
An ideal gas has pressure ‘P’, volume ‘V’ and absolute temperature ‘T’. If ‘m’ is the mass of each molecule and ‘K’ is the Boltzmann constant then density of the gas:
A. $\dfrac{Pm}{KT}$
B. $\dfrac{KT}{Pm}$
C. $\dfrac{Km}{PT}$
D. $\dfrac{PK}{Tm}$
Answer
513.6k+ views
Hint: Use the ideal gas equation to find the density of the gas. You can assume the mass of the gas and the molecular weight. Express the ideal gas equation in terms of Avogadro’s number and Boltzman constant.
Formula Used:
The density of a gas is given by,
$\rho =\dfrac{{{m}^{'}}}{v}$
Where,
${{m}^{'}}$ is the mass of the gas molecule
$v$ is the volume
The Ideal Gas equation is given by,
$PV=nRT$
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
T is the temperature
Complete step by step answer:
We can write the ideal gas equation,
$PV=nRT$………………….(1)
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
T is the temperature
$R=NK$ , N = Avogadro’s number and K = Boltzman Constant
We can replace ‘n’ with the following:
$n=\dfrac{{{m}^{'}}}{M}$
Where,
${{m}^{'}}$ is the mass of the gas
$M$ is the molecular weight of the gas.
Hence, we can write equation (1) in the following way,
$PV=(\dfrac{{{m}^{'}}}{M})RT$
$\Rightarrow P=(\dfrac{{{m}^{'}}}{V})(\dfrac{RT}{M})$
We can write the density of the gas as,
$\rho =\dfrac{{{m}^{'}}}{V}$
Hence, putting this expression we get,
$\Rightarrow P=\dfrac{\rho RT}{M}$
$\Rightarrow \dfrac{PM}{RT}=\rho $
$\Rightarrow \rho =\dfrac{PM}{NKT}$(As, R=NK)
Here, N is Avogadro's number and K is the Boltzmann Constant.
So, finally, we can write,
$\Rightarrow \rho =\dfrac{Pm}{KT}$(As, $m=\dfrac{M}{N}$)
So, the density of the gas is given by,
$\dfrac{Pm}{KT}$
Hence, the correct answer is - (A).
Note: Here, we have assumed that there is an N number of molecules in the gas. Hence, the mass of each molecule is given by,
$m=\dfrac{M}{N}$
This expression also shows that the ideal gas will always have a finite density until it reaches absolute zero temperature.
Formula Used:
The density of a gas is given by,
$\rho =\dfrac{{{m}^{'}}}{v}$
Where,
${{m}^{'}}$ is the mass of the gas molecule
$v$ is the volume
The Ideal Gas equation is given by,
$PV=nRT$
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
T is the temperature
Complete step by step answer:
We can write the ideal gas equation,
$PV=nRT$………………….(1)
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
T is the temperature
$R=NK$ , N = Avogadro’s number and K = Boltzman Constant
We can replace ‘n’ with the following:
$n=\dfrac{{{m}^{'}}}{M}$
Where,
${{m}^{'}}$ is the mass of the gas
$M$ is the molecular weight of the gas.
Hence, we can write equation (1) in the following way,
$PV=(\dfrac{{{m}^{'}}}{M})RT$
$\Rightarrow P=(\dfrac{{{m}^{'}}}{V})(\dfrac{RT}{M})$
We can write the density of the gas as,
$\rho =\dfrac{{{m}^{'}}}{V}$
Hence, putting this expression we get,
$\Rightarrow P=\dfrac{\rho RT}{M}$
$\Rightarrow \dfrac{PM}{RT}=\rho $
$\Rightarrow \rho =\dfrac{PM}{NKT}$(As, R=NK)
Here, N is Avogadro's number and K is the Boltzmann Constant.
So, finally, we can write,
$\Rightarrow \rho =\dfrac{Pm}{KT}$(As, $m=\dfrac{M}{N}$)
So, the density of the gas is given by,
$\dfrac{Pm}{KT}$
Hence, the correct answer is - (A).
Note: Here, we have assumed that there is an N number of molecules in the gas. Hence, the mass of each molecule is given by,
$m=\dfrac{M}{N}$
This expression also shows that the ideal gas will always have a finite density until it reaches absolute zero temperature.
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