Question

If the molecular weight of air is 28.9, what is the density of air at atmospheric pressure and a temperature of 354.5 K? 1 atm = 1.013 × 105 N/m2 , the mass of a proton is 1.67262 × 10−27 kg , Avogadro’s number is 6.02214 × 1023 mol−1 and k = 1.38065 × 10−23 N · m/K . Answer in units of kg/m3

Answer 1

Answer:

0.993 kg/m³

Explanation:

If the molecular weight of air is 28.9, what is the density of air at atmospheric pressure and a temperature of 354.5 K?

We can calculate the density (ρ) of air using the ideal gas equation.

$P.V=n.R.T=\frac{m}{M}.R.T$

where,

P is the pressure (P = 1.00 atm)

V is the volume

n is the number of moles

m is the mass

M is the molar mass (28.9 g/mol)

R is the ideal gas constant (0.08206 atm.L/mol.K)

T is the temperature (T = 354.5K)

$P.V=\frac{m}{M}.R.T\\\frac{P.M}{R.T} =\frac{m}{V} =\rho\\\rho=\frac{1.00atm\times 28.9g/mol}{(0.08206atm.L/mol.K)\times 354.5K } =0.993g/L\\\rho=\frac{0.993g}{L}\times \frac{1kg}{10^{3}g}\times \frac{10^{3}L }{1m^{3} } =0.993kg/m^{3}$