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Grade 10 Science · Problem type

Boyle’s law: squeezing a gas

Use this for a gas squeezed or let out at one temperature: “6 L at 1 atm is pressed into 3 L. What is the pressure?”

Diagram

Before02468LKP₁ = 1 atmV₁ = 6 LT₁ heldAfter02468LKP₂ = 2 atmV₂ = 3 LT₂ held
Before and after, P₁V₁ = P₂V₂: 1 × 6 = 2 × 3 = 6
The temperature and the amount of gas are held.
This is an example. Type your own values to start a new problem.
P₁ (atm)
×
V₁ (L)
=
P₂ (atm)
×
V₂ (L)

Formulas

P₁
×
V₁
=
P₂
×
V₂
1 × 6 = 2 × 3

Assumptions

•
The gas is ideal: its particles take up no room and do not attract each other.
•
Pressures are in atm; both volumes are in the same unit.
•
The temperature and the amount of gas stay the same.

Step-by-step

WE KNOW
P₁ = 1 atm, V₁ = 6 L, V₂ = 3 L
FIND
pressure after (P₂)
Step 1 · Find pressure after (P₂)
Use P₁ × V₁ = P₂ × V₂
At
one
temperature,
pressure
times
volume
stays
the
same.
Divide
P₁
V₁
by
the
new
volume.
P₂
=
(
P₁
×
V₁
)/
V₂
P₂
=
(1
×
6)/3
P₂
=
2
atm
Check
Put the numbers back in. Both sides should match.
1 × 6 = 2 × 3 ✓
Topic check
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