What is the estimated density of air at 300 K and 101,325 Pa using the ideal gas law?

What is the estimated density of air at 300 K and 101,325 Pa using the ideal gas law?

• A. 1.2 kg/m³
• B. 1.4 kg/m³
• C. 1.0 kg/m³
• D. 1.5 kg/m³

Answer: (A)

To calculate the estimated density of air at a temperature of 300 K and a pressure of 101,325 Pa using the ideal gas law, the formula applied is: \[ PV = nRT \] where: - P is the pressure (Pa), - V is the volume (m³), - n is the number of moles of gas, - R is the ideal gas constant (approximately 8.314 J/(mol·K)), - T is the temperature (K). Rearranging this equation to find the density (ρ) involves recognizing that density is mass (m) divided by volume (V). The number of moles can be expressed as: \[ n = \frac{m}{M} \] where M is the molar mass of air, approximately 0.029 kg/mol. Substituting for n in the ideal gas equation, we can write: \[ P = \left(\frac{m}{M}\right)RT/V \] Rearranging gives us: \[ \rho = \frac{m}{V} = \frac{PM}{RT} \] Now, substituting the known values: - P = 101,325 Pa, -

To calculate the estimated density of air at a temperature of 300 K and a pressure of 101,325 Pa using the ideal gas law, the formula applied is:

[

PV = nRT

]

where:

• P is the pressure (Pa),

• V is the volume (m³),

• n is the number of moles of gas,

• R is the ideal gas constant (approximately 8.314 J/(mol·K)),

• T is the temperature (K).

Rearranging this equation to find the density (ρ) involves recognizing that density is mass (m) divided by volume (V). The number of moles can be expressed as:

[

n = \frac{m}{M}

]

where M is the molar mass of air, approximately 0.029 kg/mol.

Substituting for n in the ideal gas equation, we can write:

[

P = \left(\frac{m}{M}\right)RT/V

]

Rearranging gives us:

[

\rho = \frac{m}{V} = \frac{PM}{RT}

]

Now, substituting the known values:

• P = 101,325 Pa,