# Combustion of Octane: Pressure & Heat Transfer Calculation ## Problem Restatement - **Fuel:** Octane (\(C_8H_{18}\)) - **Air supplied:** 200% theoretical (i.e., twice the stoichiometric oxygen) - **Initial conditions:** 25°C, 1 atm, rigid & isolated vessel - **Process:** Combustion, then cooled to 25°C at constant volume - **Tasks:** 1. Find **final pressure** after cooling 2. Find **heat transfer per kmol fuel** --- ## Step 1: **Combustion Reaction & Product Analysis** ### **1. Balanced Stoichiometric Combustion** The **stoichiometric combustion** of octane with air: \[ C_8H_{18} + 12.5O_2 + (12.5 \times 3.76)N_2 \rightarrow 8CO_2 + 9H_2O + (12.5 \times 3.76)N_2 \] With **200% theoretical air**: - Oxygen supplied: \(2 \times 12.5 = 25\) mol O\(_2\) per kmol \(C_8H_{18}\) - Nitrogen: \(25 \times 3.76 = 94\) mol N\(_2\) ### **2. Limiting Reactant & Final Products** For 1 kmol of fuel: \[ C_8H_{18} + 25O_2 + 94N_2 \rightarrow 8CO_2 + 9H_2O + (25-12.5)O_2 + 94N_2 \] \[ \rightarrow 8CO_2 + 9H_2O + 12.5O_2 + 94N_2 \] **Final product composition (per kmol \(C_8H_{18}\)):** - \(8\) kmol CO\(_2\) - \(9\) kmol H\(_2\)O - \(12.5\) kmol O\(_2\) (excess) - \(94\) kmol N\(_2\) --- ## Step 2: **Calculation of Final Pressure & Heat Transfer** ### **(a) Final Pressure at Constant Volume** #### **Initial moles:** \[ \text{Initial total moles} = 1 \text{ (fuel)} + 25 \text{ (O}_2\text{)} + 94 \text{ (N}_2\text{)} = 120 \] #### **Final moles:** \[ \text{Final total moles} = 8 \text{ (CO}_2\text{)} + 9 \text{ (H}_2\text{O)} + 12.5 \text{ (O}_2\text{)} + 94 \text{ (N}_2\text{)} = 123.5 \] #### **Using Ideal Gas Law at Constant \(T\), \(V\):** \[ \frac{P_f}{P_i} = \frac{n_f}{n_i} \] \[ P_f = P_i \times \frac{n_f}{n_i} = 1 \text{ atm} \times \frac{123.5}{120} = 1.029 \text{ atm} \] --- ### **(b) Heat Transfer per kmol Fuel** #### **Rigid, Isolated Vessel:** - **No work** (constant volume) - **Isolated:** \(Q = \Delta U\) (by first law) #### **Calculation:** \[ Q = \sum n_p u_p(T_2) - \sum n_r u_r(T_1) \] But since the system is cooled **back to initial temperature (25°C)**, the **internal energies at 25°C** cancel for the reactants and products, except for the heat of reaction. So, \[ Q = -\Delta U_{comb} \] \[ \Delta U_{comb} = \Delta H_{comb} - \Delta n_g RT \] where - \(\Delta H_{comb}\): Standard enthalpy of combustion of octane (at 25°C) - \(\Delta n_g\): Change in moles of gas (products - reactants) **For octane:** \[ \Delta H_{comb} (C_8H_{18}, l) \approx -5471~\text{kJ/mol} \] But for gas phase: \(-5114~\text{kJ/mol}\) (We'll use gas phase for this vessel problem.) #### **Calculating \(\Delta n_g\) (only count gases):** - **Reactants:** \(1\) (C8H18, assume gas) + \(25\) (O2) + \(94\) (N2) = **120** - **Products:** \(8\) (CO2) + \(9\) (H2O, assume vapor at 25°C) + \(12.5\) (O2) + \(94\) (N2) = **123.5** \[ \Delta n_g = 123.5 - 120 = 3.5~\text{kmol} \] \[ Q = -\Delta H_{comb} + \Delta n_g RT \] \[ Q = -(-5114~\text{kJ}) + (3.5~\text{kmol}) \times (8.314~\text{kJ/kmol}\cdot\text{K}) \times (298~\text{K}) \] \[ Q = 5114~\text{kJ} + (3.5 \times 8.314 \times 298) \] \[ = 5114~\text{kJ} + 8672~\text{kJ} \] \[ = 5114~\text{kJ} + 8.67~\text{kJ} \] Wait, let's check units: \(3.5 \times 8.314 \times 298 = 8672~\text{kJ}\) So, \[ Q = 5114~\text{kJ} + 8672~\text{kJ} = 13,786~\text{kJ} \] But this is not correct; the standard enthalpy is negative (\(-5114~\text{kJ}\)), so: \[ Q = -(-5114) + 3.5 \times 8.314 \times 298 \] \[ = 5114 + 8672 = 13,786~\text{kJ} \] But typically, **the heat transfer is negative** (heat released), so the sign should reflect heat loss: Thus, \[ Q = -\Delta U_{comb} = \Delta H_{comb} - \Delta n_g RT \] \[ Q = -5114~\text{kJ} - (3.5 \times 8.314 \times 298) \] \[ = -5114~\text{kJ} - 8672~\text{kJ} \] \[ = -13,786~\text{kJ} \] So the **heat that must be removed per kmol of fuel** is **13,786 kJ** (heat must be extracted to return to 25°C). --- ## **Summary Table** | Quantity | Value | |--------------------|----------------| | Final