Thermodynamics · Chapter 13
Thermodynamics · Chapter 13

Gas Mixtures

Nonreacting gas mixtures require disciplined bookkeeping: select a mass or mole basis, conserve every species, and distinguish total pressure from component partial pressure before evaluating properties.

Learning objectives

  • Convert consistently between mass and mole fractions.
  • Compute apparent molar mass and mixture gas constant.
  • Apply Dalton and Amagat models within ideal-gas limits.
  • Balance species, energy, and entropy in mixture control volumes.
  • Select authoritative ideal- or real-mixture property sources and audit uncertainty.
13–1

Composition of a Gas Mixture: Mass and Mole Fractions

Composition is a declared basis

Mass fraction \(w_i=m_i/m\) and mole fraction \(y_i=N_i/N\) each sum to one, but they are not interchangeable. Molecular-scale gas behavior uses mole fractions; mass-based energy equations usually use mass fractions.

Basis conversion
\[M_m=\sum_i y_iM_i,\qquad w_i=\frac{y_iM_i}{M_m},\qquad R_m=\frac{R_u}{M_m}\]

Physical meaning

Molar mass is the mass carried by one kmol of mixture. A light species can dominate molecule count while contributing little mass, which changes pressure and diffusional behavior.

EXAMPLE 1Converting composition bases
A calibration cylinder contains 1.60 kg of helium and 6.40 kg of nitrogen. Using supplied molar masses 4.00 and 28.0 kg/kmol, determine mole fractions and apparent molar mass.

Recognition strategy

Masses are given but ideal-mixture pressure and property relations use mole fractions, so convert each component through its supplied molar mass before averaging.

Given information

  • Component masses and molar masses are supplied; total mass is 8.00 kg.
  • The target is mole-basis composition and mixture molar mass.

Model and interpretation

This is a closed, nonreacting, single-phase gas mixture at one equilibrium state. Composition is independent of absolute pressure and temperature here; no process, heat/work, KE, PE, entropy, or property table is required.

Step-by-step solution
1. Plan. Convert masses to amounts: \(N_{He}=1.60/4.00=0.400\ \mathrm{kmol}\) and \(N_{N2}=6.40/28.0=0.2286\ \mathrm{kmol}\).
2. Formulate. Sum the amounts to obtain \(N_m=0.6286\ \mathrm{kmol}\); conserve each species because no reaction or leakage occurs.
3. Calculate. Divide by the total: \(y_{He}=0.6364\) and \(y_{N2}=0.3636\), whose sum is exactly one within rounding.
4. Resolve. Compute \(M_m=\sum y_iM_i=12.73\ \mathrm{kg/kmol}\), also confirmed by \(m_m/N_m=8.00/0.6286\).

Verification

Units: Dividing kilograms by kilograms per kmol gives kmol; mole fractions are dimensionless, and their weighted molar masses retain kilograms per kmol.

Signs and direction: Both component amounts and fractions are positive; helium has the larger mole fraction despite lower mass because its molar mass is much smaller.

Plausibility: The fractions sum to 1.0000, and the independent identity m_total/N_total=8.00/0.6286=12.73 kg/kmol matches the weighted molar mass.

Final result: \(y_{He}=0.636\), \(y_{N2}=0.364\), and \(M_m=12.7\ \mathrm{kg/kmol}\).

Engineering interpretation

A modest helium mass becomes the majority of molecules because helium is light. Gas procurement and metering must clearly specify mass, mole, or volume basis and use certified composition data.

13–2

P-v-T Behavior of Gas Mixtures: Ideal and Real Gases

For an ideal mixture, every component shares mixture temperature and volume and exerts \(P_i=y_iP\); equivalently, additive component volumes satisfy \(V_i=y_iV\). These are exact only under the ideal-gas model. At high density, use a validated mixture equation of state with interaction parameters rather than blindly averaging pure-fluid compressibility factors.

Ideal-mixture state relations
\[P=\sum_iP_i,\qquad P_iV=N_iR_uT,\qquad PV=NR_uT\]
EXAMPLE 2Pressure of a heated ideal mixture
A rigid 0.800 m³ vessel contains 0.500 kmol nitrogen and 0.200 kmol argon at 300 K. It is heated to 450 K. Treat both as ideal gases and find final absolute pressure and component partial pressures.

Recognition strategy

A fixed amount in a rigid vessel signals a closed-system state calculation; Dalton partitions the ideal-mixture pressure after the total is found.

Given information

  • Rigid closed volume; initial and final absolute temperatures are 300 K and 450 K.
  • Use supplied universal gas constant \(R_u=8.314\ \mathrm{kPa\!\cdot\!m^3/(kmol\!\cdot\!K)}\).

Model and interpretation

The boundary encloses a fixed nonreacting mixture; volume is constant and the final state is uniform. Ideal-gas behavior is approximate. Heat is positive into the vessel, boundary work is zero, and KE/PE are neglected.

