SME
Grade 10 Science · Chemistry
Gases and the gas laws (PV = nRT)
Use this for “2 mol of gas at 2 atm and 27 °C. What volume does it fill?” with PV = nRT.
Diagram
R = 0.0821 L·atm/(mol·K).
PV = nRT, which gives 2 × 24.63 = 2 × 0.0821 × 300 = 49.26.
Each particle drawn stands for 0.05 mol: 40 particles.
Temperature in °C: 27 °C
This is an example. Type your own values to start a new problem.
P (atm)
×
V (L)
=
n (mol)
×
0.0821
×
T (K)
t
Temperature in °C
calculated · °C
Formulas
P
×
V
=
n
×
0.0821
×
T
2 × 24.63 = 2 × 0.0821 × 300
T
=
t
+
273
300 = 27 + 273
Assumptions
•
Temperature must be in kelvins: add 273 to °C. R = 0.0821 L·atm/(mol·K), so P is in atm and V in liters.
•
Equal volumes of any gases at the same T and P hold the same number of particles.
•
Real gases stray from this at high pressure and low temperature.
Step-by-step
WE KNOW
n = 2 mol, T = 300 K, V = 24.63 L
FIND
pressure (P), temperature in °C (t)
Step 1 · Find pressure (P)
Use P × V = n × 0.0821 × T
Divide
both
sides
by
V
.
With
R
=
0.0821,
the
pressure
comes
out
in
atmospheres.
P
=
(
n
×
0.0821
×
T
)/
V
P
=
(2
×
0.0821
×
300)/24.63
P
=
0.1642
×
300/24.63
P
=
49.26/24.63
P
=
2
atm
Step 2 · Find temperature in °C (t)
Use T = t + 273
Take 273 off the kelvins for °C.
t
=
T
−
273
t
=
300
−
273
t
=
27
°C
Check
Put the numbers back in. Both sides should match.
2 × 24.63 = 2 × 0.0821 × 300 ✓
300 = 27 + 273 ✓
Topic check
See what you know: 5 questions on this topic
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