To find the density of air at 23 °C and normal atmospheric pressure (101.325 kPa), we can use the Ideal Gas Law, which is given by the formula:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of the gas
- \( R \) = ideal gas constant (approximately \( 8.314 , \text{J/(mol·K)} \))
- \( T \) = temperature (in Kelvin)
First, we convert the temperature from Celsius to Kelvin:
\[ T = 23 + 273.15 = 296.15 , K \]
We can rewrite the ideal gas law to express the density (\( \rho \)) of the air, which is defined as mass per unit volume:
\[ \rho = \frac{P M}{R T} \]
Where:
- \( \rho \) = density in kg/m³
- \( P \) = pressure in pascals (101325 Pa for normal atmospheric pressure)
- \( M \) = molar mass of dry air (approximately 0.029 kg/mol)
- \( R \) = specific gas constant for dry air, approximately \( 287.05 , \text{J/(kg·K)} \)
- \( T \) = temperature in Kelvin
Using the specific gas constant \( R \) for dry air, the density can be calculated as:
\[ \rho = \frac{P}{R T} \]
Now inserting the values:
\[ \rho = \frac{101325}{287.05 \times 296.15} \]
Calculating the denominator:
\[ R T = 287.05 \times 296.15 \approx 85089.9 \]
Now, substituting back into the density equation:
\[ \rho = \frac{101325}{85089.9} \approx 1.19 , \text{kg/m}^3 \]
Thus, the density of air at 23 °C and normal atmospheric pressure is approximately 1.19 kg/m³.