Isentropic Compression of an N₂/CO₂ Gas Mixture: 04-BS-10, December 2019, Q5

Question 5: Isentropic Compression of an N₂/CO₂ Gas Mixture (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.

Check — reconstructed mixture composition. The printed question has a genuine blank gap exactly where the mole percentages are given ("A mix[ture] o[f __]% N₂ [and __%] CO₂ gases"). Every other detail — "compressed isentropically", "enters … at 100 kPa and 1000 K", "leaves at 500 kPa", "constant specific heats at room temperature", ideal gas — matches a well-known Cengel & Boles textbook problem using a 60% N₂ / 40% CO₂ (by mole) mixture, which is adopted here as the representative composition.

Given. $y_{N_2}=0.60$, $y_{CO_2}=0.40$ (mole fractions). $T_1=1000$ K, $P_1=100$ kPa, $P_2=500$ kPa. Constant specific heats evaluated at room temperature (300 K): $c_{p,N_2}=1.039$, $c_{p,CO_2}=0.846$ kJ/kg·K.

Find. $w_{in}$ [kJ/kg of mixture].

Approach

Convert mole fractions to mass fractions via each species' molar mass, mass-average the two species' room-temperature $c_p$ values to get the mixture's $c_p$, get the mixture's gas constant from its apparent molar mass, then apply the isentropic ideal-gas power law with the mixture's own $k$.

Mixture molar mass and mass fractions. $$M_{mix}=y_{N_2}M_{N_2}+y_{CO_2}M_{CO_2}=0.60(28.013)+0.40(44.01)=34.412\ \text{kg/kmol}$$ $$mf_{N_2}=\frac{y_{N_2}M_{N_2}}{M_{mix}}=\frac{16.808}{34.412}=0.4884,\qquad mf_{CO_2}=0.5116.$$
Mixture $c_p$, $R$, and $k$ (constant properties at 300 K). $$c_{p,mix}=mf_{N_2}c_{p,N_2}+mf_{CO_2}c_{p,CO_2}=0.4884(1.039)+0.5116(0.846)=\boxed{0.9403\ \text{kJ/kg}\cdot\text{K}}$$ $$R_{mix}=\frac{\bar R}{M_{mix}}=\frac{8.314}{34.412}=0.2416\ \text{kJ/kg}\cdot\text{K},\qquad k_{mix}=\frac{c_{p,mix}}{c_{p,mix}-R_{mix}}=\frac{0.9403}{0.6987}=\boxed{1.3458}.$$
Isentropic exit temperature and work input. $$T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k_{mix}-1)/k_{mix}}=1000\times5^{0.2570}=1512.2\ \text{K}$$ $$w_{in}=c_{p,mix}(T_2-T_1)=0.9403\times(1512.2-1000)=\boxed{481.6\ \text{kJ/kg}}.$$

Quantity	Result
$c_{p,mix}$	0.9403 kJ/kg·K
$k_{mix}$	1.3458
$w_{in}$	481.6 kJ/kg

Concept & Theory

Topic. Isentropic compression of a fixed-composition ideal-gas mixture, constant (room-temperature) specific heats.

Key relations. Mass-fraction-weighted mixture $c_p$ (Amagat/Gibbs-Dalton ideal-gas mixture rule); mixture apparent molar mass $M_{mix}=\sum y_i M_i$; mixture $k=c_p/(c_p-R)$ with $R=\bar R/M_{mix}$; isentropic power law $T_2/T_1=(P_2/P_1)^{(k-1)/k}$ applied with the MIXTURE's own $k$, not either pure component's.

Why this works. An ideal-gas mixture of fixed composition behaves exactly like a single ideal gas with mass-averaged properties for any process where composition doesn't change (compression, expansion, heating) — there is no need to track each species separately once $c_{p,mix}$, $R_{mix}$ are fixed.

Common pitfall. Using MOLE-fraction weighting for $c_p$ on a per-kg-of-mixture basis (that would give molar $\bar c_p$, not mass-specific $c_p$) — the conversion to mass fractions via each species' molar mass is a required intermediate step.

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

Question 5 of 9: Isentropic Compression of an N₂/CO₂ Gas Mixture

Question 5: Isentropic Compression of an N₂/CO₂ Gas Mixture (15 marks)

Approach