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6.1 KMT and Gas Laws

1 min readfebruary 27, 2024

So far in previous study guides, we’ve discussed some properties of gases. Today, we’ll do a deeper dive into how gases behave and get into some gas laws! 🪂

⚛️ Kinetic Molecular Theory (KMT)

The study of gas laws provides us with a framework to predict and explain how gases will respond to changes in pressure, volume, and temperature. Gas laws are rooted in the Kinetic Molecular Theory (KMT) which describes gases as large numbers of tiny particles in constant, random motion.

The KMT of gases is based on five fundamental assumptions:

  1. 👟 Gas particles are in constant, random, straight-line motion.
  2. 🌏 Particles are separated by great distances, so gases are a lot of empty space with dispersed particles.
  3. ⚡ Collisions are rapid and elastic. When particles hit each other, no energy is lost. Total energy remains constant.
  4. 🪠 No forces operate between particles.
  5. 🌡️ The average kinetic energy of gas particles depends on the gas’ temperature.

When gases meet all these assumptions, we refer to them as “ideal gases” and when they don’t, we call them “real gases.” Although no gas can truly be ideal, it is a term that we use for gases that are closer to ideals than others. Keep in mind, that these laws are idealizations. Real gases may show deviations under certain conditions, such as high pressure or low temperature.


⛽ Fundamental Gas Laws

Now, let's go over fundamental gas laws that help us predict and explain how gases will act under different conditions.

Boyle’s Law

Boyle's Law is one of the fundamental gas laws, discovered by Robert Boyle in the 17th century. This law describes the relationship between the pressure and volume of a gas at a constant temperature.

Essentially, Boyle’s discovered that pressure and volume are inversely proportional. What does this mean? 🤔

As the volume of a gas decreases, its pressure increases, provided the temperature remains constant. Conversely, increasing the volume leads to a decrease in pressure. This relationship is crucial in understanding how gases compress and expand.

Boyle’s law can be observed in our daily lives—think about our breathing. As your diaphragm expands, your chest volume increases, leading to a decrease in pressure. This allows air to flow in your lungs!

oapLBD7.gif

GIF Courtesy of Alchaeus via Imgur.

✏️ Boyle’s Law Practice Question

A balloon has a volume of 2.0 liters at a pressure of 1.0 atm. If the balloon is compressed to a volume of 1.5 liters, what will be the new pressure inside the balloon, assuming the temperature remains constant?

Think about the initial condition of the balloon and assign the appropriate variables, V1V_1 and P1P_1. Then, do the same for the final condition of the balloon. 🎈

(1.0 atm)(2.0 L)=(P2)(1.5 L)(1.0 \ \text{atm})(2.0 \ \text{L})=(P_2)(1.5 \ \text{L})
1.33 atm1.33 \ \text{atm}

Great work! 👏

🧠 Boyle’s Law Conceptual Question

A sealed syringe contains air and is plunged into a container of water at room temperature. As the plunger is pushed in, reducing the volume of air by half, what happens to the pressure inside the syringe? Explain your answer using Boyle's Law.

Imagine you’re pressing the plunger of a syringe! The question tells you that volume is halved, and is testing your knowledge of the inverse relationship between pressure and volume.

Reducing the volume, specifically by half, means you’re increasing the pressure inside the syringe. But, by how much? Set up Boyle’s law to see what happens to P2P_2.

P1V1=P212V2P_1V_1=P_2\cdot\frac12V_2

To make this equation hold true, we have to double the pressure. Therefore, when volume is halved, pressure is doubled.

P1V1=2P212V2P_1V_1=2P_2\cdot\frac12V_2

Now both sides of the equation are equal! 🎊


Charles’ Law

Charles' Law focuses on the direct relationship between the volume and temperature of a gas, while holding pressure constant. When there is a direct relationship, both variables either increase together or decrease together. When volume increases, pressure will too, and vice versa!

In other words, Charles' Law describes how gases tend to expand and increase in volume when heated. This direct relationship shows that warmer gases occupy more space because their particles move faster and spread apart.

