6.0 L. Both are at the same temperature. If the valve is opened, what is the total pressure? What are the partial pressures of He and Ne?
Answer:
When the valve opens, each gas expands to fill the total volume (3.0 + 6.0 = 9.0 L). Since T is constant, use Boyle’s Law for each gas separately:
For He: $$P_1 V_1 = P_2 V_2 \Rightarrow (2.0)(3.0) = P_{He}(9.0) \Rightarrow P_{He} = 0.667 \text{ atm}$$
For Ne: $$(4.0)(6.0) = P_{Ne}(9.0) \Rightarrow P_{Ne} = 2.667 \text{ atm}$$
Total pressure: $$P_{\text{total}} = 0.667 + 2.667 = 3.33 \text{ atm}$$
Partial pressures: He = 0.67 atm, Ne = 2.67 atm.
This works because each gas behaves independently (Dalton’s Law).
When the valve opens, each gas expands to fill the total volume (3.0 + 6.0 = 9.0 L). Since T is constant, use Boyle’s Law for each gas separately:
For He: $$P_1 V_1 = P_2 V_2 \Rightarrow (2.0)(3.0) = P_{He}(9.0) \Rightarrow P_{He} = 0.667 \text{ atm}$$
For Ne: $$(4.0)(6.0) = P_{Ne}(9.0) \Rightarrow P_{Ne} = 2.667 \text{ atm}$$
Total pressure: $$P_{\text{total}} = 0.667 + 2.667 = 3.33 \text{ atm}$$
Partial pressures: He = 0.67 atm, Ne = 2.67 atm.
This works because each gas behaves independently (Dalton’s Law).
Q21. The density of a gas is 2.86 g/L at 27 °C and 1.5 atm. Calculate the molar mass. Identify the gas if it is one of: N₂ (28), O₂ (32), CO₂ (44), or Cl₂ (71).
Answer:
$$T = 27 + 273 = 300 \text{ K}$$
$$M = \frac{\rho RT}{P} = \frac{(2.86)(0.0821)(300)}{1.5} = \frac{70.44}{1.5} = 46.96 \text{ g/mol}$$
The molar mass ≈ 47 g/mol. This is closest to NO₂ (46 g/mol) but among the given options, none match exactly. If we check CO₂ (44 g/mol), it’s closest but not perfect. With rounding, this could be a measurement of CO₂ with slight experimental error. Among the given choices, CO₂ (44 g/mol) is the closest.
$$T = 27 + 273 = 300 \text{ K}$$
$$M = \frac{\rho RT}{P} = \frac{(2.86)(0.0821)(300)}{1.5} = \frac{70.44}{1.5} = 46.96 \text{ g/mol}$$
The molar mass ≈ 47 g/mol. This is closest to NO₂ (46 g/mol) but among the given options, none match exactly. If we check CO₂ (44 g/mol), it’s closest but not perfect. With rounding, this could be a measurement of CO₂ with slight experimental error. Among the given choices, CO₂ (44 g/mol) is the closest.
Q22. A student collects 245 mL of oxygen gas over water at 25 °C and 752 mmHg total pressure. The vapor pressure of water at 25 °C is 23.8 mmHg.
(a) Calculate the partial pressure of dry O₂.
(b) Calculate the moles of dry O₂ collected.
(c) Calculate the mass of O₂ collected.
(a) Calculate the partial pressure of dry O₂.
(b) Calculate the moles of dry O₂ collected.
(c) Calculate the mass of O₂ collected.
Answer:
(a) $$P_{O_2} = 752 – 23.8 = 728.2 \text{ mmHg}$$
(b) V = 0.245 L; T = 298 K; R = 62.36 L·mmHg/(mol·K)
$$n = \frac{PV}{RT} = \frac{(728.2)(0.245)}{(62.36)(298)} = \frac{178.4}{18,583.3} = 0.00960 \text{ mol}$$
(c) M(O₂) = 32 g/mol
$$m = n \times M = 0.00960 \times 32 = 0.307 \text{ g}$$
(a) $$P_{O_2} = 752 – 23.8 = 728.2 \text{ mmHg}$$
(b) V = 0.245 L; T = 298 K; R = 62.36 L·mmHg/(mol·K)
$$n = \frac{PV}{RT} = \frac{(728.2)(0.245)}{(62.36)(298)} = \frac{178.4}{18,583.3} = 0.00960 \text{ mol}$$
(c) M(O₂) = 32 g/mol
$$m = n \times M = 0.00960 \times 32 = 0.307 \text{ g}$$