Scientific Understanding

**What are Boyle’s Law, Charles’ Law, Gay-Lusac’s Law, Combined Gas Law…?**

I call these the dynamic gas laws because dynamic means changing and some kind of change is exactly what these gas laws are describing. These laws are meant to be useful when you have some kind of container of gas at certain conditions (of pressure, volume, moles, and temperature) and then several minutes or seconds later you change some of the conditions (of pressure, volume, moles, and temperature). These laws allow you to predict the future of what will happen to a gas if started at one set of conditions and then move to another set of conditions. This is key in fields like astrophysics, resource storage, resource mining, space exploration, and undersea exploration. These dynamic gas laws will NEVER use the gas constant of 0.08206.

How did the dynamic gas laws come about? Well, I could go into an in depth history of how all these people that lived long ago whom you have probably never heard about before worked painstakingly hard to come up with these laws that we benefit from every day. However, that will bore you and drag you away from the point I am trying to make. The point is that you only need simple mathematics manipulations to uncover the dynamic gas laws from the ideal gas law (PV = nRT).

If you take two ideal gas law equations and put them side by side of each other they look like what I have below.

P_{1}V_{1} = n_{1}R_{1}T_{1} |
P_{2}V_{2} = n_{2}R_{2}T_{2} |

1 | 1 |

Because the gas constant R is always they same number we can take it away from both sides.

P_{1}V_{1} = n_{1}T_{1} |
P_{2}V_{2} = n_{2}T_{2} |

1 | 1 |

Now we sweep all the variables to one side of the equals sign on each the right and left side.

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

Last we set them equal to each other.

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

Another way to write that equation would be.

(P_{1}V_{1} / n_{1}T_{1}) = (P_{2}V_{2} / n_{2}T_{2})

This equation in either from will help you solve any dynamic gas law problem like Boyle’s Law, Charles’ Law, Gay-Lusac’s Law, Combined Gas Law. I call the equation above the dynamic gas law equation. No one else calls it that so don’t type it in google and look for it. Keep in mind answering dynamic gas law questions is all about your problems solving skills. Make sure you know how to organize the information in the word problem well before you start in this section. Go back to other sections like……. if you are having problems here. With the following demonstrated examples I will prove my point on how to use math and problems solving skills together to achieve and answer.

**VIDEO Dynamic Gas Law Demonstrated Example 1 (Boyle’s Law)**: If a gas at 2.3 atm and 6.1L becomes 4.5L, what will be the new pressure?

**Step 1:**

What information are we given and what do they ask for?

Answer: Group them into two categories (those that came first in the sentence versus those that came next in the sentence).

Category 1

P_{1} = 2.3 atm

V_{1} = 6.1 L

Category 2

P_{2} = ?

V_{2} = 4.5 L

**Step 2:**

What gas law equation does this fall into?

Answer: Set up the dynamic gas law equation to start.

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 3:**

Which terms or variables are missing from the information of the question?

Answer: n and T, so cross them off of the equation

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 4:**

Simplify

P_{1}V_{1} = |
P_{2}V_{2} |

1 | 1 |

**Step 5:**

What can we fill in for the equation?

Answer: The information we are given (red).

2.3 atm * 6.1 L = | P_{2} * 4.5 L |

1 | 1 |

**Step 6:**

How do we rearrange the equation?

Answer: Divide both sides by 4.5 L (red).

2.3 atm * 6.1 L = | P_{2} * 4.5 L |

4.5 L | 4.5 L |

**Step 7:**

Cross off 4.5 L from right side.

2.3 atm * 6.1 L = | P_{2} * 4.5 L |

4.5 L | 4.5 L |

**Step 8:**

Simplify

2.3 atm * 6.1 L = | P_{2} |

4.5 L |

**Step 9:**

How do I do the calculations?

Answer: (2.3 * 6.1) / 4.5 = 3.12

2.3 atm * 6.1 L = | 3.12 atm |

4.5 L |

**Step 10:**

What is the complete answer?

COMPLETE ANSWER: 3.12 atm

**VIDEO Dynamic Gas Law Demonstrated Example 2 (Charles’ Law)**: What is the volume of a gas at 30 C and constant moles if its volume at 15 C is 920mL?

**Step 1:**

What information are we given and what do they ask for?

Answer: Group them into two categories (those that came first in the sentence versus those that came next in the sentence). Since moles is constant you don’t need to include n in your information.

Category 1

V_{1} = ?

T_{1} = 30 C

Category 2

V_{2} = 920 mL

T_{2} = 15 C

**Step 2:**

What conversions are needed?

Answer:

Category 1

V_{1} = ?

T_{1} = 30 C —> 303 K

Category 2

V_{2} = 920 mL

T_{2} = 15 C —> 288 K

**Step 3:**

What gas law equation does this fall into?

