If the atmospheric pressure is 101.3 kPa and the temperature is 25°C25°C, what is the density of air? (Use the ideal gas law)
Question
If the atmospheric pressure is 101.3 kPa and the temperature is 25°C25°C, what is the density of air? (Use the ideal gas law)
Solution
To find the density of air using the ideal gas law, we need to know the pressure (P), the temperature (T), and the gas constant (R). The ideal gas law is PV = nRT, where n is the number of moles and V is the volume. We can rearrange this to find the density (ρ = n/V) as ρ = P/RT.
Given:
- The atmospheric pressure P = 101.3 kPa = 101300 Pa (since 1 kPa = 1000 Pa)
- The temperature T = 25°C = 298.15 K (since you have to convert the temperature from Celsius to Kelvin by adding 273.15)
- The gas constant R = 287 J/(kg·K) for air
Substituting these values into the formula, we get:
ρ = P/RT ρ = 101300 Pa / (287 J/(kg·K) * 298.15 K) ρ = 1.184 kg/m³
So, the density of air at 25°C and 101.3 kPa is approximately 1.184 kg/m³.
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