Ideal gas law, partial pressure and gas mixture calculations

Ideal gas law with absolute units, normal and standard volumes, gas density, Dalton and Amagat laws for mixtures, and when to use a compressibility factor.

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Why it matters

Most plant gas streams – fuel gas, flue gas, reactor feeds, compressed air – are metered as volumes at some pressure and temperature, but balances are written in moles. The ideal gas law and Dalton's and Amagat's laws let you move between volume, moles, mass, density and partial pressure, and they are the starting point for humidity, combustion and vapour-liquid equilibrium calculations.

Key ideas

Ideal gas model. Molecules are treated as point masses with no intermolecular forces. The model is good (within a few per cent) at low pressure and at temperatures well above the critical temperature; it fails near condensation and at high pressure, where a compressibility factor Z or an equation of state (van der Waals, Peng–Robinson) is needed.

Absolute units only. P must be absolute pressure and T absolute temperature (K). Gauge pressure and °C give nonsense.

Standard conditions. Gas flows are often quoted at a reference state. Normal conditions (0 °C, 1 atm): 1 kmol occupies 22.414 m³ (Nm³). Other references are used (15 °C or 20 °C, 1 atm; 25 °C, 1 bar), so always check which one a data sheet uses. To convert, use the ideal gas law between the two states – moles stay the same.

Gas density. ρ = P·M / (R·T). Density rises with pressure and falls with temperature; a light gas (H₂) and a heavy gas (CO₂) differ in density in proportion to M.

Mixtures – Dalton's law. Each component exerts its partial pressure p_i = y_i·P as if it alone filled the whole volume at the mixture temperature; the partial pressures add to the total pressure.

Amagat's law. The pure-component volume v_i = y_i·V is the volume component i would occupy alone at the mixture P and T; these add to the total volume. Hence for ideal gases mole fraction = pressure fraction = volume fraction.

Average molar mass of a gas mixture, M_avg = Σ y_i·M_i, is used in the density formula.

Real gases. PV = Z·n·R·T, where Z = 1 for an ideal gas. Z is read from the generalised compressibility chart using reduced properties T_r = T / T_c and P_r = P / P_c (from your data book). The van der Waals equation (P + a·n²/V²)(V − n·b) = n·R·T corrects for attraction (a) and finite molecular volume (b).

Formulas

  • P·V = n·R·T – P absolute pressure (Pa), V volume (m³), n amount (mol), T absolute temperature (K), R = 8.314 J/(mol·K) = 8.314 kPa·m³/(kmol·K) = 0.08206 L·atm/(mol·K).
  • P₁·V₁ / T₁ = P₂·V₂ / T₂ – same amount of gas between two states.
  • ρ = P·M / (R·T) – ρ in kg/m³ with P in kPa, M in kg/kmol, R = 8.314 kPa·m³/(kmol·K).
  • p_i = y_i·P and Σ p_i = P – Dalton's law, ideal gases.
  • v_i = y_i·V and Σ v_i = V – Amagat's law, ideal gases.
  • M_avg = Σ y_i·M_i
  • V_m = 22.414 m³/kmol at 0 °C and 101.325 kPa.
  • P·V = Z·n·R·T – real gas; T_r = T / T_c, P_r = P / P_c.

Worked examples

Example 1 (standard). Find the density of CO₂ (M = 44.01 kg/kmol) at 200 kPa absolute and 50 °C, assuming ideal gas behaviour.

  1. Absolute temperature: T = 50 + 273.15 = 323.15 K.
  2. Formula: ρ = P·M / (R·T).
  3. Substitute: ρ = (200 kPa × 44.01 kg/kmol) / (8.314 kPa·m³/(kmol·K) × 323.15 K) = 8802 / 2686.7.
  4. ρ = 3.28 kg/m³.

Answer: ρ ≈ 3.28 kg/m³

Example 2 (GATE level). A natural gas (80 mol % CH₄, M_avg = 18.74 kg/kmol) flows at 1000 Nm³/h (0 °C, 1 atm). It is delivered at 5 bar gauge and 40 °C; atmospheric pressure is 1.01325 bar. Find (a) the molar and mass flow rates, (b) the actual volumetric flow at delivery conditions and (c) the partial pressure of CH₄ in the line.

  1. Molar flow: n = 1000 Nm³/h / 22.414 Nm³/kmol = 44.61 kmol/h.
  2. Mass flow: ṁ = 44.61 × 18.74 = 836 kg/h.
  3. Absolute line pressure: P = 5 + 1.01325 = 6.013 bar = 601.3 kPa; T = 313.15 K.
  4. Actual volume: V = n·R·T / P = 44.61 × 8.314 × 313.15 / 601.3 = 193.2 m³/h.
  5. Check by ratio: V = 1000 × (101.325 / 601.3) × (313.15 / 273.15) = 193.2 m³/h. ✓
  6. Partial pressure of CH₄: p = y·P = 0.80 × 6.013 bar = 4.81 bar.

