Question

The density of neon will be highest at
(A) $STP$
(B) $0^\circ C,2atm$
(C) $273^\circ C,1atm$
(D) $273^\circ C,2atm$

Answer 1

Hint – You can start the solution by describing what factors decide the density of a material. Then use the equations $PV = nRT$ and $n = \dfrac{w}{M}$. Then finally use the equation for density $d = \dfrac{w}{V}$ to find the relation between $P,d$ and $T$.

Complete step by step solution:
Density of a material is decided by two internal factors, the level of compression of the material and the kinetic energy of the material. There also exists internal factors like temperature, pressure, etc. that affect the internal factors and cause the density to change.
We know that the density of materials especially gases and liquids change with temperature. For example – The density of water is maximum at $4^\circ C$.
We also know that for an ideal gas, the following is used.
$PV = nRT$(Equation 1)
Here
$P =$Pressure,
$V =$Volume,
$n =$Number of moles,
$R =$Universal gas constant
$T =$Temperature.
We also know
$n = \dfrac{w}{M}$(Equation 2)
Here
$w =$Weight,
$M =$Molar mass.
Using the value of number of moles derived in equation 2 in equation 2, we get
$PV = \left( {\dfrac{w}{M}RT} \right)$
$\Rightarrow PM = \left( {\dfrac{w}{V}} \right)RT$
As $density(d) = \dfrac{{weight}}{{volume}}$
So,
$PM = dRT$
$\Rightarrow d = \dfrac{{PM}}{{RT}}$
As $M$and$R$are constants
$d \propto P$
$d \propto \dfrac{1}{T}$
Thus low temperature conditions and high pressure conditions will result in greater density of Neon ($Ne$).
Choosing from the available options it is clear that the density of Neon will be highest at $0^\circ C$ and $2atm$.
Hence, option B is the correct choice.

Note – You can also figure out the solution theoretically. You know that more pressure will result in the molecules of the substance being pushed closely together in compact form leading to an increase in density. Also increasing the temperature will provide more kinetic energy to the molecules further decreasing the density of the substance.