To find the density of air at 23°C (296.15 K) and normal atmospheric pressure (101.325 kPa), we can use the ideal gas law, which is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = amount of substance (in moles)
- \( R \) = universal gas constant (\( R = 8.314 , \text{J/(mol·K)} \))
- \( T \) = temperature (in kelvins)
We can rewrite the ideal gas law in terms of density. The density (\( \rho \)) can be expressed as:
\[ \rho = \frac{PM}{RT} \]
Where:
- \( \rho \) = density of the gas (in kg/m³)
- \( P \) = pressure (in pascals)
- \( M \) = molar mass of air (approximately \( 0.029 , \text{kg/mol} \) for dry air)
- \( R \) = universal gas constant (approximately \( 287.05 , \text{J/(kg·K)} \) for dry air)
- \( T \) = temperature (in kelvins)
Step 1: Convert the pressure to pascals:
\[ P = 101.325 , \text{kPa} = 101325 , \text{Pa} \]
Step 2: Use \( R \) for dry air:
For dry air, \( R \) can be used as \( 287.05 , \text{J/(kg·K)} \).
Step 3: Convert the temperature to kelvins:
\[ T = 23 , \text{°C} = 23 + 273.15 = 296.15 , \text{K} \]
Step 4: Substitute values into the density formula:
\[ \rho = \frac{PM}{RT} = \frac{101325 , \text{Pa} \times 0.029 , \text{kg/mol}}{287.05 , \text{J/(kg·K)} \times 296.15 , \text{K}} \]
However, since we are using \( R \) in \( J/(kg·K) \), we will only need the \( R \) when using the \( 0.029 , \text{kg/mol} \) value, it's more straightforward to use:
\[ \rho = \frac{P}{RT} = \frac{101325}{287.05 \times 296.15} \]
Now calculating:
\[ \rho = \frac{101325}{287.05 \times 296.15} \approx \frac{101325}{8517.901} \approx 11.89 , \text{kg/m}^3 \]
The final calculation gives the density of air at 23°C and normal atmospheric pressure:
\[ \boxed{1.19 , \text{kg/m}^3} \]
This value is consistent with standard atmospheric conditions for air.
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