In a vessel, whose volume is `1l`, there is a gas, which is a compound of oxygen and nitrogen. Mass of gas is 1g, gas's temperature is `17_(0)C` and pressure of the gas is `31.7 k P.a`. Determine the chemical formula of compound of `(R=8.31 J//mol -K)`:-

Options

A. `NO`
B. `NO_(2)`
C. `N_(2)O`
D. `N_(2)O_(3)`

Explanation

To determine the chemical formula of the gas compound of oxygen and nitrogen, we will follow these steps:

### Step 1: Convert Given Values
1. **Convert Temperature to Kelvin**:
\[ T = 17^\circ C + 273.15 = 290.15 \, K \]

2. **Convert Volume to Cubic Meters**:
\[ V = 1 \, L = \frac{1}{1000} \, m^3 = 0.001 \, m^3 \]

3. **Convert Pressure to Pascals**:
\[ P = 31.7 \, kPa = 31.7 \times 10^3 \, Pa = 31700 \, Pa \]

### Step 2: Use the Ideal Gas Equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( n \) = number of moles
- \( R = 8.31 \, J/(mol \cdot K) \)

### Step 3: Calculate Number of Moles
We can rearrange the ideal gas equation to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(31700 \, Pa)(0.001 \, m^3)}{(8.31 \, J/(mol \cdot K))(290.15 \, K)} \]
Calculating \( n \):
\[ n = \frac{31.7}{8.31 \times 290.15} \approx 0.013 \, mol \]

### Step 4: Calculate Molar Mass
We know the mass of the gas is 1g, so we can find the molar mass \( M \):
\[ M = \frac{mass}{n} = \frac{1 \, g}{0.013 \, mol} \approx 76.92 \, g/mol \]

### Step 5: Determine the Chemical Formula
Now, we need to find a combination of nitrogen (N) and oxygen (O) that gives a molar mass of approximately 76.92 g/mol. The atomic masses are:
- Nitrogen (N) = 14 g/mol
- Oxygen (O) = 16 g/mol

Let’s denote the number of nitrogen atoms as \( x \) and the number of oxygen atoms as \( y \). The equation for the molar mass is:
\[ 14x + 16y = 76.92 \]

### Step 6: Trial and Error for Integer Solutions
We can try different integer values for \( x \) and \( y \).

1. **If \( x = 2 \)**:
\[ 14(2) + 16y = 76.92 \implies 28 + 16y = 76.92 \implies 16y = 48.92 \implies y \approx 3.06 \quad \text{(not an integer)} \]

2. **If \( x = 3 \)**:
\[ 14(3) + 16y = 76.92 \implies 42 + 16y = 76.92 \implies 16y = 34.92 \implies y \approx 2.18 \quad \text{(not an integer)} \]

3. **If \( x = 2 \) and \( y = 3 \)**:
\[ 14(2) + 16(3) = 28 + 48 = 76 \quad \text{(close enough)} \]

Thus, the chemical formula is \( N_2O_3 \).

### Final Answer:
The chemical formula of the compound is \( N_2O_3 \).

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