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Chemistry

Dec 5, 2025

55

18 pages

Exploring Gas Laws and Stoichiometry Insights

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Vanessa Vasquez @van3ssa_08

Gas laws describe how gases behave when temperature, pressure, volume, and amount of gas change. These physical relationships... Show more

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Unit 9 Gas Laws Overview

This unit explores the fundamental relationships between gases and their physical properties. You'll discover how gas particles behave under different conditions and learn to apply several important gas laws.

The Kinetic Molecular Theory (KMT) explains gas behavior at the particle level. Key topics include Boyle's Law pressurevolumerelationshippressure-volume relationship, Charles' Law temperaturevolumerelationshiptemperature-volume relationship, and Gay-Lussac's Law temperaturepressurerelationshiptemperature-pressure relationship.

You'll also tackle more complex concepts like the Combined Gas Law, Ideal Gas Law, and Dalton's Law of partial pressures. These relationships help explain why gases compress, expand, and mix the way they do.

💡 All gas laws involve some combination of four main factors number of particles, temperature, pressure, and volume. Remembering this makes it easier to follow the logic behind each law!

By the end of this unit, you'll be able to solve gas law problems and explain everyday phenomena like why soda fizzes when opened or why deep-sea divers must ascend slowly.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Kinetic Molecular Theory and Standard Conditions

The Kinetic Molecular Theory (KMT) explains gas behavior at the molecular level. According to this theory, gas particles

  • Don't attract or repel each other
  • Move in constant random motion
  • Are much smaller than the distance between them
  • Don't lose energy when they collide

These properties explain why gases are compressible and fill their containers completely!

Four key factors affect gas behavior

  1. Number of gas particles (atoms, molecules, ions)
  2. Temperature (measured in °C or K)
  3. Pressure (measured in atm, torr, mmHg, or kPa)
  4. Volume (measured in L or cm³)

Standard Temperature and Pressure (STP) provides reference conditions

  • Temperature 0°C (273 K)
  • Pressure 1 atm (760 torr, 760 mmHg, or 101.3 kPa)

💡 When converting between Celsius and Kelvin, remember K = °C + 273. This conversion is crucial because gas laws require absolute temperature in Kelvin!

Temperature conversions are straightforward ex86.0°C+273=359Kex 86.0°C + 273 = 359 K, while pressure conversions use proportions ex4.5atm×101.3kPa/1atm=456kPaex 4.5 atm × 101.3 kPa/1 atm = 456 kPa.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

STP Conversions and Unit Measurements

Converting between different units is a key skill for solving gas law problems. You'll frequently need to switch between temperature scales and various pressure units.

Temperature Conversions

  • Kelvin to Celsius °C = K - 273
  • Celsius to Kelvin K = °C + 273

Examples

  • 5.5 K - 273 = -267.5°C
  • 304.8°C + 273 = 577.8 K

Pressure Conversions

  • 1 atm = 760 torr = 760 mmHg = 101.3 kPa
  • Use conversion factors and proportions to solve problems

Examples

  • 5.7 atm × 101.3kPa/1atm101.3 kPa/1 atm = 577.41 kPa
  • 8001.14 kPa × 760mmHg/101.3kPa760 mmHg/101.3 kPa = 60028.3 mmHg

💡 When solving conversion problems, setting up a proportion is often the easiest approach. Write the known relationship as a fraction, then multiply by your starting value.

Volume/Mass Conversions

  • 1 milliliter = 0.001 liter
  • 1 kilogram = 1000 grams
  • 100 centimeters = 10 decimeters

These conversions will be essential for all gas law calculations throughout this unit!

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Boyle's Law, Charles' Law, and Gay-Lussac's Law

These three gas laws form the foundation for understanding gas behavior under changing conditions.

Boyle's Law describes the relationship between pressure and volume

  • When pressure increases, volume decreases (and vice versa)
  • At constant temperature, P × V = constant
  • Formula P₁V₁ = P₂V₂
  • This is why a balloon shrinks when you squeeze it!

Charles' Law relates volume and temperature

  • When temperature increases, volume increases (and vice versa)
  • At constant pressure, V/T = constant
  • Formula V₁/T₁ = V₂/T₂ (temperature in Kelvin!)
  • This explains why balloons expand when heated

Gay-Lussac's Law connects pressure and temperature

  • When temperature increases, pressure increases (and vice versa)
  • At constant volume, P/T = constant
  • Formula P₁/T₁ = P₂/T₂ (temperature in Kelvin!)
  • This is why aerosol cans warn "Do not incinerate"

💡 Remember that all temperatures must be in Kelvin for gas law calculations. Using Celsius will give you incorrect answers!