🧠 You can also relate this to the fifth assumption of the kinetic molecular theory: the average kinetic energy of particles depends on the temperature of the gas. As you increase the temperature, the particles’ kinetic energy increases and they move faster, spreading out from one another.

Here’s an everyday example that can help you remember and understand Charles’ law. Think about hot air balloons. When the air inside the balloon is heated, the volume increases which makes the balloon rise. 🔥

charles' law.png

Image Credit to Fiveable.

✏️ Charles’ Law Practice Question

A yellow balloon inflated in a room at 27°C27\degree C has a volume of 4.00 L4.00\ L. The balloon is then heated to a temperature of 54°C54\degree C. What is the new volume if the pressure remains constant?

First, take a look at the temperature change and convert °C\degree C to KK (always remember to do this!).

T1=27°C+273=300 KT_1 = 27\degree C + 273 = 300\ K
T2=54°C+273=327 KT_2 = 54\degree C+ 273 = 327\ K

Now let’s write out the Charles’ Law equation.

V1T1=V2T2\frac{{ V_1}}{{T_1}} = \frac{{ V_2}}{{T_2}}

Assign the values given in the question to the right variables. When doing this, think about the conditions of the balloon at the initial temperature of 300 K300\ K and then the final temperature of 327 K327\ K.

4 L300 K=V2327 K\frac{{ 4\ L}}{{300\ K}} = \frac{{ V_2}}{{327\ K}}

Rearrange the equation to solve for the new volume by cross-multiplying and dividing, and calculate away!

V2=327 K4 L300 KV_2 = \frac{{327\ K* 4\ L}}{{300\ K}}
V2=4.36 LV_2 = 4.36\ L

Now you know that when the temperature increased to 54°C54\degree C, the volume also increased to 4.36 L4.36\ L. You did it! 👏


Gay-Lussac’s Law

Gay-Lussac's Law describes that if the volume stays constant, then temperature increases along with the pressure. This means that temperature and pressure are directly related.

If a gas is in a fixed container, as the temperature rises, molecules move more quickly, hitting the walls of their container with more force, which increases pressure. Let’s break this down a bit further:

  • Volume is constant, so when discussing Gay-Lussac’s, the gas must be in a fixed container. It can’t expand or contract! 🔦
  • Thinking back to the 5th assumption of the KMT, as temperature rises, particles are “running” around in the fixed container much faster.
  • Therefore, they’re hitting the rigid walls of the container much more often, and with increased force. Think of this force as pressure!

What about if the temperature was decreasing? ⛄

  • Volume is still constant, but particles are moving much slooower. They’re going to hit the walls of the container less frequently and with less force, so pressure goes down as well!

A good example where we can see Gay-Lussac’s law is with a hair spray. You might have seen the warning signs that you should not expose the hair spray to heat because it’s under pressure. So, what do you think will happen if it were to be exposed to heat? The hair spray bottle would explode because the walls are rigid and the bottle can’t take the increased pressure! 💥 

Take a look at what happens when the walls aren’t rigid, like the balloon below! Volume is able to be manipulated, so pressure won’t cause the balloon to explode. That’s why it’s important to note that volume is constant with Gay-Lussac’s law.

volume-temperature-2.png

Image Credit to Fiveable.

✏️ Gay-Lussac’s Law Practice Question

The pressure in a car tire is 3.00 atm3.00\ \text{atm} at 28°C28\degree C. After going on a trip, the tire pressure went up to 3.15 atm3.15\ \text{atm}. Assuming the volume is constant, what is the temperature of the air in the tire in °C\degree C? 🚗

Let’s write out the Gay-Lussac’s Law equation.

P1T1=P2T2 \frac{{ P_1}}{{T_1}} = \frac{{ P_2}}{{T_2}}

Now, let’s convert the temperature that’s provided from °C\degree C to KK (always remember to do this!).

