Adiabatic Gas Mixing (Part a) and an Air-Standard Otto Cycle (Part b): 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2018, Q1

Question 1: Adiabatic Gas Mixing (Part a) and an Air-Standard Otto Cycle (Part b)

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.

Part (a) — Given. Two rigid, initially isolated vessels are connected through a valve; once opened, the two gases occupy the combined volume and the FINAL mixture state ($P_2$, $T_2$) is measured directly — so the unknowns are found from the ideal-gas law applied to each species separately (Gibbs–Dalton: each gas in a mixture behaves as if it alone occupied the total volume at the mixture temperature, its own partial pressure summing to the total).

Given data
Quantity	Symbol	Value
Air vessel volume	$V_A$	0.06 m³
Air initial pressure	$P_{A1}$	40 atm
Air initial temperature	$T_{A1}$	$40\,{}^{\circ}\text{C}$
Argon mass	$m_{Ar}$	1.35 kg
Argon initial pressure	$P_{Ar1}$	7 atm
Mixture pressure	$P_2$	18 atm
Mixture temperature	$T_2$	$30\,{}^{\circ}\text{C}$
Air gas constant	$R_{air}$	0.287 kJ/kg·K

Part (a) — Find. The volume of the argon vessel $V_{Ar}$ and the argon temperature $T_{Ar1}$ before mixing.

Two rigid vessels joined by a valve; air and argon mix into the combined volume at the measured final state $P_2=18$ atm, $T_2=30\,{}^{\circ}\text{C}$.

Part (a) — Approach. Find the mass of air from its own initial state, then write the ideal gas law for EACH species at the final mixture state (same $T_2$, combined volume $V_{total}$, partial pressures summing to $P_2$) to solve for $V_{total}$ and hence $V_{Ar}$; finally apply the ideal-gas law to argon's own INITIAL state to back out $T_{Ar1}$.

Mass of air from its initial state. $$m_{air}=\frac{P_{A1}V_A}{R_{air}T_{A1}}=\frac{(40\times101.325)\times0.06}{0.287\times313.15}$$ $$\boxed{m_{air}=2.706\text{ kg}}$$
Gibbs–Dalton at the final state. Each gas occupies the SAME combined volume $V_{total}=V_A+V_{Ar}$ at the SAME final temperature $T_2$; its partial pressure is $P_i=m_iR_iT_2/V_{total}$, and the two partial pressures sum to $P_2$: $$P_2=\frac{T_2}{V_{total}}\left(m_{air}R_{air}+m_{Ar}R_{Ar}\right)$$ With $R_{Ar}=R_u/M_{Ar}=8.314/39.948=0.2081\text{ kJ/kg}\cdot\text{K}$: $$V_{total}=\frac{T_2\left(m_{air}R_{air}+m_{Ar}R_{Ar}\right)}{P_2} =\frac{303.15\times(2.706\times0.287+1.35\times0.2081)}{18\times101.325}$$ $$\boxed{V_{total}=0.1758\text{ m}^3}$$
Argon vessel volume. $$V_{Ar}=V_{total}-V_A=0.1758-0.06$$ $$\boxed{V_{Ar}=0.1158\text{ m}^3}$$
Argon temperature before mixing. Apply the ideal-gas law to argon's OWN initial state ($P_{Ar1}=7$ atm, the now-known $V_{Ar}$): $$T_{Ar1}=\frac{P_{Ar1}V_{Ar}}{m_{Ar}R_{Ar}}=\frac{(7\times101.325)\times0.1158}{1.35\times0.2081}$$ $$\boxed{T_{Ar1}=292.3\text{ K}=19.1\,{}^{\circ}\text{C}}$$

Question 1(a) — results
Quantity	Value
Mass of air	2.706 kg
Combined final volume $V_{total}$	0.1758 m³
Volume of argon vessel $V_{Ar}$	0.1158 m³
Argon temperature before mixing $T_{Ar1}$	$19.1\,{}^{\circ}\text{C}$ (292.3 K)

Topic. Adiabatic mixing of two ideal gases initially confined to separate rigid vessels (Gibbs–Dalton model).

Key relations. Ideal-gas law $PV=mRT$ per species; Dalton's law $P_{mix}=\sum P_i$ with each $P_i=m_iR_iT_{mix}/V_{total}$; specific gas constant $R_i=R_u/M_i$.

Why this works. Because BOTH the final pressure and temperature are given (measured after mixing), the problem never needs an energy balance at all — the two Gibbs–Dalton partial-pressure equations at the known final state are sufficient to pin down the one remaining geometric unknown ($V_{total}$, hence $V_{Ar}$), and argon's own ideal-gas law at its initial state then recovers $T_{Ar1}$ from the now-known volume. An energy balance would only be needed if the final temperature were UNKNOWN and had to be predicted from the two initial states.