Each of these laws describes what happens when two variables change while the others remain constant. In the real world, multiple variables often change simultaneously.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Gas Law Problem-Solving

Applying gas laws requires practice and a systematic approach. Let's look at some examples of Boyle's Law, Charles' Law, and Gay-Lussac's Law problems.

Boyle's Law Example If 1 L of gas at standard pressure (101.3 kPa) is compressed to 473 mL, what's the new pressure?

  • Set up the equation P₁V₁ = P₂V₂
  • Convert to same units 473 mL = 0.473 L
  • Solve 101.3 kPa × 1 L = P₂ × 0.473 L
  • P₂ = 214.2 kPa

Charles' Law Example If a balloon initially has volume 0.5 L at 22°C, what will be its volume at 4°C?

  • Convert temperatures to Kelvin 22°C + 273 = 295 K and 4°C + 273 = 277 K
  • Set up the equation V₁/T₁ = V₂/T₂
  • Solve 0.5 L/295 K = V₂/277 K
  • V₂ = 0.47 L (balloon shrinks when cooled)

💡 Always check that your answer makes physical sense. For example, a balloon should expand when heated, not shrink!

Mixed Law Problems may involve multiple variables. Identify which laws apply, then solve step by step. For instance, a potato chip bag "inflating" in a hot car involves Charles' Law - as temperature increases, the volume increases at constant pressure.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Combined Gas Law

The Combined Gas Law unifies Boyle's, Charles', and Gay-Lussac's laws into a single powerful equation that handles changes in pressure, volume, and temperature simultaneously

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

This formula is incredibly useful because it doesn't require any variables to remain constant. You can analyze situations where pressure, volume, and temperature all change at once.

How the Combined Gas Law relates to other gas laws

  • Boyle's Law If temperature is constant T1=T2T₁ = T₂, the formula simplifies to P₁V₁ = P₂V₂
  • Charles' Law If pressure is constant P1=P2P₁ = P₂, it simplifies to V₁/T₁ = V₂/T₂
  • Gay-Lussac's Law If volume is constant V1=V2V₁ = V₂, it simplifies to P₁/T₁ = P₂/T₂

Real-life example Imagine a helium balloon released into the air. As it rises

  • Temperature drops (T decreases)
  • Air pressure decreases (P decreases)
  • These changes affect the volume (V) of the balloon

💡 When solving Combined Gas Law problems, make sure all temperatures are in Kelvin and all other measurements use consistent units!

Using the Combined Gas Law, you can predict exactly what happens to the balloon's volume as it ascends through the atmosphere, even though multiple variables are changing.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Gas Law Problems Combined Gas Law

The Combined Gas Law allows you to solve complex problems involving changes in pressure, volume, and temperature. Here's how to approach these problems

  1. Identify your variables Label what you know for both initial and final states
  2. Convert units Ensure consistent units (especially Kelvin for temperature)
  3. Set up the equation P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
  4. Solve for the unknown variable

Example Problem If 17 liters of gas has a pressure of 2.3 atm and a temperature of 299 K, what is the new volume if the temperature rises to 350 K and pressure drops to 1.5 atm?

Step 1 Identify variables

  • V₁ = 17 L, P₁ = 2.3 atm, T₁ = 299 K
  • V₂ = ?, P₂ = 1.5 atm, T₂ = 350 K

Step 2 Set up the equation (2.3 atm)(17 L)299 K=(1.5 atm)(V2)350 K\frac{(2.3 \text{ atm})(17 \text{ L})}{299 \text{ K}} = \frac{(1.5 \text{ atm})(V_2)}{350 \text{ K}}

Step 3 Solve for V₂ V2=(2.3 atm)(17 L)(350 K)(299 K)(1.5 atm)=31 LV_2 = \frac{(2.3 \text{ atm})(17 \text{ L})(350 \text{ K})}{(299 \text{ K})(1.5 \text{ atm})} = 31 \text{ L}

💡 When pressure decreases and temperature increases, volume increases - this makes physical sense as the gas expands under these conditions!