Pressure | 1.03 atm | | Heat transfer (\(Q\)) | \(-13,786\) kJ per kmol fuel | --- ## **Graph: Moles vs. Pressure** ```plaintext | . | . | . | . | . | . |----------|----------------- Initial Final ``` - **X-axis:** State (Initial, Final) - **Y-axis:** Total moles / Pressure - **Interpretation:** Pressure increases proportional to moles (at constant V, T). --- ## **Final Answers** ### **1. Final Pressure:** \[ P_f = 1.03~\text{atm} \] ### **2. Heat Transfer per kmol Fuel:** \[ Q = -13,786~\text{kJ per kmol} \] --- **Example Calculation:** - For 2 kmol fuel, heat removed = \(2 \times 13,786 = 27,572~\text{kJ}\) - For .5 kmol fuel, heat removed = \(.5 \times 13,786 = 6,893~\text{kJ}\) --- **Let me know if you want more details or a plotted graph!**

# Combustion of Octane: Pressure & Heat Transfer Calculation ## Problem Restatement - **Fuel:** Octane (\(C_8H_{18}\)) - **Air supplied:** 200% theoretical (twice the stoichiometric oxygen) - **Initial conditions:** 25°C, 1 atm, rigid & isolated vessel - **Process:** Combustion, then cooled to 25°C at constant volume - **Tasks:** 1. Find **final pressure** after cooling 2. Find **heat transfer per kmol of fuel** --- ## Step 1: Combustion Reaction & Product Analysis ### 1. Balanced Stoichiometric Combustion The **stoichiometric combustion** of octane with air is given by: \[ C_8H_{18} + 12.5O_2 + (12.5 \times 3.76)N_2 \rightarrow 8CO_2 + 9H_2O + (12.5 \times 3.76)N_2 \] With **200% theoretical air**: - Oxygen supplied: \(2 \times 12.5 = 25\) mol O\(_2\) per kmol \(C_8H_{18}\) - Nitrogen: \(25 \times 3.76 = 94\) mol N\(_2\) ### 2. Limiting Reactant & Final Products For 1 kmol of fuel: \[ C_8H_{18} + 25O_2 + 94N_2 \rightarrow 8CO_2 + 9H_2O + (25-12.5)O_2 + 94N_2 \] \[ \rightarrow 8CO_2 + 9H_2O + 12.5O_2 + 94N_2 \] **Final product composition (per kmol \(C_8H_{18}\)):** - \(8\) kmol CO\(_2\) - \(9\) kmol H\(_2\)O - \(12.5\) kmol O\(_2\) (excess) - \(94\) kmol N\(_2\) --- ## Step 2: Calculation of Final Pressure & Heat Transfer ### (a) Final Pressure at Constant Volume #### Initial Moles \[ \text{Initial total moles} = 1 \text{ (fuel)} + 25 \text{ (O}_2\text{)} + 94 \text{ (N}_2\text{)} = 120 \] #### Final Moles \[ \text{Final total moles} = 8 \text{ (CO}_2\text{)} + 9 \text{ (H}_2\text{O)} + 12.5 \text{ (O}_2\text{)} + 94 \text{ (N}_2\text{)} = 123.5 \] #### Using Ideal Gas Law at Constant \(T\), \(V\): \[ \frac{P_f}{P_i} = \frac{n_f}{n_i} \] \[ P_f = P_i \times \frac{n_f}{n_i} = 1 \text{ atm} \times \frac{123.5}{120} = 1.029 \text{ atm} \] --- ### (b) Heat Transfer per kmol Fuel #### Rigid, Isolated Vessel - **No work** (constant volume) - **Isolated:** \(Q = \Delta U\) (by first law) #### Calculation \[ Q = \sum n_p u_p(T_2) - \sum n_r u_r(T_1) \] Since the system is cooled **back to initial temperature (25°C)**, the **internal energies at 25°C** cancel for the reactants and products, except for the heat of reaction. So, \[ Q = -\Delta U_{comb} \] #### Calculation of \(\Delta n_g\): - **Reactants:** \(1 + 25 + 94 = 120\) - **Products:** \(8 + 9 + 12.5 + 94 = 123.5\) \[ \Delta n_g = 123.5 - 120 = 3.5~\text{kmol} \] Using the standard enthalpy of combustion for gas phase: \[ \Delta H_{comb} \approx -5114~\text{kJ/mol} \] #### Heat Transfer Calculation \[ Q = -\Delta H_{comb} + \Delta n_g RT \] Substituting values: \[ Q = -(-5114) + (3.5~\text{kmol}) \times (8.314~\text{kJ/kmol}\cdot\text{K}) \times (298~\text{K}) \] \[ = 5114 + (3.5 \times 8.314 \times 298) = 5114 + 8672 \approx 13786~\text{kJ} \] The heat transfer is negative, indicating heat must be extracted: \[ Q = -13786~\text{kJ} \] --- ## Summary Table | Quantity | Value | |--------------------|----------------| | Final Pressure | 1.03 atm | | Heat Transfer (\(Q\)) | \(-13786\) kJ per kmol fuel | --- ## Final Answers ### 1. Final Pressure: \[ P_f = 1.03~\text{atm} \] ### 2. Heat Transfer per kmol Fuel: \[ Q = -13786~\text{kJ per kmol} \] --- **Note:** The negative sign indicates heat is released (exothermic reaction).