Question

find the density of air at 23°c and normal atmospheric pressure

Answer 1

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**.