T1=28°C+273=301 KT_1 = 28\degree C + 273 = \boxed{301 \ K}

Assign the values given in the question to the right variables. Our initial values are P1=3.00 atmP_1=3.00\ \text{atm} and T1=301 KT_1=301\ K. Our final values are P2=3.15 atmP_2=3.15\ \text{atm} and we are solving for T2T_2. ⬇️

3.00 atm301 K=3.15 atmT2\frac{{3.00\ \text{atm}}}{{301\ K}} = \frac{{ 3.15\ \text{atm}}}{{T_2}}

Rearrange the equation to solve for the new volume, and calculate away!

T2=3.15 atm301 K3.00 atmT_2 = \frac{{3.15\ \text{atm}* 301\ K}}{{3.00\ \text{atm}}}
T2=316.05 KT_2 = 316.05\ K

We’re almost done! The question asked for us to provide our final answer in °C\degree C. So, we have to convert the Kelvin back to Celsius to get full credit.

T2=316.05 K273 K=43.05°CT_2 = 316.05\ K - 273\ K = 43.05 \degree C

Combined Gas Law

The Combined Gas Law brings all the three gas laws we weren't over today (Boyle's, Charles's, and Gay-Lussac's laws). This means that you can derive all other gas laws from this one equation! It’s a single formula to calculate changes involving pressure, volume, and temperature. Remember that this law only applies when the number of molecules stays constant & everything else changes.

P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}

Let’s dive right into a guided practice question!

✏️Combined Gas Law Practice Question

Pep 🌶️ has a weather balloon inflated to a volume of 36.4 Liters36.4\ \text{Liters} and is prepared at a pressure of 850 mmHg850\ \text{mmHg} and a temperature of 27.0°C27.0\degree C. The balloon rises higher in the sky where the pressure is 325 mmHg325\ \text{mmHg} and the temperature is 14.0°C-14.0\degree C. If we assume the balloon can expand without popping, Pep is curious about the volume of the weather balloon at the higher altitude.

Let’s write out the Combined Gas Law equation.

P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}

Now, let’s convert the temperature that’s provided from °C\degree C to KK (always remember to do this!).

T1=27.0°C+273=300 KT_1 = 27.0\degree C + 273 = 300 \ K
T2=14.0°C+273=259 KT_2 = -14.0\degree C + 273 = 259\ K

Assign the values given in the question to the right variables. Again, think about the initial and final conditions of the weather balloon, as well as your unknown variable.

850 mmHg36.4 L300 K=325 mmHgV2259 K\frac{{850\ \text{mmHg} * 36.4\ L}}{{300\ K}} = \frac{{ 325\ \text{mmHg} * V_2}}{{259\ K}}

Rearrange the equation to solve for the new volume, and calculate away!

V2=850 mmHg36.4 L259 K300 K325 mmHgV_2 = \frac{{850\ \text{mmHg}* 36.4\ L * 259\ K}}{{300\ K * 325\ \text{mmHg}}}

Once we calculate, we get a new volume of:

V282.2LV_2 \approx 82.2L

💡 Steps to Gas Law Problems

When you’re tackling practice questions on this topic, here’s a step-by-step process to approach Gas Law-related calculations:

  1. ↕️ Separate what was given. Write down the values you are given with units.
  2. 🔍 Assign corresponding variables to each given value. Think about the initial and final condition of the object described. Are you dealing with temperature, pressure, or volume?
    1. 🌡️ if you are dealing with temperature, make sure the units are in Kelvin!
  3. 📝 Write out the appropriate equation. Once you see what you’re dealing with, write the law you are using ( P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}).
  4. 💪 Rearrange the equation to solve. You want to isolate the unknown variable so it’s easier to solve!
  5. 🔌 Substitute given values. Plug in the values given, including their units.
  6. 💯 Solve! Cancel out units and if necessary, do unit conversions to get the units to cancel out.
  7. 🏁 Calculate. Write your answer down with the correct significant figures and appropriate unit.

🥑 Avogadro’s Law

Avogadro’s Law states that equal volumes of all gases under identical temperature and pressure citcumstances contain an equal number of molecules.

In our next study guide, we will continue to explore Gas Laws and specifically go over the Ideal Gas Law. Remember, we have additional practice questions in the Practice tab so check those out to be extra extra prepared! 🎉 Happy studying!