Step-by-step solution
1. Plan. Sum amount \(N=0.700\ \mathrm{kmol}\) and compute mole fractions \(y_{N2}=0.7143\), \(y_{Ar}=0.2857\).
2. Formulate. Apply the final mixture equation \(P_2=NR_uT_2/V\), using absolute temperature and pressure.
3. Calculate. Substitution gives \(P_2=0.700(8.314)(450)/0.800=3274.9\ \mathrm{kPa}\).
4. Resolve. Use Dalton’s law: \(P_{N2}=2339.2\ \mathrm{kPa}\) and \(P_{Ar}=935.7\ \mathrm{kPa}\); their sum closes the pressure balance.

Verification

Units: In NRuT/V, kmol times kPa cubic metres per kmol kelvin times kelvin divided by cubic metres leaves absolute kilopascals.

Signs and direction: Heating at fixed amount and volume raises absolute pressure; both partial pressures are positive because both mole fractions are positive.

Plausibility: Dalton closure gives 2339.2+935.7=3274.9 kPa, and P2/P1=T2/T1=1.50 for the same rigid ideal-gas mixture.

Final result: \(P_2=3.27\ \mathrm{MPa\ abs}\), with \(P_{N2}=2.34\ \mathrm{MPa}\) and \(P_{Ar}=0.936\ \mathrm{MPa}\).

Engineering interpretation

Pressure rises in direct proportion to absolute temperature for fixed ideal-gas amount and volume. At 3.27 MPa, ideality requires verification with authoritative mixture fugacity or compressibility data.

13–3

Properties of Gas Mixtures: Ideal and Real Gases

Ideal-mixture internal energy and enthalpy are amount-weighted sums evaluated at the common temperature. Entropy also includes composition through each component's partial pressure. Mixing distinct ideal gases at common \(T,P\) is irreversible even when no heat or work crosses the outer boundary.

Ideal-mixture properties
\[h_m=\sum_i w_i h_i(T),\qquad u_m=\sum_i w_i u_i(T)\]\[\bar s_m=\sum_i y_i\left[\bar s_i^\circ(T)-R_u\ln\left(\frac{y_iP}{P^\circ}\right)\right]\]
EXAMPLE 3Adiabatic blending temperature
A steady insulated mixer receives 1.20 kg/s nitrogen at 320 K and 0.800 kg/s argon at 500 K. Supplied constant ideal-gas values are \(c_{p,N2}=1.04\) and \(c_{p,Ar}=0.520\ \mathrm{kJ/(kg\!\cdot\!K)}\). Find outlet temperature.

Recognition strategy

Two steady gas streams entering an insulated junction identify a control-volume mixing problem; conserve each species and balance inlet and outlet enthalpy rates.

Given information

  • Both inlet pressures and outlet pressure are positive absolute values low enough for the stipulated ideal-gas model.
  • Mass flow and supplied constant specific heats define a mass-basis enthalpy calculation.

Model and interpretation

A steady open control volume encloses the mixer. It is adiabatic with no shaft work and negligible KE/PE. Heat into and work out are positive. Species do not react; enthalpy depends only on absolute temperature in this approximate model.

Step-by-step solution
1. Plan. Conserve total and species mass: \(\dot m_3=1.20+0.800=2.00\ \mathrm{kg/s}\), with both species leaving in that flow.
2. Formulate. Write the steady energy balance \(\sum\dot m_i c_{p,i}T_i=(\sum\dot m_i c_{p,i})T_3\) using one common enthalpy reference.
3. Calculate. The numerator is \(1.20(1.04)(320)+0.800(0.520)(500)=607.36\ \mathrm{kW}\); heat-capacity rate is \(1.664\ \mathrm{kW/K}\).
4. Resolve. Divide to obtain \(T_3=365.0\ \mathrm{K}\); an exact release calculation would iterate temperature-dependent ideal-gas enthalpy tables.

Verification

Units: Each m-dot cp T term has kg/s times kJ/(kg K) times K, which is kW; dividing kW by the 1.664 kW/K capacity rate gives kelvins.

Signs and direction: Heat and shaft work are zero, so hotter argon loses enthalpy and cooler nitrogen gains it; the common outlet must move in those respective directions.

Plausibility: The 365 K outlet lies strictly between 320 and 500 K, and 1.664 times 365.0 equals the 607.36 kW inlet enthalpy rate within rounding.

Final result: \(T_3=365\ \mathrm{K}\) on the supplied constant-\(c_p\) model.

Engineering interpretation

The outlet lies between inlet temperatures and is weighted by heat-capacity rates, not mass alone. Mixer sizing still requires pressure-loss, mixing-uniformity, material, and measured-composition review.

Problem-solving strategy

  1. Recognize the gas-mixture problem family and identify the requested property or performance measure.
  2. Draw a closed-system or control-volume boundary and label every state with independent properties.
  3. Choose a property source and model before selecting equations; interpolate only between valid tabulated states.
  4. Write conservation and constitutive relations symbolically, count unknowns, and iterate when properties depend on the answer.
  5. Audit dimensions, signs, limiting behavior, conservation, entropy generation, and model applicability.

Chapter recap

  • Declare mass or mole basis and verify fractions sum to one.
  • Use component partial pressure, not total pressure, in component entropy.
  • Conserve each nonreacting species before closing energy and entropy balances.
  • Before moving on, be able to justify ideality, property sources, interpolation, and real-mixture uncertainty.
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