Act like a helpful tutor and exlain me :7.29 Air composition at room temperature is approxi-
mated by 21% O, and 79% N, by volume, or by mole frac-
tions, i.e., one mole of air is composed of
0.210

Question

Act like a helpful tutor and exlain me :7.29 Air composition at room temperature is approxi-
mated by 21% O, and 79% N, by volume, or by mole frac-
tions, i.e., one mole of air is composed of
0.210, +0.79N,
Strong shocks are, however, formed when aircraft fly at super-
sonic/hypersonic Mach numbers. The composition of air thus
changes as a sudden compression is encountered through a
shock. At 6000 K and one atmospheric pressure, one mole of
air assumes the following equilibrium composition:
0.18N+0310+0.51N,
Calculate
(a) the molecular weight of air in the new mixture,
MW,,, in kg/kmol
(b) the fuel-to-air ratio for the stoichiometric combus-
tion of hydrogen in air (at 6000 K)

VIPSolutions · Accepted Answer

Let's break down the problem step by step:

---

## **Given Data**

At **600 K** and 1 atm, **1 mole of air** has the following composition:
- $.18$ N (atomic nitrogen)
- $.31$ O (atomic oxygen)
- $.51$ N₂ (molecular nitrogen)

### **Atomic/Molecular Weights (kg/kmol)**
- N: 14
- O: 16
- N₂: 28

---

## **Part (a): Molecular Weight of Air in New Mixture**

The **molecular weight** ($ MW_m $) is the sum of the mole fractions multiplied by the molecular weight of each component:

$$
MW_m = \sum_{i} (n_i \times MW_i)
$$
where $ n_i $ = number of moles of component $ i $ per 1 mole of air, and $ MW_i $ = molecular weight of component $ i $.

Let's plug in the values:

$$
MW_m = (.18 \times 14) + (.31 \times 16) + (.51 \times 28)
$$

Calculate each term:
- $.18 \times 14 = 2.52$
- $.31 \times 16 = 4.96$
- $.51 \times 28 = 14.28$

Add them up:
$$
MW_m = 2.52 + 4.96 + 14.28 = \boxed{21.76 \ \text{kg/kmol}}
$$

---

## **Part (b): Fuel-to-Air Ratio for Stoichiometric Combustion of Hydrogen**
### **Step 1: Combustion Reaction with 600 K Air**

Hydrogen ($H_2$) reacts with oxygen to form water:
$$
2H_2 + O_2 \rightarrow 2H_2O
$$

But at **600 K**, air contains:
- Atomic O (.31 mol)
- Atomic N (.18 mol)
- N₂ (.51 mol)

### **Step 2: Available Oxygen Atoms**
Each **O atom** can form half an $O_2$ molecule, but here, we use **atomic oxygen** for combustion.

To combust **1 mol H₂**:
$$
H_2 + \frac{1}{2} O_2 \rightarrow H_2O
$$
1 mol H₂ needs **1 mol O atoms** (since $O_2$ is 2 O atoms).

So, for **1 mol of air**:
- **Available O atoms = .31 mol**

Thus, **.31 mol H₂** can be completely combusted with the O atoms in 1 mol air.

### **Step 3: Fuel-to-Air Ratio**
- **Fuel (H₂) required per 1 mol air** = .31 mol
- **Fuel mass:** $.31 \ \text{mol} \times 2 \ \text{kg/kmol} = .62 \ \text{kg/kmol}$
- **Air mass:** $1 \ \text{mol} \times 21.76 \ \text{kg/kmol} = 21.76 \ \text{kg/kmol}$

$$
\text{Fuel-to-Air Ratio} = \frac{\text{Fuel mass}}{\text{Air mass}} = \frac{.62}{21.76} = \boxed{.0285}
$$

---

## **Summary of Answers**

### (a) Molecular weight of air at 600 K:  
$$
\boxed{21.76 \ \text{kg/kmol}}
$$

### (b) Stoichiometric fuel-to-air ratio (hydrogen/air, 600 K):  
$$
\boxed{.0285}
$$

---

Let me know if you'd like a more detailed explanation or have questions about the steps!

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