Question

Calculate (find) the density of air at 23°C and normal atmospheric pressure

Answer 1

To calculate the density of air at a specific temperature and pressure, you can use the Ideal Gas Law:

\[
PV = nRT
\]

Where:
- \( P \) = pressure in pascals (Pa)
- \( V \) = volume in cubic meters (m³)
- \( n \) = number of moles (mol)
- \( R \) = universal gas constant (\( R \approx 8.314 \, \text{J/(mol·K)} \))
- \( T \) = temperature in kelvins (K)

The density (\( \rho \)) of an ideal gas can be expressed as:

\[
\rho = \frac{m}{V} = \frac{PM}{RT}
\]

Where:
- \( m \) = mass of the gas
- \( M \) = molar mass of the gas (for air, approximately \( 0.029 \, \text{kg/mol} \))

### Given:
- Temperature: \( T = 23^\circ C = 23 + 273.15 = 296.15 \, K \)
- Atmospheric pressure: \( P = 101325 \, \text{Pa} \)
- Molar mass of air: \( M \approx 0.029 \, \text{kg/mol} \)
- Universal gas constant: \( R \approx 8.314 \, \text{J/(mol·K)} \)

### Calculation:

We can plug the values into the density formula:

\[
\rho = \frac{PM}{RT}
\]

Substituting the values:

\[
\rho = \frac{(101325 \, \text{Pa})(0.029 \, \text{kg/mol})}{(8.314 \, \text{J/(mol·K)})(296.15 \, K)}
\]

Calculating the denominator:

\[
8.314 \times 296.15 \approx 2468.05 \, \text{J/(mol·K)}
\]

Now substituting that back into the equation for density:

\[
\rho = \frac{(101325)(0.029)}{2468.05}
\]

Calculating the numerator:

\[
101325 \times 0.029 \approx 2948.425 \, \text{Pa·kg/mol}
\]

Putting that in:

\[
\rho = \frac{2948.425}{2468.05} \approx 1.194 \, \text{kg/m}^3
\]

### Conclusion:

The density of air at 23°C and normal atmospheric pressure is approximately **1.194 kg/m³**.