6.1 KMT and Gas Laws

1 min readfebruary 27, 2024

So far in previous study guides, we’ve discussed some properties of gases. Today, we’ll do a deeper dive into how gases behave and get into some gas laws! 🪂

⚛️ Kinetic Molecular Theory (KMT)

The study of gas laws provides us with a framework to predict and explain how gases will respond to changes in pressure, volume, and temperature. Gas laws are rooted in the Kinetic Molecular Theory (KMT) which describes gases as large numbers of tiny particles in constant, random motion.

The KMT of gases is based on five fundamental assumptions:

  1. 👟 Gas particles are in constant, random, straight-line motion.
  2. 🌏 Particles are separated by great distances, so gases are a lot of empty space with dispersed particles.
  3. ⚡ Collisions are rapid and elastic. When particles hit each other, no energy is lost. Total energy remains constant.
  4. 🪠 No forces operate between particles.
  5. 🌡️ The average kinetic energy of gas particles depends on the gas’ temperature.

When gases meet all these assumptions, we refer to them as “ideal gases” and when they don’t, we call them “real gases.” Although no gas can truly be ideal, it is a term that we use for gases that are closer to ideals than others. Keep in mind, that these laws are idealizations. Real gases may show deviations under certain conditions, such as high pressure or low temperature.


⛽ Fundamental Gas Laws

Now, let's go over fundamental gas laws that help us predict and explain how gases will act under different conditions.

Boyle’s Law

Boyle's Law is one of the fundamental gas laws, discovered by Robert Boyle in the 17th century. This law describes the relationship between the pressure and volume of a gas at a constant temperature.

Essentially, Boyle’s discovered that pressure and volume are inversely proportional. What does this mean? 🤔

As the volume of a gas decreases, its pressure increases, provided the temperature remains constant. Conversely, increasing the volume leads to a decrease in pressure. This relationship is crucial in understanding how gases compress and expand.

Boyle’s law can be observed in our daily lives—think about our breathing. As your diaphragm expands, your chest volume increases, leading to a decrease in pressure. This allows air to flow in your lungs!

oapLBD7.gif

GIF Courtesy of Alchaeus via Imgur.

✏️ Boyle’s Law Practice Question

A balloon has a volume of 2.0 liters at a pressure of 1.0 atm. If the balloon is compressed to a volume of 1.5 liters, what will be the new pressure inside the balloon, assuming the temperature remains constant?

Think about the initial condition of the balloon and assign the appropriate variables, V1V_1 and P1P_1. Then, do the same for the final condition of the balloon. 🎈

(1.0 atm)(2.0 L)=(P2)(1.5 L)(1.0 \ \text{atm})(2.0 \ \text{L})=(P_2)(1.5 \ \text{L})
1.33 atm1.33 \ \text{atm}

Great work! 👏

🧠 Boyle’s Law Conceptual Question

A sealed syringe contains air and is plunged into a container of water at room temperature. As the plunger is pushed in, reducing the volume of air by half, what happens to the pressure inside the syringe? Explain your answer using Boyle's Law.

Imagine you’re pressing the plunger of a syringe! The question tells you that volume is halved, and is testing your knowledge of the inverse relationship between pressure and volume.

Reducing the volume, specifically by half, means you’re increasing the pressure inside the syringe. But, by how much? Set up Boyle’s law to see what happens to P2P_2.

P1V1=P212V2P_1V_1=P_2\cdot\frac12V_2

To make this equation hold true, we have to double the pressure. Therefore, when volume is halved, pressure is doubled.

P1V1=2P212V2P_1V_1=2P_2\cdot\frac12V_2

Now both sides of the equation are equal! 🎊


Charles’ Law

Charles' Law focuses on the direct relationship between the volume and temperature of a gas, while holding pressure constant. When there is a direct relationship, both variables either increase together or decrease together. When volume increases, pressure will too, and vice versa!

In other words, Charles' Law describes how gases tend to expand and increase in volume when heated. This direct relationship shows that warmer gases occupy more space because their particles move faster and spread apart.

🧠 You can also relate this to the fifth assumption of the kinetic molecular theory: the average kinetic energy of particles depends on the temperature of the gas. As you increase the temperature, the particles’ kinetic energy increases and they move faster, spreading out from one another.

