An unknown gas at 59.1 ∘C and 1.05 atm has a molar mass of 28.01 g/mol. Assuming ideal behavior, what is the density of the gas?

The density of this gas is 1.079 g/L

Explanation:

Step 1: Data given

Temperature of the gas = 59.1 °C = 273.15 + 59.1 = 332.25 Kelvin

Pressure of the gas = 1.05 atm

Molar mass of the gas = 28.01 g/mol

Step 2: Calculate density

Density ρ = mass / volume

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.05 atm

⇒with V = the volume of the gas = unknown

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature of the gas = 332.25 Kelvin

Since we don't know the number of moles, either the volume; we will calculate n/v

n/V = P/RT

Since ρ = m/V

and mass m = moles n * Molar mass MM

We can say that: (P*MM) / RT = (n*MM) / v = m/V = ρ

so ρ = (P*MM) / RT

ρ = (1.05 atm*28.01g/mol) / (0.08206 L*atm/K*mol * 332.25 Kelvin)

ρ = 1.079 g/L

The density of this gas is 1.079 g/L