13.1 Composition of gas mixtures

Composition of Gas Mixtures

Describing the composition of a gas mixture is the first step toward predicting how that mixture will behave thermodynamically. Three quantities do most of the heavy lifting: mass fraction, mole fraction, and volume fraction. Once you can move between these, you can calculate mixture properties like apparent molecular weight and density.

Mass, Mole, and Volume Fractions in Gas Mixtures

Defining and Differentiating Fractions

Mass fraction ($mf_i$) is the mass of component $i$ divided by the total mass of the mixture. Mole fraction ($y_i$) is the number of moles of component $i$ divided by the total moles. Volume fraction is the volume a component would occupy (at the mixture's $T$ and $P$) divided by the total volume.

All three share two properties:

• They are dimensionless.
• They must sum to 1 across all components in the mixture.

For an ideal-gas mixture, the volume fraction of a component equals its mole fraction. This is a direct consequence of the ideal gas law: at the same $T$ and $P$, volume is proportional to the number of moles. So in ideal-gas problems, you can use mole fraction and volume fraction interchangeably.

Interrelationships and Conversions

You can convert between fractions using the molecular weight ($M$) of each component.

Mass fraction → Mole fraction:

1. For each component, divide its mass fraction by its molecular weight. This gives a quantity proportional to its moles: $mf_i / M_i$.
2. Sum these values for all components: $\sum (mf_i / M_i)$.
3. The mole fraction of component $i$ is:

$y_i = \frac{mf_i / M_i}{\sum_j (mf_j / M_j)}$

Mole fraction → Mass fraction:

1. For each component, multiply its mole fraction by its molecular weight: $y_i \times M_i$.
2. Sum these products for all components: $\sum (y_j \times M_j)$.
3. The mass fraction of component $i$ is:

$mf_i = \frac{y_i \times M_i}{\sum_j (y_j \times M_j)}$

Mole fraction → Volume fraction (ideal gas):

For ideal-gas mixtures, the volume fraction equals the mole fraction. No extra calculation is needed. If the gas is not ideal, you'd need the equation of state for each component to find its partial volume, but that situation is beyond the scope of this unit.

Calculating Component Fractions in Gas Mixtures

Calculating Mass Fraction

Divide the mass of each component by the total mass of the mixture.

Example: A mixture contains 3 kg of $N_2$ and 2 kg of $O_2$.

• Total mass = 3 + 2 = 5 kg
• Mass fraction of $N_2$ = 3 / 5 = 0.60
• Mass fraction of $O_2$ = 2 / 5 = 0.40
• Check: 0.60 + 0.40 = 1.00 ✓

Make sure all masses are in the same units before dividing.

Calculating Mole Fraction

First convert each component's mass to moles using $n = m / M$, then divide by total moles.

Example: A mixture contains 50 g of $CO_2$ ($M = 44$ g/mol) and 25 g of $CH_4$ ($M = 16$ g/mol).

1. $n_{CO_2} = 50 / 44 = 1.136$ mol
2. $n_{CH_4} = 25 / 16 = 1.5625$ mol
3. Total moles = 1.136 + 1.5625 = 2.699 mol
4. $y_{CO_2} = 1.136 / 2.699 = 0.421$
5. $y_{CH_4} = 1.5625 / 2.699 = 0.579$
6. Check: 0.421 + 0.579 = 1.00 ✓

Calculating Volume Fraction

For an ideal-gas mixture, the volume fraction equals the mole fraction. You can verify this by computing each component's volume from the ideal gas law and dividing by the total.

Example: 10 g of He ($M = 4$ g/mol) and 20 g of Ar ($M = 40$ g/mol) at 300 K and 1 atm.

1. $n_{He} = 10 / 4 = 2.5$ mol

2. $n_{Ar} = 20 / 40 = 0.5$ mol

3. Using $V = nRT/P$ with $R = 0.08206$ L·atm/(mol·K):

   • $V_{He} = (2.5 \times 0.08206 \times 300) / 1 = 61.5$ L
   • $V_{Ar} = (0.5 \times 0.08206 \times 300) / 1 = 12.3$ L

4. Total volume = 61.5 + 12.3 = 73.8 L

5. Volume fraction of He = 61.5 / 73.8 = 0.833

6. Volume fraction of Ar = 12.3 / 73.8 = 0.167

Notice these match the mole fractions: $y_{He} = 2.5/3.0 = 0.833$ and $y_{Ar} = 0.5/3.0 = 0.167$. That's the ideal-gas equivalence at work.

