Question

Derive a relation between density and molar mass of a gaseous substances from ideal gas equation

Answer 1

The relationships between molar mass and density for a monoatomic gas can be easily studied because gases are compressible so you can calculate or easily compress a gas and change its density at constant temperature. By changing the pressue of the gas you can use Boyle's Law to calculate the change in volume. Solids and liquids are much less compressible and you aren't able to use the Ideal Gas Law equation to solve for molar mass. 

The Ideal Gas Law, PV=nRT can be arrange so that n(moles) equals the mass/molar mass of the gas to become 

PV= mRT/M where m is the mass and M is the molar mass you can then solve for M using algebra to get 

M= mRT/PV, if you hold the temperature of a gas constant the equation reduces to boyles law or 

M= m/PV 

The mass will be constant assuming you have a closed container where you do not allow any gas to escape and if the conditions are close to ideal the PV will be a constant. You can form a direct relationship between the mass and PV. 

If you compare two gases such as He and Ne. Ne has a much larger molar mass (5x as much) as He. Therefore if you start with the same mass of each gas the PV value for each will be different. This just means that one is more compressible than the other. 

You can also compare Density since D= m/V and M=mRT/PV 

M= DRT/P 

The higher the density of the gas the higher the molar mass, and vice versa.
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Answer 2

Answer: [/tex]\rho=\frac{PM}{RT}[/tex]

Explanation: Ideal gas law in terms of density:

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of the gas

R = gas constant

moles=$\frac{\text {given mass}}{\text { molar mass}}$

$PV=\frac{w}{M}\times RT$

M = molar mass of gas

w = mass of gas

$P=\frac {wRT}{VM}$

$P=\frac{\rho\times RT}{M}$

$\rho=density=\frac{mass}{volume}$

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