Question

Find the density of air at 23 degree celcus and normal atmosphere pressure.

Answer 1

To find the density of air at 23 degrees Celsius and normal atmospheric pressure, we can use the Ideal Gas Law, which states:

\[ PV = nRT \]

where:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (\( R = 8.314 \, \text{J/(mol·K)} \))
- \( T \) = temperature (in kelvin)

We can also express the density of the gas (\( \rho \)) in terms of its molar mass (\( M \)):

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

For air, the average molar mass \( M \) is approximately \( 0.029 \, \text{kg/mol} \).

1. Convert temperature from degrees Celsius to Kelvin:

\[ T = 23 \, °C + 273.15 = 296.15 \, K \]

2. Use normal atmospheric pressure, which is approximately \( 101325 \, \text{Pa} \).

3. Now, we can substitute the values into the formula for density:

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

Calculating this gives:

\[
\rho = \frac{2956.425 \, \text{Pa·kg/mol}}{2468.0321 \, \text{J/mol}} \approx 1.198 \, \text{kg/m}^3
\]

Thus, the density of air at 23 degrees Celsius and normal atmospheric pressure is approximately:

\[
\boxed{1.19 \, \text{kg/m}^3}
\]