Question 5 of 9: Fixed-Composition O₂/N₂ Ideal-Gas Mixture, Constant-Pressure Heating

Question 5: Fixed-Composition 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. $m_{O_2}=6$ kg, $m_{N_2}=21$ kg, $T_1=300$ K, $T_2=500$ K, constant total pressure throughout (5 MPa, though the ideal-gas assumption makes the property changes independent of the actual pressure level). Composition is fixed (no mixing/unmixing occurs during the process).

Find. (a) $Q$ [kJ]; (b) $\Delta S_{mix}$ [kJ/K].

Approach

Treat each species independently using its own temperature-dependent ideal-gas properties, weighted by its actual mass (not mole fraction, since masses are already given directly). At constant total pressure with unchanged composition, each species' own partial pressure is also unchanged, so its entropy change is purely the temperature-dependent term — no additional Gibbs mixing-entropy term appears, because the gases are already mixed and stay mixed throughout.

1. Heat transfer, constant-pressure process (part a). For a constant-pressure process, $Q=\Delta H$. Summing each species' mass times its own specific enthalpy change: $$Q=m_{O_2}\big[h_{O_2}(500)-h_{O_2}(300)\big]+m_{N_2}\big[h_{N_2}(500)-h_{N_2}(300)\big]=\boxed{5526.2\ \text{kJ}}.$$
2. Entropy change of the mixture (part b). Each species' partial pressure is unchanged (fixed composition, fixed total $P$), so its entropy change reduces to the temperature term alone, $\Delta s_i=s_i^\circ(T_2)-s_i^\circ(T_1)$: $$\Delta S_{mix}=m_{O_2}\Delta s_{O_2}+m_{N_2}\Delta s_{N_2}=\boxed{14.101\ \text{kJ/K}}.$$

Cross-check against the supplied tables. The appendix provides Table A-17 (ideal-gas properties of N₂) and Table A-18 (ideal-gas properties of O₂) on a molar basis, with molar masses from Table A-1 ($M_{N_2}=28.013$, $M_{O_2}=31.999$ kg/kmol). Reading $\bar h_{N_2}(300)=8723$, $\bar h_{N_2}(500)=14{,}581$, $\bar h_{O_2}(300)=8736$ and $\bar h_{O_2}(500)=14{,}770$ kJ/kmol gives $$Q=\frac{21(14{,}581-8723)}{28.013}+\frac{6(14{,}770-8736)}{31.999}=4391.4+1131.4=5522.8\ \text{kJ},$$ within 0.06% of the value above; the same reading of $\bar s^\circ$ ($191.682\to206.630$ for N₂, $205.213\to220.589$ for O₂ kJ/kmol·K) gives $\Delta S=11.206+2.883=14.089$ kJ/K, within 0.09%. The agreement also confirms the ideal-gas assumption was applied as the question directs: the stated 5 MPa never enters the property evaluation. Evaluating the same enthalpies at 5 MPa instead — i.e. letting the real-gas departure enthalpy in — would return $Q=5755$ kJ, about 4% high and inconsistent with the tables the examiners supplied.