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Question -
Derive the following expression by using the ideal gas equation [ or PV = nRT ] :
[tex]\longrightarrow[/tex] [tex]\large{\sf{Vapour\:density\:=\:\dfrac{Molecular\:Mass}{2}}}[/tex]
[tex]\rule{200}2[/tex]
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Hey there!
Question -
Derive the following expression by using the ideal gas equation [ or PV = nRT ] :
[tex]\longrightarrow[/tex] [tex]\large{\sf{Vapour\:density\:=\:\dfrac{Molecular\:Mass}{2}}}[/tex]
[tex]\rule{200}2[/tex]
Thank You.
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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.
Verified answer
Derivation
We have from ideal gas law
[tex]PV = nRT [/tex]
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 ,
[tex]d = \frac{m}{V} [/tex]
Where d is the density
m is the given mass
And
[tex]n = \frac{m}{M} [/tex]
where m is the given mass
M is the molecular mass
So using this value in the ideal gas law we have
[tex]PV = \frac{m}{M} RT \\ \implies PM = \frac{m}{V} RT \\ \implies \frac{PM}{RT} = d \\ \implies d = \frac{PM}{RT} [/tex]
Now taking the density of Hydrogen gas we have
[tex] d_{H} = \frac{PM_{H} }{RT} \\ \implies d_{H} = \frac{P \times 2}{RT} [/tex]
Since the molecular mass of hydrogen is 2
Now we know that ,
Vapour density = density of a gas/density of hydrogen gas
[tex]d_{v} = \frac{d}{d_{H} } \\ \implies d_{v} = \frac{ \frac{PM}{RT} }{ \frac{P \times 2}{RT} } \\ \\ \implies d_{v} = \frac{M}{2} [/tex]
Thus Vapour density = Molecular mass/2
[tex]Derived[/tex]