How to Solve: Boyle’s Law

How to Solve: Boyle’s Law

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

Introduction

"Imagine your bike tire goes flat on a cold morning—why does it feel firmer when you pump it up? Or why does a scuba diver’s tank pressure change as they descend? Master Boyle’s Law, and you’ll solve these real-life problems AND crush exam questions on gas behavior—guaranteed."

What You Need To Know First

Before diving into Boyle’s Law, you must understand: 1. Pressure (P): Force per unit area (e.g., Pascals, atm, mmHg). Think of it as how hard gas molecules push against their container. 2. Volume (V): The space a gas occupies (e.g., liters, mL, cm³). Smaller volume = molecules packed tighter. 3. Temperature (T): Must stay constant for Boyle’s Law to apply. If temperature changes, you’ll need Charles’s or Gay-Lussac’s Law instead.

If you’re shaky on these, pause and review them first—Boyle’s Law won’t make sense otherwise!

Key Vocabulary

Formulas To Know

1. Boyle’s Law Equation

Formula: [ P_1 V_1 = P_2 V_2 ]

Variables: - ( P_1 ) = Initial pressure (any unit, but must match ( P_2 )) - ( V_1 ) = Initial volume (any unit, but must match ( V_2 )) - ( P_2 ) = Final pressure - ( V_2 ) = Final volume

MEMORISE THIS? ✅ Yes! It’s the core of every Boyle’s Law problem.

2. Pressure Unit Conversions (Given on Exam Sheet, but Practice!)

• 1 atm = 101.3 kPa = 760 mmHg = 760 torr
• 1 kPa = 1000 Pa

Pro Tip: If the exam gives you mixed units (e.g., atm and kPa), convert to the same unit first before plugging into ( P_1 V_1 = P_2 V_2 ).

Step-by-Step Method

Follow these exact steps for every Boyle’s Law problem. No shortcuts!

Step 1: Identify What’s Given and What’s Asked

• Read the problem twice.
• Underline or highlight:
• Initial pressure (( P_1 ))
• Initial volume (( V_1 ))
• Final pressure (( P_2 )) or final volume (( V_2 ))
• What’s missing (the unknown you’re solving for).

Example: A gas has a volume of 3.0 L at 2.0 atm. What’s its volume if pressure increases to 4.0 atm? → Given: ( P_1 = 2.0 ) atm, ( V_1 = 3.0 ) L, ( P_2 = 4.0 ) atm → Asked: ( V_2 = ? )

Step 2: Check Units

• Pressure units must match (e.g., both in atm or both in kPa).
• Volume units must match (e.g., both in L or both in mL).
• If they don’t match, convert now.

Example: If ( P_1 = 101.3 ) kPa and ( P_2 = 1.5 ) atm, convert ( P_2 ) to kPa: ( 1.5 ) atm × ( 101.3 ) kPa/atm = 152 kPa.

Step 3: Write Boyle’s Law Equation

[ P_1 V_1 = P_2 V_2 ]

Pro Tip: Write it big and clear at the top of your work. This keeps you from mixing up variables.

Step 4: Plug in Known Values

• Substitute the numbers with units into the equation.
• Leave the unknown as a variable (e.g., ( V_2 )).

Example: ( (2.0 \text{ atm})(3.0 \text{ L}) = (4.0 \text{ atm})(V_2) )

Step 5: Solve for the Unknown

• Isolate the unknown by dividing or multiplying.
• Show every step—examiners love this!

Example: ( 6.0 \text{ atm·L} = 4.0 \text{ atm} \times V_2 ) ( V_2 = \frac{6.0 \text{ atm·L}}{4.0 \text{ atm}} ) ( V_2 = 1.5 \text{ L} )

Step 6: Check Your Answer

• Does it make sense?
• If pressure increases, volume should decrease (and vice versa).
• Units should cancel correctly (e.g., atm cancels, leaving L).
• Estimate: If ( P ) doubles, ( V ) should halve. Does your answer match?

Example: Pressure went from 2.0 atm → 4.0 atm (doubled), so volume should halve (3.0 L → 1.5 L). ✅ Correct!

Step 7: Box Your Final Answer

• Write the answer clearly with units.
• No units = zero marks!

Example: Final Answer: ( V_2 = 1.5 \text{ L} )

Worked Examples

Example 1 – Basic (No Tricks)

Problem: A gas occupies 500 mL at 1.2 atm. What’s its volume if pressure drops to 0.8 atm?

