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.
1 answer