Ensuring Consistent Units

• Convert all masses to the same unit (all grams or all kilograms) before starting.
• Use matching units for $R$, $P$, $T$, and $V$ in the ideal gas law. If $R = 0.08206$ L·atm/(mol·K), then pressure must be in atm, temperature in K, and volume in L.
• A common mistake is mixing °C and K. Always convert to absolute temperature (K) for gas law calculations.

Apparent Molecular Weight of Gas Mixtures

Definition and Significance

The apparent (or average) molecular weight of a gas mixture, $M_{mix}$, is the molecular weight of a hypothetical single gas that would behave the same as the mixture at the same $T$ and $P$. It lets you use the ideal gas law on the mixture as if it were a pure substance:

$PV = n_{total} R T = \frac{m_{total}}{M_{mix}} RT$

Calculating Apparent Molecular Weight Using Mole Fractions

This is the more common and straightforward formula:

$M_{mix} = \sum_i y_i \, M_i$

Example: 75 mol% $CH_4$ ($M = 16$ g/mol) and 25 mol% $C_2H_6$ ($M = 30$ g/mol).

$M_{mix} = 0.75 \times 16 + 0.25 \times 30 = 12.0 + 7.5 = 19.5 \text{ g/mol}$

Calculating Apparent Molecular Weight Using Mass Fractions

When you're given mass fractions instead, the formula is:

$\frac{1}{M_{mix}} = \sum_i \frac{mf_i}{M_i}$

Example: 60% $N_2$ ($M = 28$ g/mol) and 40% $O_2$ ($M = 32$ g/mol) by mass.

$\frac{1}{M_{mix}} = \frac{0.60}{28} + \frac{0.40}{32} = 0.02143 + 0.01250 = 0.03393$

$M_{mix} = 1 / 0.03393 = 29.47 \text{ g/mol}$

Application in Density and Specific Volume Calculations

Once you have $M_{mix}$, the ideal gas law gives you mixture density and specific volume directly.

Density:

$\rho = \frac{P \, M_{mix}}{R_u \, T}$

where $R_u = 8.314$ kJ/(kmol·K) is the universal gas constant (use consistent units).

Specific volume:

$v = \frac{R_u \, T}{P \, M_{mix}}$

Example: $M_{mix} = 29.47$ g/mol at 300 K and 1 atm.

$\rho = \frac{1 \text{ atm} \times 29.47 \text{ g/mol}}{0.08206 \text{ L·atm/(mol·K)} \times 300 \text{ K}} = \frac{29.47}{24.62} = 1.197 \text{ g/L}$

$v = 1/\rho = 0.835 \text{ L/g}$

Analyzing Gas Mixture Composition with Psychrometrics

Psychrometric Charts and Tables

Psychrometrics deals specifically with moist air, which is a binary gas mixture of dry air and water vapor. This is a direct application of gas mixture composition concepts.

Two key terms (they mean the same thing):

• Specific humidity (or humidity ratio), $\omega$: mass of water vapor per unit mass of dry air (kg water / kg dry air).

Psychrometric charts plot $\omega$ on the vertical axis against dry-bulb temperature on the horizontal axis. Overlaid on the chart are curves of constant:

• Relative humidity
• Wet-bulb temperature
• Specific volume
• Enthalpy

These charts let you read off multiple moist-air properties from just two known values.

Psychrometric tables provide the same information in numerical form for various conditions.

Analyzing Moist Air Composition

To find the state of moist air on a psychrometric chart:

1. Identify two known properties (e.g., dry-bulb temperature and relative humidity).
2. Locate the intersection of those two lines on the chart.
3. Read off the remaining properties at that point: humidity ratio, wet-bulb temperature, specific volume, and enthalpy.

Example: Moist air at 25°C dry-bulb and 50% relative humidity.

• Humidity ratio: $\omega \approx 0.0098$ kg water / kg dry air
• Wet-bulb temperature: ≈ 18.5°C
• Specific volume: ≈ 0.855 m³/kg dry air

Applications of Psychrometric Analysis

Psychrometric analysis shows up in two major areas:

• HVAC systems: Determining how much cooling and dehumidification is needed to reach target indoor conditions. Equipment selection depends on knowing the state of the incoming and desired air.
• Industrial drying: Moist air is used as a drying medium. Psychrometric data helps optimize temperature and humidity for efficient moisture removal and estimate energy requirements.