O₂/N₂ Ideal-Gas Mixture — Constant-Pressure Heating: 04-BS-10, December 2014, Q5

Question 5: O₂/N₂ Ideal-Gas Mixture — Constant-Pressure Heating (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $m_{O_2}=10$ kg, $m_{N_2}=15$ kg, $P=5$ MPa (constant), $T_1=300$ K, $T_2=450$ K.

Find. $Q$ [kJ].

Approach

Since composition and total pressure are both fixed, the mixture's total enthalpy change is simply the mass-weighted sum of each pure species' own enthalpy change (no mixing enthalpy for ideal gases); because the process is constant-pressure with no other work than boundary work, the heat transfer equals that total enthalpy change directly.

Per-species molar enthalpy change, 300→450 K. The question directs that the mixture be treated as an ideal gas, so each species' enthalpy is a function of temperature alone and is read from the ideal-gas property tables supplied with the paper (Tables A-17 nitrogen and A-18 oxygen), not from a real-fluid chart at the 5 MPa cylinder pressure: $\text{O}_2$: $\bar h_1=8{,}736$ kJ/kmol, $\bar h_2=13{,}228$ kJ/kmol, $\Delta\bar h_{O_2}=4{,}492$ kJ/kmol. $\text{N}_2$: $\bar h_1=8{,}723$ kJ/kmol, $\bar h_2=13{,}105$ kJ/kmol, $\Delta\bar h_{N_2}=4{,}382$ kJ/kmol.
Convert to a mass basis using the molar masses from Table A-1 ($M_{O_2}=31.999$, $M_{N_2}=28.013$ kg/kmol): $$\Delta h_{O_2}=\frac{4492}{31.999}=140.38\ \text{kJ/kg},\qquad \Delta h_{N_2}=\frac{4382}{28.013}=156.43\ \text{kJ/kg}.$$
Total heat transfer. Constant $P$ ⇒ $Q=\Delta H$ (no other work terms; the mixing enthalpy is zero for ideal gases at fixed composition): $$Q=m_{O_2}\Delta h_{O_2}+m_{N_2}\Delta h_{N_2}=10\times140.38+15\times156.43=1403.8+2346.4=\boxed{3750\ \text{kJ}}.$$

The ideal-gas treatment the question specifies is what makes the 5 MPa pressure irrelevant to the answer: it appears only in the statement of the state, never in the enthalpy evaluation. Evaluating the same two enthalpy changes from a real-fluid equation of state at 5 MPa instead would give 147.57 and 163.81 kJ/kg and a total near 3933 kJ — about 4.9% higher, because at 5 MPa and 300 K both gases carry a non-negligible residual (departure) enthalpy that the ideal-gas model, and therefore the marking scheme, excludes. The supplied tables are the authority here.

Quantity	Result
$Q$	3750 kJ

Concept & Theory

Topic. Constant-pressure heating of a fixed-composition ideal-gas mixture (Amagat/Dalton mixture model).

Key relations. Ideal-gas mixture enthalpy: $H_{mix}=\sum m_i h_i(T)$, with no mixing-enthalpy correction. At constant $P$ with only boundary work: $Q=\Delta H$.

Why this works. For an ideal gas, enthalpy depends on temperature alone, and the Gibbs–Dalton model states that a mixture's total enthalpy is exactly the sum of what each component would have alone at the mixture temperature — this holds regardless of the (high, 5 MPa) total pressure, since the ideal-gas enthalpy assumption is a property of each pure species considered alone, not of the total mixture pressure.

Common pitfall. Evaluating the two enthalpy changes from a real-fluid chart or equation of state at the stated 5 MPa. The question explicitly says to assume ideal-gas behaviour, and the paper supplies ideal-gas tables for both species; a 5 MPa real-gas evaluation adds a departure-enthalpy term and inflates the answer by roughly 5% (3933 kJ instead of 3750 kJ). A second, smaller pitfall is defining a single blended "mixture $c_p$" and multiplying by total mass and $\Delta T$ — workable for a constant-$c_p$ estimate, but it discards the temperature variation the supplied $\bar h(T)$ tables already carry.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 13 (Gas Mixtures).

Question 5 of 9: O₂/N₂ Ideal-Gas Mixture — Constant-Pressure Heating

Question 5: O₂/N₂ Ideal-Gas Mixture — Constant-Pressure Heating (15 marks)

Approach