Answer: Set up the dynamic gas law equation to start.

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 4:**

Which terms or variables are missing from the information of the question?

Answer: P and n, so cross them off of the equation

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 5:**

Simplify

V_{1} = |
V_{2} |

T_{1} |
T_{2} |

**Step 6:**

What can we fill in for the equation?

Answer: The information we are given (red).

V_{1} = |
920 mL |

303 K | 288 K |

**Step 7:**

How do we rearrange the equation?

Answer: Multiply both sides by 303 K (red).

303 K * V_{1} = |
920 mL * 303 K |

303 K | 288 K |

**Step 8:**

Cross out the 303 K on the left side

303 K * V_{1} = |
920 mL * 303 K |

303 K | 288 K |

**Step 9:**

Simplify

V_{1} = |
920 mL * 303 K |

288 K |

**Step 10:**

How do I do the calculations?

Answer: (920 * 303) / 288 = 968

968 mL = | 920 mL * 303 K |

288 K |

**Step 11:**

What is the complete answer?

COMPLETE ANSWER: 968 mL

**VIDEO Dynamic Gas Law Demonstrated Example 3 (Combined Gas Law)**: What is the temperature of a gas at 500 mm Hg and 750 mL if it was at 55 C, 0.8L, and 1.2 atm?

**Step 1:**

What information are we given and what do they ask for?

Answer: Group them into two categories (those that came first in the sentence versus those that came next in the sentence).

Category 1

P_{1} = 500 mm Hg

V_{1} = 750 mL

T_{1} = ?

Category 2

P_{2} = 1.2 atm

V_{2} = 0.8 L

T_{2} = 55 C

**Step 2:**

What conversions are needed?

Answer:

Category 1

P_{1} = 500 mm Hg

V_{1} = 750 mL

T_{1} = ?

Category 2

P_{2} = 1.2 atm —> 912 mm Hg

V_{2} = 0.8 L —> 800 mL

T_{2} = 55 C —> 328 K

**Step 3:**

What gas law equation does this fall into?

Answer: Set up the dynamic gas law equation to start.

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 4:**

Which terms or variables are missing from the information of the question?

Answer: n, so cross that off of the equation

P_{1}V_{1} = |
P_{2}V_{2} |

n_{1}T_{1} |
n_{2}T_{2} |

**Step 5:**

Simplify

P_{1}V_{1} = |
P_{2}V_{2} |

T_{1} |
T_{2} |

**Step 6:**

What can we fill in for the equation?

Answer: The information we are given (red).

500 * 750 = | 912 * 800 |

T_{1} |
328 |

**Step 7:**

How do we rearrange the equation?

Answer: Multiply both sides by T_{1} and 328 (red).

500 * 750 * 328 * T_{1} = |
912 * 800 * 328 * T_{1} |

T_{1} |
328 |

**Step 8:**

Simplify

500 * 750 * 328 = | 912 * 800 * T_{1} |

1 | 1 |

**Step 9:**

How do we further rearrange the equation?

Answer: Divide both sides by 912 and 800 (red).

500 * 750 * 328 = | 912 * 800 * T_{1} |

912 * 800 | 912 * 800 |

**Step 10:**

Cross out the 912 and 800 on the right side.

500 * 750 * 328 = | 912 * 800 * T_{1} |

912 * 800 | 912 * 800 |

**Step 11:**

Simplify

500 * 750 * 328 = | T_{1} |

912 * 800 |

**Step 12:**

How do I do the calculations?

Answer: (500 * 750 * 328) / (912 * 800) = 168.6

500 * 750 * 328 = | 168.6 K |

912 * 800 |

**Step 13:**

What is the complete answer?

COMPLETE ANSWER: 169 K

**PRACTICE PROBLEMS**: Solve the dynamic gas law equations below. Remember to always convert Celsius temperature to Kelvin.

If a gas at 3.2 atm and 9.4L is allowed to expand to 17.8L, what will be the new pressure?

Answer: 1.69 atm

What is the volume of a gas at 960 mm Hg if it was 210 mm Hg and 840 mL?

Answer: 184 mL

What is the volume of a gas at 65 C and constant moles if its volume at 25 C is 76 mL?

Answer: 86 mL

If a gas at 21 C is 31 L and we change the volume to 65L. What will be the new temperature in Celcius?

Answer: 343 C

If a gas at 3.6 atm and 400 K is compressed to 2.4 atm what will its temperature be?

Answer: 267 K

If 58 mol of a gas takes up 90L. How many moles would the same gas take up if it is 56L?

Answer: 36 mol

What is the temperature of a gas at 420 mm Hg and 890 mL if it was at 108 C, 2.3 L, and 2.5 atm?

Answer: 32.6 K