Question 8 of 9: Adiabatic Compression of a Mass-Basis N₂/CO₂/O₂ Mixture
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — May 2017. 3 hours, Closed-Book Exam (approved
calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and charts supplied
in an appendix, interpolation not required). Part A: answer 2 of Questions 1-3 (20 marks each). Part
B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the first two Part-A and
first four Part-B questions as they appear in the answer book are marked. All nine questions (Part A
complete, Part B complete) are solved below for completeness.
Reference texts: Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering
Thermodynamics, 8th ed. All state properties (water/steam, R-134a, air, N₂, CO₂,
O₂, moist air) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric
result.
Question 8: Adiabatic Compression of a Mass-Basis N₂/CO₂/O₂ Mixture (15 marks)
Given. Closed system, $m=2$ kg, mass fractions $mf_{N_2}=0.30$,
$mf_{CO_2}=0.40$, $mf_{O_2}=0.30$ (fixed composition throughout). State 1: 1 bar, 300 K. State 2:
4 bar, 500 K. Process: adiabatic ($Q=0$), not necessarily reversible/isentropic (both end states are
given independently).
Gas
Mass frac.
$M$ (kg/kmol)
Mole frac. $y_i$
$R_i$ (kJ/kg·K)
N₂
0.30
28.013
0.3671
0.29679
CO₂
0.40
44.010
0.3115
0.18891
O₂
0.30
31.999
0.3214
0.25982
Find. (a) $W$ [kJ]; (b) $S_{gen}$ [kJ/K].
Approach
Because the mixture's composition is fixed (a closed system, no mixing occurring during the
process), the mole fractions cancel out of every isentropic-style Gibbs term and each species can be
treated independently: compute each pure component's specific internal-energy and entropy change
between the two given states, then combine by mass fraction. The
adiabatic closed-system first law gives the work directly from $\Delta U$; the adiabatic condition
also means entropy generation equals the system's own entropy change (no heat-transfer entropy term
to net out).
Per-species internal-energy change. For each species, $u=h-R_iT$ (ideal-gas
limit, evaluated at each state's own partial pressure $P_i=y_iP$, whose ratio equals the
total-pressure ratio $P_2/P_1$ since composition — hence every $y_i$ — is unchanged):
$$\Delta u_{N_2}=149.90,\quad\Delta u_{CO_2}=150.05,\quad\Delta u_{O_2}=136.66\ \text{kJ/kg}.$$
$$\Delta u_{mix}=\sum mf_i\,\Delta u_i=0.30(149.90)+0.40(150.05)+0.30(136.66)=145.99\ \text{kJ/kg}.$$
Work (part a), from the closed-system first law. Adiabatic ($Q=0$), so
$\Delta U=-W_{by}=W_{on}$:
$$W_{on}=m\,\Delta u_{mix}=2\times145.99=\boxed{291.97\ \text{kJ}}\ \text{(work done ON the gas, compression)}.$$
Per-species entropy change. Since $y_i$ is unchanged between states (fixed
composition), the Gibbs mixing-entropy term is identical at inlet and exit and cancels in the
difference, leaving $\Delta s_i=s_i^\circ(T_2)-s_i^\circ(T_1)-R_i\ln(P_2/P_1)$ (the $\ln(P_2/P_1)$ term
uses the TOTAL pressure ratio directly, because $P_{2,i}/P_{1,i}=y_iP_2/(y_iP_1)=P_2/P_1$):
$$\Delta s_{N_2}=0.12272,\quad\Delta s_{CO_2}=0.21430,\quad\Delta s_{O_2}=0.12043\ \text{kJ/kg}\cdot\text{K}.$$
$$\Delta s_{mix}=\sum mf_i\,\Delta s_i=0.30(0.12272)+0.40(0.21430)+0.30(0.12043)=0.15867\ \text{kJ/kg}\cdot\text{K}.$$
Entropy generated (part b). Adiabatic system: $S_{gen}=\Delta S_{system}$
(no surroundings entropy-transfer term, since $Q=0$):
$$S_{gen}=m\,\Delta s_{mix}=2\times0.15867=\boxed{0.3173\ \text{kJ/K}}\quad(>0\ \checkmark,\text{consistent with an irreversible adiabatic compression}).$$