Here’s an everyday example that can help you remember and understand Charles’ law. Think about hot air balloons. When the air inside the balloon is heated, the volume increases which makes the balloon rise. 🔥

charles' law.png

Image Credit to Fiveable.

✏️ Charles’ Law Practice Question

A yellow balloon inflated in a room at 27°C27\degree C has a volume of 4.00 L4.00\ L. The balloon is then heated to a temperature of 54°C54\degree C. What is the new volume if the pressure remains constant?

First, take a look at the temperature change and convert °C\degree C to KK (always remember to do this!).

T1=27°C+273=300 KT_1 = 27\degree C + 273 = 300\ K
T2=54°C+273=327 KT_2 = 54\degree C+ 273 = 327\ K

Now let’s write out the Charles’ Law equation.

V1T1=V2T2\frac{{ V_1}}{{T_1}} = \frac{{ V_2}}{{T_2}}

Assign the values given in the question to the right variables. When doing this, think about the conditions of the balloon at the initial temperature of 300 K300\ K and then the final temperature of 327 K327\ K.

4 L300 K=V2327 K\frac{{ 4\ L}}{{300\ K}} = \frac{{ V_2}}{{327\ K}}

Rearrange the equation to solve for the new volume by cross-multiplying and dividing, and calculate away!

V2=327 K4 L300 KV_2 = \frac{{327\ K* 4\ L}}{{300\ K}}
V2=4.36 LV_2 = 4.36\ L

Now you know that when the temperature increased to 54°C54\degree C, the volume also increased to 4.36 L4.36\ L. You did it! 👏


Gay-Lussac’s Law

Gay-Lussac's Law describes that if the volume stays constant, then temperature increases along with the pressure. This means that temperature and pressure are directly related.

If a gas is in a fixed container, as the temperature rises, molecules move more quickly, hitting the walls of their container with more force, which increases pressure. Let’s break this down a bit further:

  • Volume is constant, so when discussing Gay-Lussac’s, the gas must be in a fixed container. It can’t expand or contract! 🔦
  • Thinking back to the 5th assumption of the KMT, as temperature rises, particles are “running” around in the fixed container much faster.
  • Therefore, they’re hitting the rigid walls of the container much more often, and with increased force. Think of this force as pressure!

What about if the temperature was decreasing? ⛄

  • Volume is still constant, but particles are moving much slooower. They’re going to hit the walls of the container less frequently and with less force, so pressure goes down as well!

A good example where we can see Gay-Lussac’s law is with a hair spray. You might have seen the warning signs that you should not expose the hair spray to heat because it’s under pressure. So, what do you think will happen if it were to be exposed to heat? The hair spray bottle would explode because the walls are rigid and the bottle can’t take the increased pressure! 💥 

Take a look at what happens when the walls aren’t rigid, like the balloon below! Volume is able to be manipulated, so pressure won’t cause the balloon to explode. That’s why it’s important to note that volume is constant with Gay-Lussac’s law.

volume-temperature-2.png

Image Credit to Fiveable.

✏️ Gay-Lussac’s Law Practice Question

The pressure in a car tire is 3.00 atm3.00\ \text{atm} at 28°C28\degree C. After going on a trip, the tire pressure went up to 3.15 atm3.15\ \text{atm}. Assuming the volume is constant, what is the temperature of the air in the tire in °C\degree C? 🚗

Let’s write out the Gay-Lussac’s Law equation.

P1T1=P2T2 \frac{{ P_1}}{{T_1}} = \frac{{ P_2}}{{T_2}}

Now, let’s convert the temperature that’s provided from °C\degree C to KK (always remember to do this!).

T1=28°C+273=301 KT_1 = 28\degree C + 273 = \boxed{301 \ K}

Assign the values given in the question to the right variables. Our initial values are P1=3.00 atmP_1=3.00\ \text{atm} and T1=301 KT_1=301\ K. Our final values are P2=3.15 atmP_2=3.15\ \text{atm} and we are solving for T2T_2. ⬇️

3.00 atm301 K=3.15 atmT2\frac{{3.00\ \text{atm}}}{{301\ K}} = \frac{{ 3.15\ \text{atm}}}{{T_2}}

Rearrange the equation to solve for the new volume, and calculate away!

