Question

Hey there!

Question -

Derive the following expression by using the ideal gas equation [ or PV = nRT ] :

$\longrightarrow$ $\large{\sf{Vapour\:density\:=\:\dfrac{Molecular\:Mass}{2}}}$


$\rule{200}2$

Thank You. ​

Answer 1

Answer:

From ideal gas equation we have :

P V = n R T

= > n = P V / R T

We know :

No. of moles n = Given mass W / Molar mass M

i.e. n = W / M

= > W / M =  P V / R T

= > W / V =  P M / R T

But mass per unit volume is equal to density d :

= > n = W / V = d

= > d =  P M / R T

Since V.D. of gas is given as :

V.D. of gas = Density of gas / Density of H₂

= > V.D. of gas = ( P M / R T ) / ( P × 2 / R T )

As we know molar mass of H₂ is 2 g / mol , we used in above expression and P , R & T get cancel out :

Hence we left with :

= >  V.D. of gas = M / 2

Therefore , V.D. is equal of half of molecular mass.

Answer 2

Derivation

We have from ideal gas law

$PV = nRT$

Where

• P is the pressure of the gas

• V is the volume

• n is the number of moles

• R is the ideal gas constant

• T is the temperature

Now we know that ,

$d = \frac{m}{V}$

Where d is the density

m is the given mass

And

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

where m is the given mass

M is the molecular mass

So using this value in the ideal gas law we have

$PV = \frac{m}{M} RT \\ \implies PM = \frac{m}{V} RT \\ \implies \frac{PM}{RT} = d \\ \implies d = \frac{PM}{RT}$

Now taking the density of Hydrogen gas we have

$d_{H} = \frac{PM_{H} }{RT} \\ \implies d_{H} = \frac{P \times 2}{RT}$

Since the molecular mass of hydrogen is 2

Now we know that ,

Vapour density = density of a gas/density of hydrogen gas

$d_{v} = \frac{d}{d_{H} } \\ \implies d_{v} = \frac{ \frac{PM}{RT} }{ \frac{P \times 2}{RT} } \\ \\ \implies d_{v} = \frac{M}{2}$

Thus Vapour density = Molecular mass/2

$Derived$