Question

Helium is the lightest gas.
(A) True
(B) False

Answer 1

Hint: The helium gas $\text{(H}{{\text{e}}_{\text{2}}}\text{)}$ has an atomic molar mass as $\text{4 g mo}{{\text{l}}^{\text{-1}}}$ while the hydrogen gas molar mass is$2\text{ g mo}{{\text{l}}^{\text{-1}}}$. Therefore at constant temperature and pressure (STP) conditions, the two gases have the same number of moles but different weights at the constant volume. As density is equal to the mass divided by volume, the larger the mass density is the gas. Hydrogen is less dense and lighter gas compared to helium.

Complete step by step solution:
The gas which is less dense than the air can be called the lightest gas. The density of the gas depends on the volume and the mass of the gas.
The hydrogen and helium are considered to be the lightest gas. Let us find out which of them is the lightest gas.
Let us use the ideal gas law.
The gas law is given as:
$\text{PV=nRT}$
Where P is the pressure in pascal, V is the volume of gas in ${{\text{m}}^{\text{3}}}$, n is the number of moles, R is the ideal gas constant and T is the absolute temperature in kelvin.
Applying the standard temperature condition (STP)
 Where,
\[\begin{align}
  & \text{ P = 1}\text{.013 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{Pa} \\
 & \text{ R = 8}\text{.314 J mo}{{\text{l}}^{\text{-1}}}{{\text{K}}^{\text{-1}}} \\
 & \text{ T = 298}\text{.15 K} \\
 & \text{And consider the volume as:} \\
 & \text{V=0}\text{.001 }{{\text{m}}^{\text{3}}} \\
\end{align}\]
For hydrogen gas:
\[\text{PV=}{{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}\text{RT}\]
Let us substitute the values we get,
\[\begin{align}
  & \text{(1}\text{.013 }\times \text{1}{{\text{0}}^{\text{5}}}\text{ Pa) (0}\text{.001 }{{\text{m}}^{\text{3}}}\text{)= }{{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}(8.314\text{ J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\text{)(298}\text{.15 K)} \\
 & \Rightarrow {{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}=\text{ }\dfrac{\text{(1}\text{.013 }\times \text{1}{{\text{0}}^{\text{5}}}\text{ Pa) (0}\text{.001 }{{\text{m}}^{\text{3}}}\text{)}}{(8.314\text{ J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\text{)(298}\text{.15 K)}} \\
 & \Rightarrow {{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}=\text{ }\dfrac{\text{(101}\text{.3)}}{(2478.82\text{ mo}{{\text{l}}^{\text{-1}}}\text{ )}} \\
 & \Rightarrow {{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}=\text{ 0}\text{.0408 mol} \\
\end{align}\]
Since the molecular weight of ${{\text{H}}_{\text{2}}}$ the gas is $2\text{ g mo}{{\text{l}}^{\text{-1}}}$
Thus the mass of ${{\text{H}}_{\text{2}}}$gas is,
$\begin{align}
  & {{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}\text{=}\dfrac{\text{m}}{\text{M}} \\
 & \Rightarrow \text{m}=\text{ }{{\text{n}}_{\text{(}{{\text{H}}_{\text{2}}}\text{)}}}\text{ }\times \text{ M} \\
 & \Rightarrow \text{m}=\text{ 0}\text{.0408 }\times \text{ 2 g} \\
 & \Rightarrow \text{m}=\text{ 0}\text{.0817 g} \\
\end{align}$
Thus the mass of the ${{\text{H}}_{\text{2}}}$gas is$\text{0}\text{.0817 g}$. (1)

For helium gas,
\[\text{PV=}{{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}\text{RT}\]
Let us substitute the values we get,
\[\begin{align}
  & \text{(1}\text{.013 }\times \text{1}{{\text{0}}^{\text{5}}}\text{ Pa) (0}\text{.001 }{{\text{m}}^{\text{3}}}\text{)= }{{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}(8.314\text{ J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\text{)(298}\text{.15 K)} \\
 & \Rightarrow {{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}=\text{ }\dfrac{\text{(1}\text{.013 }\times \text{1}{{\text{0}}^{\text{5}}}\text{ Pa) (0}\text{.001 }{{\text{m}}^{\text{3}}}\text{)}}{(8.314\text{ J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\text{)(298}\text{.15 K)}} \\
 & \Rightarrow {{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}=\text{ }\dfrac{\text{(101}\text{.3)}}{(2478.82\text{ mo}{{\text{l}}^{\text{-1}}}\text{ )}} \\
 & \Rightarrow {{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}=\text{ 0}\text{.0408 mol} \\
\end{align}\]
Since the molecular weight of $\text{H}{{\text{e}}_{\text{2}}}$ the gas is $\text{4 g mo}{{\text{l}}^{\text{-1}}}$
Thus the mass of $\text{H}{{\text{e}}_{\text{2}}}$gas is,
$\begin{align}
  & {{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}\text{=}\dfrac{\text{m}}{\text{M}} \\
 & \Rightarrow \text{m}=\text{ }{{\text{n}}_{\text{(H}{{\text{e}}_{\text{2}}}\text{)}}}\text{ }\times \text{ M} \\
 & \Rightarrow \text{m}=\text{ 0}\text{.0408 }\times \text{ 4 g} \\
 & \Rightarrow \text{m}=\text{ 0}\text{.1632 g} \\
\end{align}$
Thus the mass of the $\text{H}{{\text{e}}_{\text{2}}}$gas is$\text{0}\text{.1632 g}$. (2)
The density is directly related to the mass of the gas. Thus from (1) and (2), we get that the density of helium is greater than the hydrogen. Therefore hydrogen is the lightest gas.
$\begin{align}
  & \text{Density=}\dfrac{\text{mass}}{\text{Volume}} \\
 & \text{ or} \\
 & \text{Density }\propto \text{ mass } \\
 & \text{(when volume =constant)} \\
\end{align}$
Hydrogen is so light because its molar mass is only $2\text{ g mo}{{\text{l}}^{\text{-1}}}$ and a fixed amount of gas occupies a fixed volume regardless of the species of the gas. Hydrogen is the lightest element.
Thus, the helium is not the lightest gas. But it is the second lightest element.

The given statement is not true. Hence, (B) is the correct option.

Note: The air weighs 1 gram then compared to air the hydrogen and helium gases are less dense. Thus the balloon filled with hydrogen gas or helium gas has a higher tendency to float than the balloon filled with the air. This can be also explained based on bouncy, the liquid or air tends to exert the upward force on the body floating in it.