Practice with different scenarios, paying special attention to which variable is unknown. The Combined Gas Law can solve problems involving any combination of these three variables.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Ideal Gas Law

The Ideal Gas Law takes gas behavior understanding to the next level by incorporating the number of gas particles into the equation

PV = nRT

This powerful formula connects all four gas variables

  • P = pressure (in atm)
  • V = volume (in L)
  • n = number of moles of gas
  • T = temperature (in K)
  • R = ideal gas constant 0.0821Latm/molK0.0821 L·atm/mol·K

The ideal gas constant (R) has different values depending on the units you're using

  • 0.0821 L·atm/mol·K
  • 62.4 L·mmHg/mol·K
  • 8.31 L·kPa/mol·K

Using the Ideal Gas Law When given any three of the four variables (P, V, n, T), you can solve for the fourth. For example, if you have 5.0 g of neon gas at 256 mmHg and 35°C, you could find its volume.

💡 When using the Ideal Gas Law, you'll often need to convert between grams and moles using the molar mass of the gas forneon,20.18g/molfor neon, 20.18 g/mol.

The Ideal Gas Law assumes gases behave "ideally" - that gas particles have no volume and no interactions with each other. While real gases don't behave perfectly this way, the law works very well under most normal conditions.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Ideal Gas Law Problems

The Ideal Gas Law PV=nRTPV = nRT allows you to solve problems involving all four gas variables. Here's how to approach these problems

Step 1 Identify what you know and what you need to find Step 2 Make sure units are compatible with your R value Step 3 Rearrange the equation to solve for your unknown Step 4 Calculate and check if your answer makes physical sense

Example Problem Types

Finding Temperature If you have 4 moles of gas at 5.6 atm in a 12 L container, what's the temperature?

  • Rearrange to T = PV/nR = (5.6 atm × 12 L)/4mol×0.0821Latm/molK4 mol × 0.0821 L·atm/mol·K
  • Solve for T in Kelvin

Finding Number of Moles If you have oxygen gas at 1.2 atm, 31 L, and 87°C, how many moles do you have?

  • Convert 87°C to 360 K
  • Rearrange to n = PV/RT = (1.2 atm × 31 L)/0.0821Latm/molK×360K0.0821 L·atm/mol·K × 360 K
  • Then calculate grams using molar mass of O₂ 32g/mol32 g/mol

💡 When using the Ideal Gas Law to find moles, you can convert to mass by multiplying by the molar mass mass (g) = moles × molar mass g/molg/mol

Mixed Law Problems can involve any combination of gas laws. First identify which variables are changing and which remain constant, then select the appropriate law to solve the problem.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure of a gas mixture equals the sum of the pressures each gas would exert if it were alone in the container. For a mixture of gases

P_{total} = P_1 + P_2 + P_3 + ...

This law works because gas particles move independently of each other and don't interact significantly.

Example 1 Air contains nitrogen (594 mmHg), argon (7.10 mmHg), carbon dioxide (0.27 mmHg), and oxygen. If total air pressure is 760 mmHg at sea level, what's the partial pressure of oxygen?

Ptotal=PN2+PAr+PCO2+PO2P_{total} = P_{N_2} + P_{Ar} + P_{CO_2} + P_{O_2} 760=594+7.10+0.27+PO2760 = 594 + 7.10 + 0.27 + P_{O_2} PO2=158.63=160P_{O_2} = 158.63 = 160 mmHg (with proper significant figures)

Example 2 A 50 L tank contains 5210 g N₂ and 4490 g O₂. What's the total pressure (in atm) at 24°C?

Step 1 Convert grams to moles

  • N₂ 5210 g ÷ 28.01 g/mol = 186.0 mol
  • O₂ 4490 g ÷ 32.00 g/mol = 140.3 mol

💡 You can find each gas's partial pressure using PV = nRT, then add them together for the total pressure.

Step 2 Calculate each partial pressure using the Ideal Gas Law Step 3 Add the partial pressures for the total pressure (approximately 159 atm)

Dalton's Law explains why deep-sea divers must be careful about nitrogen levels in their breathing mixture!

