Question

A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?

Answer 1

Hint The total mass of a gas is equal to the number of molecules multiplied by the mass of the molecule. Pressure is directly proportional to temperature and inversely proportional to volume, the constant of proportionality, in one of its forms can be given as the Boltzmann constant multiplied by the number of molecules of the gas.
Formula used: $PV = NkT$ where P, T, V are the pressure, absolute temperature, and volume, respectively, N is the number of molecules and k is the Boltzmann constant.

Complete step by step answer
According to Charles’s law, the volume of a fixed pressure of gas is directly proportional to its absolute temperature i.e. $V \propto T$. And Boyle’s law states that the pressure of a gas is inversely proportional to its volume when the temperature is constant i.e. $P \propto \dfrac{1}{V}$.
An ideal gas is any gas, irrespective of type, which obeys both Boyle’s and Charles’s law. Combining both laws we have that
$\Rightarrow P \propto \dfrac{T}{V} $
$\Rightarrow P = K\dfrac{T}{V} $
where K is a constant. In reality, an ideal gas doesn’t exist but at low pressures and temperature many gases exceptionally approach the ideal behavior.
To solve the above equation, we recall the ideal gas relation i.e. PV = NkT.
Multiplying both sides by m and dividing both sides by V, we get
$\Rightarrow Pm = \dfrac{{mNkT}}{V}$
Also, mN = M where M is the mass of the gas. Thus, we get,
$\Rightarrow Pm = \dfrac{{MkT}}{V}$
Dividing both sides by kT we get,
$\Rightarrow \dfrac{M}{V} = \dfrac{{Pm}}{{kT}} $
$\Rightarrow \rho = \dfrac{{Pm}}{{kT}} $
Where $\rho$ is the density.
Hence, the correct option is C.

Note
Often, questions of this kind can be solved by a simple dimensional analysis or unit check. Thus, we could check which of the options corresponds with the unit of density.
For option A,
$\Rightarrow mkT = kg \cdot \dfrac{J}{K} \cdot K = Jkg$
But
$\Rightarrow J = Nm = Kgm{s^{ - 2}} \cdot m = kg{m^2}{s^{ - 2}}$
Thus, the unit of mkT is
$\Rightarrow kg{m^2}{s^{ - 2}} \cdot kg = k{g^2}{m^2}{s^{ - 2}}$
Which is not the unit of density.
For option B,
$\Rightarrow \dfrac{P}{{kT}} = \dfrac{{kgm{s^{ - 2}}}}{{{m^2}}} \div \left( {\dfrac{{kgm{s^{ - 2}} \cdot m}}{K}K} \right)$
Simplifying further we get,
$\Rightarrow \dfrac{1}{{{m^3}}} = {m^{ - 3}}$ which is also not the unit of density.
For option C, we see that this is the expression of option B multiplied by mass.
Therefore, we multiply the unit of option B by kg which gives
$\Rightarrow \dfrac{1}{{{m^3}}} \cdot kg = \dfrac{{kg}}{{{m^3}}} = kg{m^{ - 3}}$ which is the unit of density.
Thus, option C could be the solution.
To ensure it is certainly the solution, we investigate option D.
In option D, the unit of \dfrac{P}{{kTV}} is simply the unit of option B divided by the unit of volume.
Hence,
$\Rightarrow \dfrac{1}{{{m^3}}} \div {m^3} = \dfrac{1}{{{m^3}}} \cdot \dfrac{1}{{{m^3}}} = {m^{ - 6}}$
This is not the unit of density. Thus, it can be established that option is certainly the solution, since it is the only option whose expression as a unit of density.