13.3 First and second-law analysis of gas mixtures

Thermodynamics of Ideal Gas Mixtures

First Law of Thermodynamics for Ideal Gas Mixtures

The first law for a gas mixture works the same way as for a pure gas: the change in internal energy equals the heat added minus the work done by the system. The difference is that mixture properties must be built from the individual component properties.

For an ideal gas mixture, internal energy depends only on temperature. You calculate the mixture's specific heats from the mass fractions (or mole fractions) of each component:

$c_{v,\text{mix}} = \sum_i m f_i \, c_{v,i} \qquad c_{p,\text{mix}} = \sum_i m f_i \, c_{p,i}$

where $mf_i$ is the mass fraction of component $i$. If you're working on a molar basis, use mole fractions $y_i$ with molar specific heats $\bar{c}_v$ and $\bar{c}_p$ instead.

Once you have the mixture specific heats, the first law applies to all the standard processes: isothermal, isobaric, isochoric, and adiabatic.

Analyzing Processes with Ideal Gas Mixtures

The general approach for any process is the same three steps:

1. Find the change in internal energy. Use $\Delta U = m \, c_{v,\text{mix}} \, (T_2 - T_1)$ for the mixture.

2. Find the work done. This depends on the process path and the pressure-volume relationship.

3. Apply the first law to get the heat transfer: $Q = \Delta U + W$ (sign convention: $W$ is work done by the system).

Here's how each standard process simplifies:

• Isothermal (constant $T$): $\Delta U = 0$, so $Q = W$.
• Isobaric (constant $P$): $W = P(V_2 - V_1)$, and $Q = m \, c_{p,\text{mix}} \, (T_2 - T_1)$.
• Isochoric (constant $V$): $W = 0$, so $Q = \Delta U = m \, c_{v,\text{mix}} \, (T_2 - T_1)$.
• Adiabatic ($Q = 0$): $W = -\Delta U$, and the process follows $PV^{k_\text{mix}} = \text{const}$ for a reversible path, where $k_\text{mix} = c_{p,\text{mix}} / c_{v,\text{mix}}$.

Work, Heat, and Internal Energy Changes

Calculating Work Done by Ideal Gas Mixtures

Work is found by integrating $W = \int P \, dV$ along the process path. For reversible processes with an ideal gas mixture of total moles $n$:

Example: A 2-component mixture (0.5 kmol total) expands isothermally at 300 K from 1 L to 2 L.

$W = (0.5)(8.314)(300)\ln\!\left(\frac{2}{1}\right) = 864 \;\text{J}$

Because $\Delta U = 0$ for an isothermal ideal-gas process, the heat transfer equals 864 J as well.

Determining Internal Energy Changes and Heat Transfer

The change in internal energy for the mixture is always:

$\Delta U = m \, c_{v,\text{mix}} \, (T_2 - T_1)$

Once you know $\Delta U$ and $W$, the first law gives $Q$. The table below summarizes which quantity vanishes in each standard process, making the algebra straightforward:

Entropy Changes and Second Law Implications

Entropy Changes in Ideal Gas Mixtures

Entropy measures the microscopic disorder of a system. For a gas mixture, the entropy change of each component is calculated independently, then summed.

For an ideal gas component $i$ going from state 1 to state 2, the entropy change per unit mass is:

$\Delta s_i = c_{p,i} \ln\!\left(\frac{T_2}{T_1}\right) - R_i \ln\!\left(\frac{P_{i,2}}{P_{i,1}}\right)$

where $P_{i}$ is the partial pressure of component $i$ (not the total mixture pressure). The total mixture entropy change is:

$\Delta S_\text{mix} = \sum_i m_i \, \Delta s_i$

For the standard reversible processes, this simplifies:

• Isothermal: $\Delta S = Q / T$ (temperature is constant, so entropy change comes entirely from heat transfer).
• Isobaric: $\Delta S = m \, c_{p,\text{mix}} \ln(T_2 / T_1)$.
• Isochoric: $\Delta S = m \, c_{v,\text{mix}} \ln(T_2 / T_1)$.
• Adiabatic and reversible (isentropic): $\Delta S = 0$.

Second Law Analysis of Ideal Gas Mixture Processes

The second law requires that the total entropy of an isolated system can never decrease:

$S_\text{gen} = \Delta S_\text{system} + \Delta S_\text{surroundings} \geq 0$

• If $S_\text{gen} = 0$, the process is reversible.
• If $S_\text{gen} > 0$, the process is irreversible.

Entropy of mixing is a particularly important result for gas mixtures. When two ideal gases at the same $T$ and $P$ are allowed to mix, the entropy change is always positive, even though there's no heat transfer:

$\Delta S_\text{mixing} = -n R_u \sum_i y_i \ln(y_i)$

Each $\ln(y_i)$ term is negative (since $y_i < 1$), so the overall expression is positive. This confirms that mixing is inherently irreversible.

Reversibility vs. Irreversibility of Processes

Reversible Processes

A reversible process passes through a continuous series of equilibrium states and can be completely undone without any net effect on the system or surroundings. In practice, no real process is truly reversible, but the concept sets an upper bound on performance.

You can identify a reversible process by checking that $S_\text{gen} = 0$. This means none of the following dissipative effects are present:

• Friction
• Heat transfer across a finite temperature difference
• Unrestrained expansion
• Mixing of different substances

Reversible versions of the four standard processes (isothermal, isobaric, isochoric, adiabatic) serve as idealized benchmarks. For example, a reversible adiabatic expansion produces the maximum possible work output for given initial and final pressures.

Irreversible Processes

Any real process has some degree of irreversibility. The amount of irreversibility is quantified by the entropy generation $S_\text{gen}$, or equivalently by the exergy destruction:

X_\text{destroyed} = T_0 \, S_\text{gen}}

where $T_0$ is the dead-state (surroundings) temperature. Exergy destruction represents useful work potential that's permanently lost.

Common irreversible processes in gas mixtures:

• Throttling (unrestrained expansion): Enthalpy stays constant for an ideal gas, but entropy increases. No work is recovered.
• Mixing at different temperatures or pressures: Entropy of mixing is always positive, and additional entropy is generated if the components start at different temperatures.
• Heat transfer through a finite $\Delta T$: The larger the temperature difference, the greater the entropy generation.
• Friction in pipe flow: Converts organized kinetic energy into internal energy, increasing entropy.

To reduce irreversibility in practice, you minimize entropy generation: use counterflow heat exchangers to reduce $\Delta T$, avoid throttling where an expander could recover work, and reduce friction through proper duct sizing.