SME
Engineering Thermodynamics · Problem type
An ideal-gas mixture by mass
Use this for “2 kg of N₂ and 1 kg of CO₂ share a tank at 200 kPa. Find the mole fractions, the mixture’s molar mass and R, and each partial pressure.”
Chart
151.7 + 48.28 = 200 kPa.
This is an example. Type your own values to start a new problem.
y₁
=
m₁ (kg)
M₁ (kg/kmol)
÷
m₁ (kg)
M₁ (kg/kmol)
+
m₂ (kg)
M₂ (kg/kmol)
Mₘ (kg/kmol)
=
(
m₁ (kg)
+
m₂ (kg)
)
÷
m₁ (kg)
M₁ (kg/kmol)
+
m₂ (kg)
M₂ (kg/kmol)
P₁ (kPa)
=
y₁
×
P (kPa)
y₂
Second mole fraction
worked out (not typed)
Rₘ
Mixture gas constant
calculated · kJ/(kg·K)
P₂
Second partial pressure
worked out (not typed) · kPa
Formulas
y₁
=
m₁
÷
M₁
m₁
÷
M₁
+
m₂
÷
M₂
0.7586
=
2 ÷ 28
2 ÷ 28 + 1 ÷ 44
y₂
=
1
−
y₁
0.2414 = 1 − 0.7586
Mₘ
=
m₁
+
m₂
m₁
÷
M₁
+
m₂
÷
M₂
31.8621
=
2 + 1
2 ÷ 28 + 1 ÷ 44
Rₘ
=
8.314
Mₘ
0.2609
=
8.314
31.8621
P₁
=
y₁
×
P
151.7241 = 0.7586 × 200
P₂
=
P
−
P₁
48.2759 = 200 − 151.7241
Assumptions
•
Each gas is ideal and acts as if it filled the whole volume alone (Dalton’s law).
•
Moles of each gas: N = m ÷ M; the mole fractions add to 1.
•
The universal gas constant is Rᵤ = 8.314 kJ/(kmol·K).
Step-by-step
WE KNOW
m₁ = 2 kg, M₁ = 28 kg/kmol, m₂ = 1 kg, M₂ = 44 kg/kmol, P = 200 kPa
FIND
first mole fraction (y₁), second mole fraction (y₂), mixture molar mass (Mₘ), mixture gas constant (Rₘ), first partial pressure (P₁), second partial pressure (P₂)
Step 1 · Find first mole fraction (y₁)
Use y₁ = (m₁ ÷ M₁) ÷ (m₁ ÷ M₁ + m₂ ÷ M₂)
Turn
each
mass
into
kilomoles
(N
=
m
M
);
y₁
is
the
first
gas’s
share
of
all
the
moles.
y₁
=
2 ÷ 28
2 ÷ 28 + 1 ÷ 44
y₁
=
2
28
÷
(0.07143
+
0.02273)
y₁
=
2
28
÷
0.09416
y₁
=
0.07143
0.09416
y₁
=
0.7586207
Step 2 · Find second mole fraction (y₂)
Use y₂ = 1 − y₁
The two mole fractions add to 1.
y₂
=
1
−
0.7586207
y₂
=
0.2414
Step 3 · Find mixture molar mass (Mₘ)
Use Mₘ = (m₁ + m₂) ÷ (m₁ ÷ M₁ + m₂ ÷ M₂)
The mixture’s molar mass is its total mass over its total kilomoles.
Mₘ
=
2 + 1
2 ÷ 28 + 1 ÷ 44
Mₘ
=
3
0.07143 + 0.02273
Mₘ
=
3
0.09416
Mₘ = 31.8621 kg/kmol
Step 4 · Find mixture gas constant (Rₘ)
Use Rₘ = 8.314 ÷ Mₘ
Like any one gas, the mixture’s R is the universal Rᵤ over its molar mass.
Rₘ
=
8.314
31.8621
Rₘ = 0.2609 kJ/(kg·K)
Step 5 · Find first partial pressure (P₁)
Use P₁ = y₁ × P
Dalton’s law: each gas’s partial pressure is its mole fraction of the total.
P₁
=
0.7586207
×
200
P₁
=
151.7241
kPa
Step 6 · Find second partial pressure (P₂)
Use P₂ = P − P₁
The
partial
pressures
add
to
the
total,
so
P₂
is
what
P₁
leaves.
P₂
=
200
−
151.7241
P₂
=
48.2759
kPa
Check
Put the numbers back in. Both sides should match.
0.7586207
=
2 ÷ 28
2 ÷ 28 + 1 ÷ 44
✓
0.2414 = 1 − 0.7586207 ✓
31.8621
=
2 + 1
2 ÷ 28 + 1 ÷ 44
✓
0.2609
=
8.314
31.8621
✓
151.7241 = 0.7586207 × 200 ✓
48.2759 = 200 − 151.7241 ✓
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