Step-by-Step Solution: 1. Given:
( P_1 = 1.2 ) atm, ( V_1 = 500 ) mL, ( P_2 = 0.8 ) atm
Asked: ( V_2 = ? ) 2. Units match? ✅ (both atm, both mL) 3. Equation: ( P_1 V_1 = P_2 V_2 ) 4. Plug in:
( (1.2 \text{ atm})(500 \text{ mL}) = (0.8 \text{ atm})(V_2) ) 5. Solve:
( 600 \text{ atm·mL} = 0.8 \text{ atm} \times V_2 )
( V_2 = \frac{600}{0.8} )
( V_2 = 750 \text{ mL} ) 6. Check:
Pressure decreased (1.2 → 0.8 atm), so volume should increase (500 → 750 mL). ✅ 7. Final Answer: ( V_2 = 750 \text{ mL} )

What we did and why: We used Boyle’s Law to show that when pressure decreases, volume increases (inverse proportion). The math was straightforward because units matched, and we verified the answer made sense.

Example 2 – Medium (Unit Conversion)

Problem: A scuba tank has a volume of 12 L at 200 atm. If the tank is opened to the atmosphere (1 atm), what volume will the gas occupy?

Step-by-Step Solution: 1. Given:
( P_1 = 200 ) atm, ( V_1 = 12 ) L, ( P_2 = 1 ) atm
Asked: ( V_2 = ? ) 2. Units match? ✅ (both atm, both L) 3. Equation: ( P_1 V_1 = P_2 V_2 ) 4. Plug in:
( (200 \text{ atm})(12 \text{ L}) = (1 \text{ atm})(V_2) ) 5. Solve:
( 2400 \text{ atm·L} = 1 \text{ atm} \times V_2 )
( V_2 = 2400 \text{ L} ) 6. Check:
Pressure decreased (200 → 1 atm), so volume should increase (12 → 2400 L). ✅
Note: This is why scuba tanks release a huge volume of air when opened! 7. Final Answer: ( V_2 = 2400 \text{ L} )

What we did and why: We applied Boyle’s Law to a real-world scenario (scuba diving). The key was recognizing that atmospheric pressure is much lower than tank pressure, leading to a massive volume increase.

Example 3 – Exam Style (Disguised Problem)

Problem: A syringe contains 30 cm³ of air at 100 kPa. If the plunger is pushed in until the volume is 10 cm³, what’s the new pressure? Assume temperature is constant.

Step-by-Step Solution: 1. Given:
( V_1 = 30 ) cm³, ( P_1 = 100 ) kPa, ( V_2 = 10 ) cm³
Asked: ( P_2 = ? ) 2. Units match? ✅ (both cm³, both kPa) 3. Equation: ( P_1 V_1 = P_2 V_2 ) 4. Plug in:
( (100 \text{ kPa})(30 \text{ cm}³) = (P_2)(10 \text{ cm}³) ) 5. Solve:
( 3000 \text{ kPa·cm}³ = P_2 \times 10 \text{ cm}³ )
( P_2 = \frac{3000}{10} )
( P_2 = 300 \text{ kPa} ) 6. Check:
Volume decreased (30 → 10 cm³), so pressure should increase (100 → 300 kPa). ✅ 7. Final Answer: ( P_2 = 300 \text{ kPa} )

What we did and why: This was a "syringe problem," a common exam question. The trick was recognizing that pushing the plunger decreases volume, which increases pressure. We used Boyle’s Law directly and verified the answer made sense.

Common Mistakes

Avoid these 5 errors that cost marks!

Exam Traps

Examiners love these 3 traps. Spot them and avoid losing marks!

1-Minute Recap

"Alright, let’s lock this in. Boyle’s Law is simple: for a gas at constant temperature, pressure and volume are inversely proportional. That means if pressure goes up, volume goes down—and vice versa. The formula is ( P_1 V_1 = P_2 V_2 ). Always start by writing down what you know, check your units, and plug into the equation. If pressure or volume changes, just solve for the missing piece. Watch out for unit traps—convert everything to the same unit first. And remember: if temperature changes, Boyle’s Law doesn’t apply. Double-check your answer—does it make sense? If pressure doubled, volume should halve. That’s it. Practice a few problems tonight, and you’ll own this on exam day. You’ve got this!