Density of Air

Introduction

If you take the same size of raw cotton and iron piece in your hands, then you will feel that you are able to compress raw cotton but not the iron piece. This is because the atoms of the cotton are having a lot of gap between them but the atoms of the iron piece are closely packed. So we can say that iron is denser than raw cotton. We know that the atoms or molecules of gasses are much farther than that in solids. So gasses are generally less dense than solid.

From here we can conclude that the density is nothing but the mass occupied by the material in a unit volume and it can be represented as -

$$\mathrm{\rho\:=\:\frac{m}{v}}$$

where, $\mathrm{\rho}$ = density of the material, m = mass of the material and v = volume of the material. From this we can calculate the density of a solid very easily.

Suppose the density of any solid cube on earth is 1 $\mathrm{kg/m^3}$ then its density will be the same on any other planet too or its density will be the same at a height of 100 kms from the earth surface. But this is not the case with gas.

So in this tutorial we will understand about density of air, effects of different factors on the density of air.

Density of Air

The density of any gas can be calculated by the ideal gas equation after assuming that gas as an ideal gas. Now what is the ideal gas equation? Ideal gas equation is nothing but giving a relationship between three variables named as Pressure, Volume and Temperature. It can be written as -

$$\mathrm{P.v \:= \:n.\bar{R}.T}$$

Where, P = Pressure of the gas, v = volume of the gas, n = number of moles of the gas, $\mathrm{\bar{R}}$= Universal gas constant, T = Temperature of the gas.

As we know that the number of moles can be written as -

$$\mathrm{n\:=\:\frac{m}{M}}$$

Where, m = mass of the gas and M = Molar mass of the gas.

So the ideal gas equation can be written as -

$$\mathrm{P.v\:=\:\frac{m}{M}.\bar{R}.T}$$

$$\mathrm{\Rightarrow\:P.v\:=\:m.R.T.....…. (1)}$$

Where, R = Specific gas constant which can be calculated by dividing the $\mathrm{\bar{R}}$ with molar mass(M) and it depends on the molar mass of any particular gas.

Now if we want to calculate the density of that gas then we can write the equation (1) as -

$$\mathrm{P.v\:=\:m.R.T}$$

$$\mathrm{\Rightarrow\:P\:=\:\frac{m}{v}.{R}.T}$$

$$\mathrm{\Rightarrow\:P=\:\rho.R.T\:\:….. (2)}$$

Now, we can calculate the density at any given condition by using equation (2). WIth the help of the equation we can calculate the density of air at STP. For the general calculation we take the density of air as 1.29 $\mathrm{kg/m^3}$, but this density is at 0^oC and 760 mm of Hg (mercury). Barometer is used to measure the pressure of the air.

Density of Dry Air and Humid Air

Dry air: As we have discussed above, we can consider the air as an ideal gas. So we can calculate the density of air with the help of an ideal gas equation at any particular condition.

$$\mathrm{P\:=\:\rho.R.T}$$

$$\mathrm{\rho\:dry\:air\:=\:\frac{P}{RT}}$$

Where, R is the specific gas constant of dry air.

Humid air: Now we know that the air is the mixture of many gasses and water vapor. So the specific gas constant will depend on all the gasses and water vapor. Also, we have assumed that the water vapor is an ideal gas for our convenience in real life calculation. So we can apply the ideal gas equation to the water vapor too. So the density of the humid air will be the addition of density of the dry air and density of the water vapor. It can be represented as –

$$\mathrm{\Rightarrow\:\rho\:air\:=\:\rho\:dry\:air\:+\:\rho\:water\:vapor}$$

$$\mathrm{\Rightarrow\:\rho\:air\:=\:(\frac{p}{RT})dry\:air\:+(\:\frac{p}{RT})water\:vapor}$$

Calculation of Air Density

We will use the ideal gas equation for the calculation at STP (Temperature = 273.15 K and Pressure = 1 atmospheric pressure = $\mathrm{1.01325\:\times\:10^5}$ Pascal).

$$\mathrm{P \:=\: \rho.R.T }$$

For the air, the value of specific gas constant = 287.05 J/kg-K.

This comes from the equation -

$$\mathrm{R\:=\:\frac{R}{M}}$$

Where $\mathrm{R\:=\:8.314459\:J-mol^{-1}-K^{-1}}$ and $\mathrm{M_{air}\:=\:28.97}$ gram/mol (Combining all the composition of air for the molar mass)

$$\mathrm{\Rightarrow\:R\:=\:\frac{8.3144598}{29.87}}$$

$$\mathrm{\Rightarrow\:R\:=\:287\: J/kg-K}$$

Now at STP, we can write –

$$\mathrm{\Rightarrow\:1.01325\:\times\:10^5\:=\:\rho\:\times\:287\:\times\:273.15}$$

$$\mathrm{\Rightarrow\:\rho\:=\:1.29\:kg/m^3}$$

So this is the value of the density of air at STP.

Causes of Variation in the Density of Air

• Pressure: Take a piston cylinder arrangement and put the air inside it. Start compressing the air with the piston. Due to this the molecules of air start coming closer and more molecules will come within a small volume. It means that the same mass will be occupied in a small volume. So the density will increase. It means as the pressure increases the density will also increase. Also from the ideal gas equation - P = $\rho$.R.T , we can see that the pressure is directly proportional to density.

• Temperature: Let us again take a vertical piston cylinder arrangement and fill the air inside it. Start heating the base of the cylinder. After some time, we will observe that the piston starts moving upward. It happens because the molecules of air will get energy from the heating source. And after that the molecules start colliding with each other and the wall of the piston too. It will make the piston start moving upward. So the volume will increase after increasing the temperature and since the same mass will occupy more volume then we can say that the density decreases. Also from the ideal gas equation P = $\rho$.R.T, we can see that the $\rho$ is inversely proportional to T.

FAQs

Q1. Does the density of an iron block change when we move it from the surface of the earth to a height of 10 km from the surface of the earth?

Ans. No, the density of the iron block will not change. This is because the molecules of iron are very closely packed and do not get affected due to this much change in pressure.

Q2. What is the value of density of air at STP?

Ans. The value of density of air is 1.29 $\mathrm{kg/m^3}$ at STP.

Q3. Write the ideal gas equation.

Ans. $\mathrm{P.v = n.\bar{R}.T}$

Where, P = Pressure of the gas, v = volume of the gas, n = number of moles of the gas, $\mathrm{\bar{R}}$ = Universal gas constant, T = Temperature of the gas. As we know that the number of moles

Q4. What is the relationship between the density of air and the pressure?

Ans. The density of air is directly proportional to the pressure. So if the pressure increases the density of the air also increases and vice-versa.

Q5. How does the density of air depend on the temperature?

Ans. The density of air is inversely proportional to the temperature. So if the temperature increases the density of air decreases and vice-versa.