Question

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Use a molecular weight of 28.9 kg/kmol for air. Give answer in kg/m3

Answer 1

`d=0.92(kg)/(m^(3))`

Step-by-step explanation:

Using the Ideal Gas Law we have
`PV=nRT` and the number of moles n could be expressed as
`n=(m)/(M)`, where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

`PV=(m)/(M)RT`

If we pass the V to divide:

`P=(m)/(V)(RT)/(M)`

As the density is expressed as
`d=(m)/(V)`, we have:

`P=d(RT)/(M)`

Solving for the density:

`d=(PM)/(RT)`

Then we need to convert the units to the S.I.:

`T=100^(o)C+273.15`

`T=373.15K`

`P=1bar*(0.98atm)/(1bar)`

`P=0.98atm`

`M=28.9(kg)/(kmol)*(1kmol)/(1000mol)`

`M=0.0289(kg)/(mol)`

Finally we replace the values:

`d=((0.98atm)(0.0289(kg)/(mol)))/((0.082(atm.L)/(mol.K))(373.15K))`

`d=9.2*10^(-4)(kg)/(L)`

`d=9.2*10^(-4)(kg)/(L)*(1L)/(0.001m^(3))`

`d=0.92(kg)/(m^(3))`