How to Solve: Charles’ Law

How to Solve: Charles’ Law

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"Imagine a hot-air balloon rising into the sky—why does the air inside expand as it heats up? Master Charles’ Law, and you’ll solve this AND crush any exam question on gas behavior in seconds."

What You Need To Know First

Before diving into Charles’ Law, ensure you understand: 1. Temperature Scales – Know the difference between Celsius (°C) and Kelvin (K), and how to convert between them. 2. Direct vs. Inverse Proportionality – If two variables are directly proportional, as one increases, the other increases by the same factor. 3. Ideal Gas Basics – Gases expand when heated (if pressure is constant).

Key Vocabulary

Formulas To Know

1. Charles’ Law Formula

Formula: [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]

Variables: - ( V_1 ) = Initial volume (any unit, but must match ( V_2 )) - ( T_1 ) = Initial temperature (must be in Kelvin) - ( V_2 ) = Final volume - ( T_2 ) = Final temperature (must be in Kelvin)

MEMORISE THIS? ✅ Yes! (Not always given on exam sheets.)

2. Temperature Conversion (Celsius to Kelvin)

Formula: [ T(K) = T(°C) + 273.15 ]

MEMORISE THIS? ✅ Yes! (Critical for Charles’ Law.)

Step-by-Step Method

Follow these steps exactly for every Charles’ Law problem.

1. Identify the given values – Write down ( V_1, T_1, V_2, T_2 ) (some may be missing).
2. Convert temperatures to Kelvin – If given in °C, add 273.15.
3. Check if pressure is constant – If not, Charles’ Law does not apply.
4. Rearrange the formula – Solve for the missing variable.
5. Plug in the numbers – Substitute values into the equation.
6. Calculate and simplify – Cross-multiply if needed.
7. Check units – Volume units must match; temperature must be in Kelvin.
8. Write the final answer – Include units and round if necessary.

Worked Example (Using Steps Above)

Problem: A gas occupies 3.0 L at 25°C. What will its volume be if heated to 100°C (pressure remains constant)?

Solution: 1. Given:
- ( V_1 = 3.0 \, \text{L} )
- ( T_1 = 25°C )
- ( T_2 = 100°C )
- ( V_2 = ? )

1. Convert to Kelvin:
2. ( T_1 = 25 + 273.15 = 298.15 \, \text{K} )

3. ( T_2 = 100 + 273.15 = 373.15 \, \text{K} )

4. Pressure is constant – Charles’ Law applies.

5. Rearrange formula:
   [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]
   [ V_2 = \frac{V_1 \times T_2}{T_1} ]

6. Plug in numbers:
   [ V_2 = \frac{3.0 \, \text{L} \times 373.15 \, \text{K}}{298.15 \, \text{K}} ]

7. Calculate:
   [ V_2 = \frac{1119.45}{298.15} \approx 3.75 \, \text{L} ]

8. Check units: All temperatures in Kelvin, volumes in liters.

9. Final answer:
   [ V_2 = 3.75 \, \text{L} ]

What we did and why: - Converted °C to K because Charles’ Law only works with absolute temperature. - Used the direct proportion formula to find the new volume. - Ensured units were consistent to avoid errors.

Worked Examples

Example 1 – Basic

Problem: A gas has a volume of 500 mL at 300 K. What is its volume at 450 K?

Solution: 1. Given:
- ( V_1 = 500 \, \text{mL} )
- ( T_1 = 300 \, \text{K} )
- ( T_2 = 450 \, \text{K} )
- ( V_2 = ? )

1. Already in Kelvin – no conversion needed.

2. Rearrange:
   [ V_2 = \frac{V_1 \times T_2}{T_1} ]

3. Plug in:
   [ V_2 = \frac{500 \times 450}{300} ]

4. Calculate:
   [ V_2 = \frac{225,000}{300} = 750 \, \text{mL} ]

Answer: ( V_2 = 750 \, \text{mL} )

What we did and why: - Recognized temperatures were already in Kelvin. - Used direct proportion to find the new volume.

Example 2 – Medium (Missing Initial Temperature)

Problem: A gas expands from 2.0 L to 3.0 L when heated. If the final temperature is 400 K, what was the initial temperature?

Solution: 1. Given:
- ( V_1 = 2.0 \, \text{L} )
- ( V_2 = 3.0 \, \text{L} )
- ( T_2 = 400 \, \text{K} )
- ( T_1 = ? )

1. Rearrange formula:
   [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]
   [ T_1 = \frac{V_1 \times T_2}{V_2} ]

2. Plug in:
   [ T_1 = \frac{2.0 \times 400}{3.0} ]

3. Calculate:
   [ T_1 = \frac{800}{3} \approx 266.67 \, \text{K} ]

Answer: ( T_1 = 267 \, \text{K} ) (rounded to 3 significant figures)

What we did and why: - Solved for the missing initial temperature by rearranging the formula. - Ensured volume units matched before calculating.

Example 3 – Exam Style (Disguised Problem)

Problem: A student inflates a balloon to 1.5 L at room temperature (20°C). The balloon is then placed in a freezer at -10°C. What is the new volume? Assume pressure remains constant.

Solution: 1. Given:
- ( V_1 = 1.5 \, \text{L} )
- ( T_1 = 20°C )
- ( T_2 = -10°C )
- ( V_2 = ? )

1. Convert to Kelvin:
2. ( T_1 = 20 + 273.15 = 293.15 \, \text{K} )

3. ( T_2 = -10 + 273.15 = 263.15 \, \text{K} )

4. Rearrange formula:
   [ V_2 = \frac{V_1 \times T_2}{T_1} ]

5. Plug in:
   [ V_2 = \frac{1.5 \times 263.15}{293.15} ]

6. Calculate:
   [ V_2 = \frac{394.725}{293.15} \approx 1.35 \, \text{L} ]

Answer: ( V_2 = 1.35 \, \text{L} )

What we did and why: - Recognized the "room temperature" and "freezer" as temperature changes. - Converted °C to K before using Charles’ Law. - Calculated the volume decrease due to cooling.

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in—tonight, before your exam. Charles’ Law says: At constant pressure, volume and temperature are directly proportional. That means if temperature goes up, volume goes up. But here’s the catch: temperature must be in Kelvin. So, if you see °C, add 273.15. The formula is ( \frac{V_1}{T_1} = \frac{V_2}{T_2} ). Rearrange it to solve for what’s missing. Always check: Is pressure constant? Are units consistent? Did I convert to Kelvin? If you nail these three things, you’ll get every Charles’ Law question right. Now go practice—you’ve got this!