Question

What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3

Answer 1

Answer : The density of air in
`lbm/ft^3` and
`kg/m^3` is,
`0.0743lbm/ft^3` and
`1.19kg/m^3` respectively.

Explanation :

`PV=nRT\\\\PV=(m)/(M)RT\\\\P=(m)/(V)(RT)/(M)\\\\P=\rho (RT)/(M)\\\\\rho=(PM)/(RT)`

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

`^oC=(^oF-32)* (5)/(9)`

`^oC=(75-32)* (5)/(9)=24^oC`

The conversion used for the temperature from degree Celsius to Kelvin is:

`K=273+^oC`

`K=273+24=297K`

n = number of moles

m = mass of air

M = average molar mass of air = 28.97 g/mole

`\rho` = density of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above formula, we get:

`\rho=(PM)/(RT)`

`\rho=((1atm)* (28.97g/mole))/((0.0821L.atm/mol.K)* (297K))`

`\rho=1.19g/L`

Now we have to calculate density in
`lbm/ft^3`.

Conversion used :

`1g/L=0.0624lbm/ft^3`

So,

`1.19g/L=(1.19g/L)/(1g/L)* 0.0624lbm/ft^3=0.0743lbm/ft^3`

The density of air in
`lbm/ft^3` is,
`0.0743lbm/ft^3`

Now we have to calculate density in
`kg/m^3`.

Conversion used :

`1g/L=1kg/m^3`

So,

`1.19g/L=1.19kg/m^3`

The density of air in
`kg/m^3` is,
`1.19kg/m^3`