The approximate number of air molecules in a 1 `m^3` volume at room temperature (300 K) and atmospheric pressure is (Use `R = 8.2 xx 10^(-5 ) m^3` atm/mol K and `N_A =6.02 xx 10^(23)` mol^(-1)`)

Options

A. 41
B. 450
C. `2.4xx 10^(25)`
D. `2.7 xx 10^(26)`

Answer

To find the approximate number of air molecules in a 1 m³ volume at room temperature (300 K) and atmospheric pressure, we can use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in m³)
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature (in K)

### Step 1: Identify the known values
- Volume \( V = 1 \, \text{m}^3 \)
- Temperature \( T = 300 \, \text{K} \)
- Pressure \( P = 1 \, \text{atm} \) (atmospheric pressure)
- Gas constant \( R = 8.2 \times 10^{-5} \, \text{m}^3 \, \text{atm} / \text{mol} \cdot \text{K} \)
- Avogadro's number \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)

### Step 2: Calculate the number of moles \( n \)
Using the ideal gas equation, we can rearrange it to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the known values:

\[ n = \frac{(1 \, \text{atm}) \times (1 \, \text{m}^3)}{(8.2 \times 10^{-5} \, \text{m}^3 \, \text{atm} / \text{mol} \cdot \text{K}) \times (300 \, \text{K})} \]

### Step 3: Calculate the denominator
Calculating the denominator:

\[ RT = (8.2 \times 10^{-5}) \times (300) = 2.46 \times 10^{-2} \, \text{m}^3 \, \text{atm} / \text{mol} \]

### Step 4: Calculate the number of moles \( n \)
Now, substituting back into the equation for \( n \):

\[ n = \frac{1}{2.46 \times 10^{-2}} \approx 40.65 \, \text{mol} \]

### Step 5: Calculate the number of molecules
To find the number of molecules \( N \), we can use the relation:

\[ N = n \times N_A \]

Substituting the values:

\[ N = 40.65 \, \text{mol} \times 6.02 \times 10^{23} \, \text{mol}^{-1} \]

### Step 6: Calculate the total number of molecules
Calculating \( N \):

\[ N \approx 40.65 \times 6.02 \times 10^{23} \approx 2.448 \times 10^{25} \, \text{molecules} \]

Thus, the approximate number of air molecules in a 1 m³ volume at room temperature and atmospheric pressure is:

\[ \boxed{2.448 \times 10^{25}} \]