T2=3.15 atm301 K3.00 atmT_2 = \frac{{3.15\ \text{atm}* 301\ K}}{{3.00\ \text{atm}}}
T2=316.05 KT_2 = 316.05\ K

We’re almost done! The question asked for us to provide our final answer in °C\degree C. So, we have to convert the Kelvin back to Celsius to get full credit.

T2=316.05 K273 K=43.05°CT_2 = 316.05\ K - 273\ K = 43.05 \degree C

Combined Gas Law

The Combined Gas Law brings all the three gas laws we weren't over today (Boyle's, Charles's, and Gay-Lussac's laws). This means that you can derive all other gas laws from this one equation! It’s a single formula to calculate changes involving pressure, volume, and temperature. Remember that this law only applies when the number of molecules stays constant & everything else changes.

P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}

Let’s dive right into a guided practice question!

✏️Combined Gas Law Practice Question

Pep 🌶️ has a weather balloon inflated to a volume of 36.4 Liters36.4\ \text{Liters} and is prepared at a pressure of 850 mmHg850\ \text{mmHg} and a temperature of 27.0°C27.0\degree C. The balloon rises higher in the sky where the pressure is 325 mmHg325\ \text{mmHg} and the temperature is 14.0°C-14.0\degree C. If we assume the balloon can expand without popping, Pep is curious about the volume of the weather balloon at the higher altitude.

Let’s write out the Combined Gas Law equation.

P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}

Now, let’s convert the temperature that’s provided from °C\degree C to KK (always remember to do this!).

T1=27.0°C+273=300 KT_1 = 27.0\degree C + 273 = 300 \ K
T2=14.0°C+273=259 KT_2 = -14.0\degree C + 273 = 259\ K

Assign the values given in the question to the right variables. Again, think about the initial and final conditions of the weather balloon, as well as your unknown variable.

850 mmHg36.4 L300 K=325 mmHgV2259 K\frac{{850\ \text{mmHg} * 36.4\ L}}{{300\ K}} = \frac{{ 325\ \text{mmHg} * V_2}}{{259\ K}}

Rearrange the equation to solve for the new volume, and calculate away!

V2=850 mmHg36.4 L259 K300 K325 mmHgV_2 = \frac{{850\ \text{mmHg}* 36.4\ L * 259\ K}}{{300\ K * 325\ \text{mmHg}}}

Once we calculate, we get a new volume of:

V282.2LV_2 \approx 82.2L

💡 Steps to Gas Law Problems

When you’re tackling practice questions on this topic, here’s a step-by-step process to approach Gas Law-related calculations:

  1. ↕️ Separate what was given. Write down the values you are given with units.
  2. 🔍 Assign corresponding variables to each given value. Think about the initial and final condition of the object described. Are you dealing with temperature, pressure, or volume?
    1. 🌡️ if you are dealing with temperature, make sure the units are in Kelvin!
  3. 📝 Write out the appropriate equation. Once you see what you’re dealing with, write the law you are using ( P1V1T1=P2V2T2\frac{{P_1 V_1}}{{T_1}} = \frac{{P_2 V_2}}{{T_2}}).
  4. 💪 Rearrange the equation to solve. You want to isolate the unknown variable so it’s easier to solve!
  5. 🔌 Substitute given values. Plug in the values given, including their units.
  6. 💯 Solve! Cancel out units and if necessary, do unit conversions to get the units to cancel out.
  7. 🏁 Calculate. Write your answer down with the correct significant figures and appropriate unit.

🥑 Avogadro’s Law

Avogadro’s Law states that equal volumes of all gases under identical temperature and pressure citcumstances contain an equal number of molecules.

In our next study guide, we will continue to explore Gas Laws and specifically go over the Ideal Gas Law. Remember, we have additional practice questions in the Practice tab so check those out to be extra extra prepared! 🎉 Happy studying!



© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.