Rigid Tank, N₂/O₂ Ideal-Gas Mixture: 04-BS-10, Undated paper, Q8

Question 8: Rigid Tank, N₂/O₂ Ideal-Gas Mixture (Part B – 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. Rigid tank, $V=0.1\ \text{m}^3$. $m_{N_2}=0.7$ kg, $m_{O_2}=1.1$ kg. $T_1=27\ ^\circ\text{C}=300.15$ K, $T_2=127\ ^\circ\text{C}=400.15$ K. $M_{N_2}=28.013$, $M_{O_2}=31.999$ kg/kmol; $c_{v,N_2}=0.743$, $c_{v,O_2}=0.658$ kJ/kg·K (room-temperature ideal-gas values).

Find. (a) mass fractions; (b) mole fractions; (c) partial pressures (at state 1); (d) the mixture pressure; (e) $Q$ to reach $127\ ^\circ\text{C}$.

Approach

Mass fractions come directly from the given masses; converting each to moles ($n=m/M$) gives the mole fractions. Each species' partial pressure follows from its own ideal-gas law using the shared tank volume and temperature (Dalton's model). Heat transfer at constant volume uses the mixture's mass-weighted $c_v$.

Mass fractions — part (a). $m_{tot}=0.7+1.1=1.8$ kg. $$mf_{N_2}=\frac{0.7}{1.8}=\boxed{0.3889\ (38.9\%)},\qquad mf_{O_2}=\frac{1.1}{1.8}=\boxed{0.6111\ (61.1\%)}.$$
Mole fractions — part (b). $$n_{N_2}=\frac{0.7}{28.013}=0.02499\ \text{kmol},\qquad n_{O_2}=\frac{1.1}{31.999}=0.03438\ \text{kmol},\qquad n_{tot}=0.05936\ \text{kmol}.$$ $$y_{N_2}=\frac{0.02499}{0.05936}=\boxed{0.4209\ (42.1\%)},\qquad y_{O_2}=\frac{0.03438}{0.05936}=\boxed{0.5791\ (57.9\%)}.$$
Partial pressures at state 1 — part (c). Each species occupies the full tank volume at $T_1$ (Dalton's model), $R_i=\bar R/M_i$: $$P_{N_2}=\frac{m_{N_2}R_{N_2}T_1}{V}=\frac{0.7\times0.2968\times300.15}{0.1}=\boxed{623.6\ \text{kPa}}$$ $$P_{O_2}=\frac{m_{O_2}R_{O_2}T_1}{V}=\frac{1.1\times0.2598\times300.15}{0.1}=\boxed{857.8\ \text{kPa}}$$
Mixture pressure — part (d). By Dalton's law the mixture pressure is the sum of the partial pressures (equivalently $P=n_{tot}\bar RT_1/V=0.05936\times8.314\times300.15/0.1$): $$P_{tot}=P_{N_2}+P_{O_2}=623.6+857.8=\boxed{1481.4\ \text{kPa}}.$$
Heat transfer to $127\ ^\circ$C — part (e). Constant volume, so all the heat goes to raising internal energy; use the mass-weighted mixture $c_v$: $$c_{v,mix}=mf_{N_2}c_{v,N_2}+mf_{O_2}c_{v,O_2}=0.3889\times0.743+0.6111\times0.658=0.6911\ \text{kJ/kg}\cdot\text{K}.$$ $$Q=m_{tot}\,c_{v,mix}(T_2-T_1)=1.8\times0.6911\times(400.15-300.15)=\boxed{124.4\ \text{kJ}}.$$

Quantity	Value
(a) Mass fractions N₂ / O₂	38.89% / 61.11%
(b) Mole fractions N₂ / O₂	42.09% / 57.91%
(c) $P_{N_2}$ / $P_{O_2}$	623.6 / 857.8 kPa
(d) $P_{tot}$ (mixture pressure)	1481.4 kPa
(e) $Q$	124.4 kJ

Topic. Ideal-gas mixture composition and constant-volume heating (Dalton's model).

Key relations. $mf_i=m_i/m_{tot}$; $y_i=n_i/n_{tot}$, $n_i=m_i/M_i$; Dalton: $P_i=m_iR_iT/V$; mixture $c_{v,mix}=\sum mf_i\,c_{v,i}$; $Q=m_{tot}c_{v,mix}\Delta T$ (rigid tank).

Why this works. Dalton's model treats each ideal-gas component as if it alone filled the entire container at the mixture temperature, so its individual ideal-gas law applies directly with the shared $V$ and $T$ — the partial pressures then simply sum to the total pressure, which is the mixture's defining property.

Common pitfall. Using MOLE-fraction weighting for $c_v$ (which requires molar $\bar c_v$ in kJ/kmol·K) when the given specific heats are already on a per-kg basis — mixing the two conventions silently gives the wrong $c_{v,mix}$. Mass-fraction weighting of per-kg $c_v$ values is correct here.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §13-2/13-3 (ideal-gas mixture composition and P-v-T behaviour).

Question 8 of 9: Rigid Tank, N₂/O₂ Ideal-Gas Mixture

Question 8: Rigid Tank, N₂/O₂ Ideal-Gas Mixture (Part B – 15 marks)

Approach