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Chemistry

55

Dec 5, 2025

18 pages

Exploring Gas Laws and Stoichiometry Insights

user profile picture

Vanessa Vasquez

@van3ssa_08

Gas laws describe how gases behave when temperature, pressure, volume, and amount of gas change. These physical relationships are crucial for understanding everything from weather patterns to chemical reactions, and they explain why balloons expand when heated and why your... Show more

Unit 9 Stamp Sheet

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Unit 9: Gas Laws Overview

This unit explores the fundamental relationships between gases and their physical properties. You'll discover how gas particles behave under different conditions and learn to apply several important gas laws.

The Kinetic Molecular Theory (KMT) explains gas behavior at the particle level. Key topics include Boyle's Law pressurevolumerelationshippressure-volume relationship, Charles' Law temperaturevolumerelationshiptemperature-volume relationship, and Gay-Lussac's Law temperaturepressurerelationshiptemperature-pressure relationship.

You'll also tackle more complex concepts like the Combined Gas Law, Ideal Gas Law, and Dalton's Law of partial pressures. These relationships help explain why gases compress, expand, and mix the way they do.

💡 All gas laws involve some combination of four main factors: number of particles, temperature, pressure, and volume. Remembering this makes it easier to follow the logic behind each law!

By the end of this unit, you'll be able to solve gas law problems and explain everyday phenomena like why soda fizzes when opened or why deep-sea divers must ascend slowly.

Unit 9 Stamp Sheet

| Date & Pg | Title | Full Credit | Half Credit | Date & Pg | Title | Full Credit | Half Credit |
|---|---|---|---|---|-

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Kinetic Molecular Theory and Standard Conditions

The Kinetic Molecular Theory (KMT) explains gas behavior at the molecular level. According to this theory, gas particles:

  • Don't attract or repel each other
  • Move in constant random motion
  • Are much smaller than the distance between them
  • Don't lose energy when they collide

These properties explain why gases are compressible and fill their containers completely!

Four key factors affect gas behavior:

  1. Number of gas particles (atoms, molecules, ions)
  2. Temperature (measured in °C or K)
  3. Pressure (measured in atm, torr, mmHg, or kPa)
  4. Volume (measured in L or cm³)

Standard Temperature and Pressure (STP) provides reference conditions:

  • Temperature: 0°C (273 K)
  • Pressure: 1 atm (760 torr, 760 mmHg, or 101.3 kPa)

💡 When converting between Celsius and Kelvin, remember: K = °C + 273. This conversion is crucial because gas laws require absolute temperature in Kelvin!

Temperature conversions are straightforward ex:86.0°C+273=359Kex: 86.0°C + 273 = 359 K, while pressure conversions use proportions ex:4.5atm×101.3kPa/1atm=456kPaex: 4.5 atm × 101.3 kPa/1 atm = 456 kPa.

Unit 9 Stamp Sheet

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STP Conversions and Unit Measurements

Converting between different units is a key skill for solving gas law problems. You'll frequently need to switch between temperature scales and various pressure units.

Temperature Conversions:

  • Kelvin to Celsius: °C = K - 273
  • Celsius to Kelvin: K = °C + 273

Examples:

  • 5.5 K - 273 = -267.5°C
  • 304.8°C + 273 = 577.8 K

Pressure Conversions:

  • 1 atm = 760 torr = 760 mmHg = 101.3 kPa
  • Use conversion factors and proportions to solve problems

Examples:

  • 5.7 atm × 101.3kPa/1atm101.3 kPa/1 atm = 577.41 kPa
  • 8001.14 kPa × 760mmHg/101.3kPa760 mmHg/101.3 kPa = 60028.3 mmHg

💡 When solving conversion problems, setting up a proportion is often the easiest approach. Write the known relationship as a fraction, then multiply by your starting value.

Volume/Mass Conversions:

  • 1 milliliter = 0.001 liter
  • 1 kilogram = 1000 grams
  • 100 centimeters = 10 decimeters

These conversions will be essential for all gas law calculations throughout this unit!

Unit 9 Stamp Sheet

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Boyle's Law, Charles' Law, and Gay-Lussac's Law

These three gas laws form the foundation for understanding gas behavior under changing conditions.

Boyle's Law describes the relationship between pressure and volume:

  • When pressure increases, volume decreases (and vice versa)
  • At constant temperature, P × V = constant
  • Formula: P₁V₁ = P₂V₂
  • This is why a balloon shrinks when you squeeze it!

