
Is there a gas lighter than helium?
Answer
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Hint: Helium is an inert gas with atomic number $2$. It is a colourless and odourless gas and due to its inert nature, it exists as an inert monatomic gas while the hydrogen atom which has atomic number $1$ exists as a diatomic gas. At standard temperature and pressure conditions, these two gases have the same number of moles and the gas with less density at STP will be considered as the lightest gas.
Complete answer:
Assuming the helium and hydrogen gas to be ideal, so at standard pressure and temperature, the density of each gas can be estimated as follows:
Ideal gas equation in terms of density is as follows:
$PM = DRT\;\; - (1)$
Where, P is the pressure, M is the molar mass of the gas, D is the density, R is the universal gas constant and T is absolute temperature.
We know that at standard temperature and pressure i.e., at STP, the pressure considered is 1 atm at 273 K temperature. Therefore, the density of hydrogen and helium gas can be calculated using ideal gas equation at STP as follows:
For hydrogen gas:
Molecular mass $ = 2\;gmo{l^{ - 1}}$
Substituting values in equation (1):
$1 \times 2 = D \times 0.0821 \times 273$
$ \Rightarrow {D_{{H_2}}} = \dfrac{2}{{0.0821 \times 273}}$
$ \Rightarrow {D_{{H_2}}} = 0.089\;g{L^{ - 1}}$
For helium gas:
Molecular mass $ = 4\;gmo{l^{ - 1}}$
Substituting values in equation (1):
$1 \times 4 = D \times 0.0821 \times 273$
$ \Rightarrow {D_{He}} = \dfrac{4}{{0.0821 \times 273}}$
$ \Rightarrow {D_{He}} = 0.178\;g{L^{ - 1}}$
As the density of the hydrogen gas is less than helium gas at STP. Therefore, we can conclude that hydrogen gas is lighter than helium gas.
Note:
It is important to note that helium gas is used in balloons instead of hydrogen gas because helium is an inert gas and due to its incombustible properties, it does not catch fire while hydrogen gas is very reactive and has a tendency to undergo combustion readily.
Complete answer:
Assuming the helium and hydrogen gas to be ideal, so at standard pressure and temperature, the density of each gas can be estimated as follows:
Ideal gas equation in terms of density is as follows:
$PM = DRT\;\; - (1)$
Where, P is the pressure, M is the molar mass of the gas, D is the density, R is the universal gas constant and T is absolute temperature.
We know that at standard temperature and pressure i.e., at STP, the pressure considered is 1 atm at 273 K temperature. Therefore, the density of hydrogen and helium gas can be calculated using ideal gas equation at STP as follows:
For hydrogen gas:
Molecular mass $ = 2\;gmo{l^{ - 1}}$
Substituting values in equation (1):
$1 \times 2 = D \times 0.0821 \times 273$
$ \Rightarrow {D_{{H_2}}} = \dfrac{2}{{0.0821 \times 273}}$
$ \Rightarrow {D_{{H_2}}} = 0.089\;g{L^{ - 1}}$
For helium gas:
Molecular mass $ = 4\;gmo{l^{ - 1}}$
Substituting values in equation (1):
$1 \times 4 = D \times 0.0821 \times 273$
$ \Rightarrow {D_{He}} = \dfrac{4}{{0.0821 \times 273}}$
$ \Rightarrow {D_{He}} = 0.178\;g{L^{ - 1}}$
As the density of the hydrogen gas is less than helium gas at STP. Therefore, we can conclude that hydrogen gas is lighter than helium gas.
Note:
It is important to note that helium gas is used in balloons instead of hydrogen gas because helium is an inert gas and due to its incombustible properties, it does not catch fire while hydrogen gas is very reactive and has a tendency to undergo combustion readily.
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