SME
Grade 10 Science · Problem type
Charles’s law: heating a gas
Use this for a gas warmed or cooled at one pressure: “2.00 L at 300 K is heated to 450 K. What is its volume?”
Diagram
Before and after, V₁/T₁ = V₂/T₂: 2/300 = 3/450.
The pressure and the amount of gas are held.
At
450
K
the
particles
move
√
450
300
≈
1.22
times
as
fast.
Temperature before in °C: 27 °C
Temperature after in °C: 177 °C
This is an example. Type your own values to start a new problem.
V₁ (L)
T₁ (K)
=
V₂ (L)
T₂ (K)
t₁
Temperature before in °C
calculated · °C
t₂
Temperature after in °C
calculated · °C
Formulas
V₁
T₁
=
V₂
T₂
2
300
=
3
450
T₁
=
t₁
+
273
300 = 27 + 273
T₂
=
t₂
+
273
450 = 177 + 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 pressure and the amount of gas stay the same.
Step-by-step
WE KNOW
V₁ = 2 L, T₁ = 300 K, T₂ = 450 K
FIND
volume after (V₂), temperature before in °C (t₁), temperature after in °C (t₂)
Step 1 · Find volume after (V₂)
Use V₁/T₁ = V₂/T₂
At
one
pressure,
volume
grows
in
step
with
the
kelvin
temperature.
Scale
V₁
by
T₂
T₁
.
V₂
=
(
V₁
×
T₂
)/
T₁
V₂
=
(2
×
450)/300
V₂
=
900
300
V₂
=
3
L
Step 2 · Find temperature before in °C (t₁)
Use T₁ = t₁ + 273
Take 273 off the kelvins for °C.
t₁
=
T₁
−
273
t₁
=
300
−
273
t₁
=
27
°C
Step 3 · Find temperature after in °C (t₂)
Use T₂ = t₂ + 273
Take 273 off the kelvins for °C.
t₂
=
T₂
−
273
t₂
=
450
−
273
t₂
=
177
°C
Check
Put the numbers back in. Both sides should match.
2
300
=
3
450
✓
300 = 27 + 273 ✓
450 = 177 + 273 ✓
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Boyle’s law: squeezing a gas
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Gay-Lussac’s law: a sealed, rigid container
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