Charles' Law relates volume and temperature:

  • When temperature increases, volume increases (and vice versa)
  • At constant pressure, V/T = constant
  • Formula: V₁/T₁ = V₂/T₂ (temperature in Kelvin!)
  • This explains why balloons expand when heated

Gay-Lussac's Law connects pressure and temperature:

  • When temperature increases, pressure increases (and vice versa)
  • At constant volume, P/T = constant
  • Formula: P₁/T₁ = P₂/T₂ (temperature in Kelvin!)
  • This is why aerosol cans warn "Do not incinerate"

💡 Remember that all temperatures must be in Kelvin for gas law calculations. Using Celsius will give you incorrect answers!

Each of these laws describes what happens when two variables change while the others remain constant. In the real world, multiple variables often change simultaneously.

Unit 9 Stamp Sheet

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Gas Law Problem-Solving

Applying gas laws requires practice and a systematic approach. Let's look at some examples of Boyle's Law, Charles' Law, and Gay-Lussac's Law problems.

Boyle's Law Example: If 1 L of gas at standard pressure (101.3 kPa) is compressed to 473 mL, what's the new pressure?

  • Set up the equation: P₁V₁ = P₂V₂
  • Convert to same units: 473 mL = 0.473 L
  • Solve: 101.3 kPa × 1 L = P₂ × 0.473 L
  • P₂ = 214.2 kPa

Charles' Law Example: If a balloon initially has volume 0.5 L at 22°C, what will be its volume at 4°C?

  • Convert temperatures to Kelvin: 22°C + 273 = 295 K and 4°C + 273 = 277 K
  • Set up the equation: V₁/T₁ = V₂/T₂
  • Solve: 0.5 L/295 K = V₂/277 K
  • V₂ = 0.47 L (balloon shrinks when cooled)

💡 Always check that your answer makes physical sense. For example, a balloon should expand when heated, not shrink!

Mixed Law Problems may involve multiple variables. Identify which laws apply, then solve step by step. For instance, a potato chip bag "inflating" in a hot car involves Charles' Law - as temperature increases, the volume increases at constant pressure.

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Combined Gas Law

The Combined Gas Law unifies Boyle's, Charles', and Gay-Lussac's laws into a single powerful equation that handles changes in pressure, volume, and temperature simultaneously:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

This formula is incredibly useful because it doesn't require any variables to remain constant. You can analyze situations where pressure, volume, and temperature all change at once.

How the Combined Gas Law relates to other gas laws:

  • Boyle's Law: If temperature is constant T1=T2T₁ = T₂, the formula simplifies to P₁V₁ = P₂V₂
  • Charles' Law: If pressure is constant P1=P2P₁ = P₂, it simplifies to V₁/T₁ = V₂/T₂
  • Gay-Lussac's Law: If volume is constant V1=V2V₁ = V₂, it simplifies to P₁/T₁ = P₂/T₂

Real-life example: Imagine a helium balloon released into the air. As it rises:

  • Temperature drops (T decreases)
  • Air pressure decreases (P decreases)
  • These changes affect the volume (V) of the balloon

💡 When solving Combined Gas Law problems, make sure all temperatures are in Kelvin and all other measurements use consistent units!

Using the Combined Gas Law, you can predict exactly what happens to the balloon's volume as it ascends through the atmosphere, even though multiple variables are changing.

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Gas Law Problems: Combined Gas Law

The Combined Gas Law allows you to solve complex problems involving changes in pressure, volume, and temperature. Here's how to approach these problems:

  1. Identify your variables: Label what you know for both initial and final states
  2. Convert units: Ensure consistent units (especially Kelvin for temperature)
  3. Set up the equation: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
  4. Solve for the unknown variable

Example Problem: If 17 liters of gas has a pressure of 2.3 atm and a temperature of 299 K, what is the new volume if the temperature rises to 350 K and pressure drops to 1.5 atm?

Step 1: Identify variables

  • V₁ = 17 L, P₁ = 2.3 atm, T₁ = 299 K
  • V₂ = ?, P₂ = 1.5 atm, T₂ = 350 K

Step 2: Set up the equation (2.3 atm)(17 L)299 K=(1.5 atm)(V2)350 K\frac{(2.3 \text{ atm})(17 \text{ L})}{299 \text{ K}} = \frac{(1.5 \text{ atm})(V_2)}{350 \text{ K}}

Step 3: Solve for V₂ V2=(2.3 atm)(17 L)(350 K)(299 K)(1.5 atm)=31 LV_2 = \frac{(2.3 \text{ atm})(17 \text{ L})(350 \text{ K})}{(299 \text{ K})(1.5 \text{ atm})} = 31 \text{ L}

💡 When pressure decreases and temperature increases, volume increases - this makes physical sense as the gas expands under these conditions!

