Air Infiltration Rates into Homes with Windows Closed

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Windows or exterior doors on one wall 1.0
Windows or exterior doors on two walls 1.5
Windows or exterior doors on three walls 2.0

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed

A stack gas contains 7.5 ppm of a compound with molecular weight 64 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 7.5 ppm$
$MW = 64 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 7.5(64)/24.45$
Evaluate — 19.63 mg/m³
Reverse check
$ppm = 19.63(24.45)/64 = 7.5 ✓$

Answer:

≈ 19.6 mg/m³

Why the other options are there

2.9 mg/m³ (ratio inverted)
21.4 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 2

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (2)

A stack gas contains 56.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 56.5 ppm$
$MW = 46 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 56.5(46)/24.45$
Evaluate — 106.3 mg/m³
Reverse check
$ppm = 106.3(24.45)/46 = 56.5 ✓$

Answer:

≈ 106.3 mg/m³

Why the other options are there

30.0 mg/m³ (ratio inverted)
116.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 3

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (3)

A stack gas contains 44.5 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 44.5 ppm$
$MW = 28 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 44.5(28)/24.45$
Evaluate — 50.96 mg/m³
Reverse check
$ppm = 50.96(24.45)/28 = 44.5 ✓$

Answer:

≈ 51.0 mg/m³

Why the other options are there

38.9 mg/m³ (ratio inverted)
55.6 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 4

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (4)

A stack gas contains 51.0 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 51.0 ppm$
$MW = 28 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 51.0(28)/24.45$
Evaluate — 58.40 mg/m³
Reverse check
$ppm = 58.40(24.45)/28 = 51.0 ✓$

Answer:

≈ 58.4 mg/m³

Why the other options are there

44.5 mg/m³ (ratio inverted)
63.8 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 5

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (5)

A stack gas contains 34.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 34.5 ppm$
$MW = 46 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 34.5(46)/24.45$
Evaluate — 64.91 mg/m³
Reverse check
$ppm = 64.91(24.45)/46 = 34.5 ✓$

Answer:

≈ 64.9 mg/m³

Why the other options are there

18.3 mg/m³ (ratio inverted)
70.8 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 6

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (6)

A stack gas contains 11.0 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 11.0 ppm$
$MW = 28 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 11.0(28)/24.45$
Evaluate — 12.60 mg/m³
Reverse check
$ppm = 12.60(24.45)/28 = 11.0 ✓$

Answer:

≈ 12.6 mg/m³

Why the other options are there

9.6 mg/m³ (ratio inverted)
13.8 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 7

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (7)

A stack gas contains 15.0 ppm of a compound with molecular weight 44 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 15.0 ppm$
$MW = 44 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 15.0(44)/24.45$
Evaluate — 26.99 mg/m³
Reverse check
$ppm = 26.99(24.45)/44 = 15.0 ✓$

Answer:

≈ 27.0 mg/m³

Why the other options are there

8.3 mg/m³ (ratio inverted)
29.5 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 8

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (8)

A stack gas contains 23.0 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 23.0 ppm$
$MW = 46 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 23.0(46)/24.45$
Evaluate — 43.27 mg/m³
Reverse check
$ppm = 43.27(24.45)/46 = 23.0 ✓$

Answer:

≈ 43.3 mg/m³

Why the other options are there

12.2 mg/m³ (ratio inverted)
47.2 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 9

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (9)

A stack gas contains 55.5 ppm of a compound with molecular weight 64 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 55.5 ppm$
$MW = 64 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 55.5(64)/24.45$
Evaluate — 145.3 mg/m³
Reverse check
$ppm = 145.3(24.45)/64 = 55.5 ✓$

Answer:

≈ 145.3 mg/m³

Why the other options are there

21.2 mg/m³ (ratio inverted)
158.6 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed

Example 10

Converting ppm to mg/m³ — Air Infiltration Rates into Homes with Windows Closed (10)

A stack gas contains 21.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

$C = 21.5 ppm$
$MW = 46 g/mol$
$Molar volume = 24.45 L/mol$

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$mg/m^{3} = ppm \times MW / 24.45$
Substituting
$mg/m^{3} = 21.5(46)/24.45$
Evaluate — 40.45 mg/m³
Reverse check
$ppm = 40.45(24.45)/46 = 21.5 ✓$

Answer:

≈ 40.4 mg/m³

Why the other options are there

11.4 mg/m³ (ratio inverted)
44.2 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Air Infiltration Rates into Homes with Windows Closed