What is the density of air at 1.00 atm and 30 degrees Celsius given an average molar mass of 29.0 g/mol?

To calculate the density of air at the specified conditions, we can use the ideal gas law, which is given by the formula:

PV = nRT

Where:

  • P = pressure in atm
  • V = volume in liters
  • n = number of moles of gas
  • R = ideal gas constant (0.0821 L·atm/(K·mol))
  • T = temperature in Kelvin

First, we need to convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15 = 30 + 273.15 = 303.15 K

Next, we express the number of moles (n) in terms of density (ρ) and molar mass (M):

n = ρV / M

Substituting for n in the ideal gas law gives us:

P * V = (ρV / M) * R * T

We can simplify this by canceling V:

P = (ρ / M) * R * T

Rearranging to find the density (ρ) yields:

ρ = (PM) / (RT)

Now, we can substitute the values:

  • P = 1.00 atm
  • M = 29.0 g/mol
  • R = 0.0821 L·atm/(K·mol)
  • T = 303.15 K

Plugging in these values:

ρ = (1.00 atm * 29.0 g/mol) / (0.0821 L·atm/(K·mol) * 303.15 K)

Calculating the right side gives:

ρ ≈ (29.0 g/mol) / (24.916 L/mol) ≈ 1.16 g/L

Thus, rounding to appropriate significant figures, the density of air at 1.00 atm and 30 degrees Celsius is approximately 1.17 g/L.

The correct answer is e) 1.17 g/L.

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