Practice with different scenarios, paying special attention to which variable is unknown. The Combined Gas Law can solve problems involving any combination of these three variables.

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Ideal Gas Law

The Ideal Gas Law takes gas behavior understanding to the next level by incorporating the number of gas particles into the equation:

PV = nRT

This powerful formula connects all four gas variables:

  • P = pressure (in atm)
  • V = volume (in L)
  • n = number of moles of gas
  • T = temperature (in K)
  • R = ideal gas constant 0.0821Latm/molK0.0821 L·atm/mol·K

The ideal gas constant (R) has different values depending on the units you're using:

  • 0.0821 L·atm/mol·K
  • 62.4 L·mmHg/mol·K
  • 8.31 L·kPa/mol·K

Using the Ideal Gas Law: When given any three of the four variables (P, V, n, T), you can solve for the fourth. For example, if you have 5.0 g of neon gas at 256 mmHg and 35°C, you could find its volume.

💡 When using the Ideal Gas Law, you'll often need to convert between grams and moles using the molar mass of the gas forneon,20.18g/molfor neon, 20.18 g/mol.

The Ideal Gas Law assumes gases behave "ideally" - that gas particles have no volume and no interactions with each other. While real gases don't behave perfectly this way, the law works very well under most normal conditions.

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Ideal Gas Law Problems

The Ideal Gas Law PV=nRTPV = nRT allows you to solve problems involving all four gas variables. Here's how to approach these problems:

Step 1: Identify what you know and what you need to find Step 2: Make sure units are compatible with your R value Step 3: Rearrange the equation to solve for your unknown Step 4: Calculate and check if your answer makes physical sense

Example Problem Types:

Finding Temperature: If you have 4 moles of gas at 5.6 atm in a 12 L container, what's the temperature?

  • Rearrange to T = PV/nR = (5.6 atm × 12 L)/4mol×0.0821Latm/molK4 mol × 0.0821 L·atm/mol·K
  • Solve for T in Kelvin

Finding Number of Moles: If you have oxygen gas at 1.2 atm, 31 L, and 87°C, how many moles do you have?

  • Convert 87°C to 360 K
  • Rearrange to n = PV/RT = (1.2 atm × 31 L)/0.0821Latm/molK×360K0.0821 L·atm/mol·K × 360 K
  • Then calculate grams using molar mass of O₂ 32g/mol32 g/mol

💡 When using the Ideal Gas Law to find moles, you can convert to mass by multiplying by the molar mass: mass (g) = moles × molar mass g/molg/mol

Mixed Law Problems can involve any combination of gas laws. First identify which variables are changing and which remain constant, then select the appropriate law to solve the problem.

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Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure of a gas mixture equals the sum of the pressures each gas would exert if it were alone in the container. For a mixture of gases:

P_{total} = P_1 + P_2 + P_3 + ...

This law works because gas particles move independently of each other and don't interact significantly.

Example 1: Air contains nitrogen (594 mmHg), argon (7.10 mmHg), carbon dioxide (0.27 mmHg), and oxygen. If total air pressure is 760 mmHg at sea level, what's the partial pressure of oxygen?

Ptotal=PN2+PAr+PCO2+PO2P_{total} = P_{N_2} + P_{Ar} + P_{CO_2} + P_{O_2} 760=594+7.10+0.27+PO2760 = 594 + 7.10 + 0.27 + P_{O_2} PO2=158.63=160P_{O_2} = 158.63 = 160 mmHg (with proper significant figures)

Example 2: A 50 L tank contains 5210 g N₂ and 4490 g O₂. What's the total pressure (in atm) at 24°C?

Step 1: Convert grams to moles

  • N₂: 5210 g ÷ 28.01 g/mol = 186.0 mol
  • O₂: 4490 g ÷ 32.00 g/mol = 140.3 mol

💡 You can find each gas's partial pressure using PV = nRT, then add them together for the total pressure.

Step 2: Calculate each partial pressure using the Ideal Gas Law Step 3: Add the partial pressures for the total pressure (approximately 159 atm)

Dalton's Law explains why deep-sea divers must be careful about nitrogen levels in their breathing mixture!

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