SME
Grade 10 Science · Problem type
Gay-Lussac’s law: a sealed, rigid container
Use this for a gas heated in a rigid container: “A tire at 2.00 atm and 280 K warms to 308 K. What is the pressure?”
Diagram
Before and after, P₁/T₁ = P₂/T₂: 2/280 = 2.2/308.
The volume and the amount of gas are held.
At
308
K
the
particles
move
√
308
280
≈
1.05
times
as
fast.
Temperature before in °C: 7 °C
Temperature after in °C: 35 °C
This is an example. Type your own values to start a new problem.
P₁ (atm)
T₁ (K)
=
P₂ (atm)
T₂ (K)
t₁
Temperature before in °C
calculated · °C
t₂
Temperature after in °C
calculated · °C
Formulas
P₁
T₁
=
P₂
T₂
2
280
=
2.2/308
T₁
=
t₁
+
273
280 = 7 + 273
T₂
=
t₂
+
273
308 = 35 + 273
Assumptions
•
The gas is ideal: its particles take up no room and do not attract each other.
•
Temperatures are in kelvins: add 273 to °C.
•
The volume and the amount of gas stay the same.
Step-by-step
WE KNOW
P₁ = 2 atm, T₁ = 280 K, T₂ = 308 K
FIND
pressure after (P₂), temperature before in °C (t₁), temperature after in °C (t₂)
Step 1 · Find pressure after (P₂)
Use P₁/T₁ = P₂/T₂
In a sealed, rigid container the pressure grows in step with the kelvin temperature.
P₂
=
(
P₁
×
T₂
)/
T₁
P₂
=
(2
×
308)/280
P₂
=
616
280
P₂
=
2.2
atm
Step 2 · Find temperature before in °C (t₁)
Use T₁ = t₁ + 273
Take 273 off the kelvins for °C.
t₁
=
T₁
−
273
t₁
=
280
−
273
t₁
=
7
°C
Step 3 · Find temperature after in °C (t₂)
Use T₂ = t₂ + 273
Take 273 off the kelvins for °C.
t₂
=
T₂
−
273
t₂
=
308
−
273
t₂
=
35
°C
Check
Put the numbers back in. Both sides should match.
2
280
=
2.2/308
✓
280 = 7 + 273 ✓
308 = 35 + 273 ✓
Topic check
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