Piston-Cylinder H₂/N₂ Mixture, Constant-Pressure Heating: 04-BS-10, December 2017, Q7

Question 7: Piston-Cylinder H₂/N₂ 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. Closed piston-cylinder, $m_{H_2}=0.2$ kg, $m_{N_2}=1.6$ kg, $P=100$ kPa (constant), $T_1=300$ K. Heated at constant pressure until $V_2=2V_1$.

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

Approach

Since pressure and the number of moles of each species are both fixed (closed system, no reaction), the ideal-gas law gives $V\propto T$ at constant $P$ — doubling the volume therefore doubles the absolute temperature directly, with no property lookup needed to find $T_2$. Heat transfer at constant pressure for a closed system equals the enthalpy change (boundary work is already folded into $\Delta H=Q-W_{other}$ for a $P=\text{const}$ process); the mixture's total $\Delta H$ and $\Delta S$ are each the mass-weighted sum of the two species' own property changes, evaluated here with high-accuracy temperature-dependent (variable-$c_p$) properties in place of the constant-$c_p$-at-$T_{avg}$ approximation suggested in the question, for a more accurate result over this wide a temperature span.

Final temperature. Constant $P$, constant $n$ (closed, non-reacting mixture) $\Rightarrow V/T=\text{const}$: $$T_2=T_1\times\frac{V_2}{V_1}=300\times2=\boxed{600\ \text{K}}.$$
Heat transfer (part a). At constant pressure, $Q=\Delta H$ for the closed system (boundary work exactly offsets the internal-energy/enthalpy bookkeeping): $$Q=m_{H_2}\big[h_{H_2}(T_2)-h_{H_2}(T_1)\big]+m_{N_2}\big[h_{N_2}(T_2)-h_{N_2}(T_1)\big].$$ $H_2$: $h(300\,\text{K})=3958.3$, $h(600\,\text{K})=8303.0$ kJ/kg, $\Delta h_{H_2}=4344.7$ kJ/kg. $N_2$: $h(300\,\text{K})=311.2$, $h(600\,\text{K})=627.0$ kJ/kg, $\Delta h_{N_2}=315.8$ kJ/kg. $$Q=0.2\times4344.7+1.6\times315.8=868.9+505.3=\boxed{1374.2\ \text{kJ}}.$$
Entropy change of the mixture (part b). At constant pressure, each species' specific entropy change is $\Delta s_i=c_{p,i}\ln(T_2/T_1)$ in the constant-$c_p$ limit, or more precisely the variable-$c_p$ ideal-gas entropy function evaluated at the same pressure for both states (the $-R\ln(P_2/P_1)$ term vanishes since $P$ is constant): $$\Delta S_{mix}=m_{H_2}\big[s^\circ_{H_2}(T_2)-s^\circ_{H_2}(T_1)\big]+m_{N_2}\big[s^\circ_{N_2}(T_2)-s^\circ_{N_2}(T_1)\big].$$ $H_2$: $s^\circ(300\,\text{K})=53.519$, $s^\circ(600\,\text{K})=63.550$ kJ/kg·K, $\Delta s_{H_2}=10.031$ kJ/kg·K. $N_2$: $s^\circ(300\,\text{K})=6.8457$, $s^\circ(600\,\text{K})=7.5740$ kJ/kg·K, $\Delta s_{N_2}=0.7283$ kJ/kg·K. $$\Delta S_{mix}=0.2\times10.031+1.6\times0.7283=2.0062+1.1653=\boxed{3.172\ \text{kJ/K}}.$$

Quantity	Result
(a) $Q$	1374.2 kJ
(b) $\Delta S_{mix}$	3.172 kJ/K

Concept & Theory

Topic. Constant-pressure heating of a fixed-composition, mass-basis ideal-gas mixture (H₂/N₂) in a closed piston-cylinder device.

Key relations. Ideal-gas law at constant $P$, $n$: $V\propto T$. Constant-pressure closed-system heat transfer: $Q=\Delta H$. Mass-weighted mixture property change: $\Delta X_{mix}=\sum m_i\Delta x_i$.

Why this works. "Volume doubles at constant pressure" is a direct proportionality statement via the ideal-gas law that requires no iteration once $P$ and $n$ are both fixed — this is what makes $T_2=2T_1$ immediate, unlike Q4's isentropic-compression problem where $T_2$ had to be root-solved. Because H₂ has a per-kilogram specific heat roughly 14× that of N₂ (its molar mass is 14× smaller, for a similar molar $c_p$), the small 0.2 kg of H₂ contributes nearly as much to $Q$ as the much larger 1.6 kg of N₂ — a mass-only intuition badly underestimates H₂'s share.

Common pitfall. Assuming mass fractions directly give the entropy/enthalpy mixing rule the way mole fractions do for the isentropic-compression archetype (Q4) — here there is no composition change and no partial-pressure bookkeeping at all, since it's a purely mass-weighted sum of each species' own (single-substance) property change, not a mixing-entropy calculation.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 13 (Gas Mixtures) & Ch. 4 (Energy Analysis of Closed Systems).

Question 7 of 9: Piston-Cylinder H₂/N₂ Mixture, Constant-Pressure Heating

Question 7: Piston-Cylinder H₂/N₂ Mixture, Constant-Pressure Heating (15 marks)

Approach