Question

consider air as a 4:1 mixture of n2 and o2,what is the density of air at 28°c and 1atm​

Answer 1

hey mate....

here is your answer...

PV = nRT 

n = m/M (mass/molarmass) 

PV = mRT/M 

PM/RT = m/V = D 

The effective molar mass of air is 0.8x28 + 0.2x32 = 22.4 + 6.4 = 28.8g/mole. 

so D = 1atm x 28.8g/mole / (0.0821Latm/moleK x 301K) = 1.17g/L 

Another way to think about it is: 

At STP the density of air is 28.8g/22.4L = 1.29g/L 

Gases expand when heated so the density will decrease if T increases from 273 to 301 

so applying a temperature correction factor: 

1.29g/L x 273/301 = 1.17g/L

Answer 2

Answer:

Explanation:

As we know PM = dRT

d = PM / RT

where P = Pressure,

M = molar mass,

R = gas constant

T = temperature (in K )

The effective molar mass of air is 0.8x28 + 0.2x32 = 22.4 + 6.4 = 28.8g/mole.

So, d = 1atm x 28.8g/mole / (0.0821Latm/moleK x 301K) = 1.17g/L