Question
Find the density of air at 23 degree celsius and normal atmospheric pressure
Answers
GPT-4o mini
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.
\[
PV = nRT
\]
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.