Common pitfall. Assuming $T_{Ar1}=T_{A1}$ (equal initial temperatures) is not justified — the two vessels are separate and independently at whatever temperature their own gas happens to be; also forgetting that $R_{Ar}\neq R_{air}$ (argon is monatomic, larger $M$ than air's effective 28.97 kg/kmol) when writing the partial-pressure sum.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 13 (gas mixtures, Amagat's and Dalton's laws).

Part (b) — Given. An air-standard OTTO cycle (isentropic compression 1→2, constant-volume heat addition 2→3, isentropic expansion 3→4, constant-volume heat rejection 4→1) with compression ratio $r=8$, $\gamma=1.4$, $R=0.287\text{ kJ/kg}\cdot\text{K}$, state 1 at $P_1=138\text{ kPa}$, $T_1=37\,{}^{\circ}\text{C}$, and cycle maximum $T_3=1772\text{ K}$.

Part (b) — Find. $T_2,P_2,P_3,T_4,P_4$; the thermal efficiency $\eta_{th}$; and the power output for a heat-addition rate of 370 W.

Otto cycle state points (schematic, not to scale) on $P$–$v$ (left) and $T$–$s$ (right) coordinates — values in the steps and results table below.

Part (b) — Approach. Use the isentropic $Pv^\gamma$/$Tv^{\gamma-1}$ relations across the compression (1→2) and expansion (3→4) legs (both spanning the same volume ratio $r$), the constant-volume relation $P/T={\rm const}$ across heat addition (2→3) to reach $P_3$, then the Otto-cycle efficiency formula and $\dot W=\eta_{th}\dot Q_{in}$.

Compression 1→2 (isentropic). $$T_2=T_1r^{\gamma-1}=310.15\times8^{0.4}=310.15\times2.2967$$ $$\boxed{T_2=712.5\text{ K}=439.4\,{}^{\circ}\text{C}}$$ $$P_2=P_1r^{\gamma}=138\times8^{1.4}=138\times18.379$$ $$\boxed{P_2=2536\text{ kPa}}$$
Heat addition 2→3 (constant volume, $P/T$ const, $T_3=1772$ K given). $$P_3=P_2\frac{T_3}{T_2}=2536\times\frac{1772}{712.5}$$ $$\boxed{P_3=6308\text{ kPa}}$$
Expansion 3→4 (isentropic, same volume ratio $r$). $$T_4=\frac{T_3}{r^{\gamma-1}}=\frac{1772}{2.2967}$$ $$\boxed{T_4=771.3\text{ K}=498.2\,{}^{\circ}\text{C}}$$ $$P_4=\frac{P_3}{r^{\gamma}}=\frac{6308}{18.379}$$ $$\boxed{P_4=343.2\text{ kPa}}$$ A quick check: heat rejection 4→1 is constant-volume, so $v_4=v_1$ automatically since both equal $r\,v_2=r\,v_3$ — the cycle closes exactly, confirming the state points are consistent.
Thermal efficiency. For an Otto cycle the efficiency depends only on $r$ and $\gamma$: $$\eta_{th}=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{2.2967}$$ $$\boxed{\eta_{th}=0.5647=56.5\%}$$
Power output for $\dot Q_{in}=370\text{ W}$. $$\dot W_{net}=\eta_{th}\dot Q_{in}=0.5647\times370$$ $$\boxed{\dot W_{net}=208.9\text{ W}}$$

Question 1(b) — results
State	$T$	$P$
1 (BDC, start of compression)	310.15 K	138 kPa
2 (TDC, end of compression)	712.5 K	2536 kPa
3 (TDC, end of heat addition, max)	1772 K (given)	6308 kPa
4 (BDC, end of expansion)	771.3 K	343.2 kPa
Thermal efficiency $\eta_{th}$	56.5%
Power output $\dot W_{net}$ (at $\dot Q_{in}=370$ W)	208.9 W

Topic. Air-standard Otto cycle — the idealised spark-ignition engine cycle.

Key relations. Isentropic process: $Tv^{\gamma-1}={\rm const}$, $Pv^{\gamma}={\rm const}$; constant-volume process: $P/T={\rm const}$; Otto efficiency $\eta_{th}=1-r^{1-\gamma}$; power $\dot W=\eta_{th}\dot Q_{in}$.

Why this works. Compression and expansion span the identical volume ratio $r$ (TDC to BDC), so both isentropic legs use the same exponent $r^{\gamma-1}$; this is exactly what collapses the efficiency formula to a function of $r$ alone, independent of the actual heat added or the absolute temperatures — a hallmark of the ideal Otto cycle that makes the last two parts solvable without re-deriving $T_2,T_4$ from scratch.

Common pitfall. Using $r^{\gamma}$ instead of $r^{\gamma-1}$ for the temperature ratio (that exponent belongs to the PRESSURE ratio); also treating the given 370 W as the NET work rather than the heat-addition rate the problem explicitly says it is.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (air-standard Otto cycle).

Question 1 of 8: Adiabatic Gas Mixing (Part a) and an Air-Standard Otto Cycle (Part b)

Question 1: Adiabatic Gas Mixing (Part a) and an Air-Standard Otto Cycle (Part b)