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 :

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

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:

$^oC=(^oF-32)\times \frac{5}{9}$

$^oC=(75-32)\times \frac{5}{9}=24^oC$

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:

$\rho=\frac{PM}{RT}$

$\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}$

$\rho=1.19g/L$

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

Conversion used :

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

So,

$1.19g/L=\frac{1.19g/L}{1g/L}\times 0.0624lbm/ft^3=0.0743lbm/ft^3$

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, $1.19kg/m^3$