Answer: 44.6 kmol/h, 836 kg/h; 193 m³/h actual; p_CH₄ = 4.81 bar

Common mistakes

  • Using gauge pressure or °C in PV = nRT.
  • Using the wrong R for the unit set (8.314 with kPa and L gives mol × 10⁻³ errors).
  • Confusing Nm³ (0 °C, 1 atm) with standard m³ at 15 °C or 20 °C – a 5–7 % error.
  • Assuming the partial pressure of a component equals its mass fraction × P; it is the mole fraction.
  • Applying the ideal gas law to a vapour close to saturation or a gas at very high pressure without checking Z.
  • Using an overdetermined problem's data inconsistently: fixing n, V and T already fixes P.

For GATE CH

Ideal gas relations appear inside almost every process-calculation numerical: converting Nm³ to kmol, finding actual volumes after compression or heating, partial pressures for humidity and VLE, and gas densities. Expect NAT questions where the trap is gauge pressure, the reference state or a unit mismatch in R. Practise solving with the ratio form P₁V₁/T₁ = P₂V₂/T₂ as a quick check.

Quick check

  1. What volume does 1 kmol of ideal gas occupy at 0 °C and 1 atm?
  2. Air at 2 bar absolute contains 21 mol % O₂. What is the O₂ partial pressure?
  3. If T (in K) doubles at constant P, what happens to the gas density?
  4. A gas at 1 bar is compressed isothermally to 4 bar. By what factor does its volume change?
  5. In the real-gas equation PV = ZnRT, what value of Z means ideal behaviour?

Answers: 1. 22.414 m³; 2. 0.42 bar; 3. it halves; 4. it becomes one quarter; 5. Z = 1.

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is the ideal gas law and how is it expressed mathematically?Concept

    The ideal gas law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. It is expressed mathematically as PV = nRT, where P is the pressure of the gas, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

  2. 2.Explain the concept of partial pressure in a gas mixture.Concept

    Partial pressure is the pressure that a single gas in a mixture of gases would exert if it occupied the entire volume by itself at the same temperature. It is an important concept in gas mixtures and is used to calculate the total pressure of the mixture using Dalton's Law of Partial Pressures, which states that the total pressure is the sum of the partial pressures of all individual gases.

  3. 3.How does Dalton's Law of Partial Pressures apply to gas mixtures?Concept

    Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. This law is useful for calculating the total pressure of a gas mixture when the partial pressures of the components are known, or vice versa.

  4. 4.Why is the ideal gas law used in chemical engineering processes?Application

    The ideal gas law is used in chemical engineering processes because it provides a simple relationship between pressure, volume, temperature, and moles of gas, which are critical parameters in designing and analyzing processes involving gases. Although real gases deviate from ideal behavior, the ideal gas law is often a good approximation under many conditions, especially at high temperatures and low pressures.

  5. 5.What happens to the pressure of a gas if its volume is halved while keeping the temperature constant?Application

    If the volume of a gas is halved while keeping the temperature constant, the pressure of the gas will double. This is based on Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature is held constant (P1V1 = P2V2).

  6. 6.How would you calculate the partial pressure of a gas in a mixture if you know its mole fraction and the total pressure?Application

    The partial pressure of a gas in a mixture can be calculated using the formula: Partial Pressure = Mole Fraction × Total Pressure. The mole fraction is the ratio of the number of moles of the gas to the total number of moles in the mixture.

  7. 7.A 10 L container holds a mixture of oxygen and nitrogen at a total pressure of 2 atm. If the mole fraction of oxygen is 0.3, what is the partial pressure of oxygen?Numerical

    To find the partial pressure of oxygen, use the formula: Partial Pressure = Mole Fraction × Total Pressure. Here, the mole fraction of oxygen is 0.3 and the total pressure is 2 atm. Therefore, the partial pressure of oxygen is 0.3 × 2 atm = 0.6 atm.

  8. 8.Calculate the number of moles of an ideal gas contained in a 5 L vessel at 300 K and 1 atm pressure.Numerical

    Using the ideal gas law PV = nRT, where P = 1 atm, V = 5 L, R = 0.0821 L·atm/mol·K, and T = 300 K, we can solve for n (number of moles): n = PV / RT = (1 atm × 5 L) / (0.0821 L·atm/mol·K × 300 K) = 0.203 moles.

  9. 9.Explain why real gases deviate from ideal gas behavior.Concept

    Real gases deviate from ideal gas behavior due to intermolecular forces and the finite volume occupied by gas molecules. At high pressures and low temperatures, these factors become significant, causing deviations from the ideal gas law. The ideal gas law assumes no intermolecular forces and that gas molecules occupy no volume, which is not true for real gases.

  10. 10.What modifications are made to the ideal gas law to account for real gas behavior?Application

    To account for real gas behavior, the Van der Waals equation is used, which modifies the ideal gas law by including terms for intermolecular forces and molecular volume. The equation is (P + a(n/V)²)(V - nb) = nRT, where a and b are constants specific to each gas that account for intermolecular forces and the volume occupied by gas molecules, respectively.

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