Question

If the ratio of vapour density for hydrogen and oxygen is $\dfrac{1}{{16}}$ , then under constant pressure, what is the ratio of their root mean square velocities?
A. $\dfrac{4}{1}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{{16}}$
D. $\dfrac{{16}}{1}$

Answer 1

Hint:Root mean square velocity of individual gas molecules is given by ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ , where T is the temperature of gas molecules in Kelvin, M is the molar mass and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . In terms of density, it is given by ${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $ , where $P$ is pressure exerted by the gas molecules. To solve the above question, use this formula of root mean square velocity.



Formula used:
Root mean square velocity:
 ${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $

Complete answer:
RMS velocity of gas molecules is given by:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ … (1)
Here, T is the temperature of gas molecules in Kelvin,
M is the molar mass and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$.
From ideal gas law, we know that:
$PV = nRT$ … (2)
We also know that Molar mass of a gas molecule is given by the ratio of molecular mass, $m$ of the gas and number of moles, $n$ , that is,
$M = \dfrac{m}{n}$
Substituting the value of $n$ from the above equation in (2),
$PVM = mRT$
Rearranging,
$PM = \dfrac{m}{V}RT$
But we also know that density, \[\rho = \dfrac{m}{V}\] .
Thus, $PM = \rho RT$ which gives $\dfrac{{RT}}{M} = \dfrac{P}{\rho }$ .
Substituting this in equation (1), we get:
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $ … (3)
At constant pressure the RMS velocity is inversely proportional to the square root of the densities of the gas molecules.
Therefore,
$\dfrac{{{{\left( {{v_{rms}}} \right)}_1}}}{{{{\left( {{v_{rms}}} \right)}_2}}} = \sqrt {\dfrac{{{\rho _2}}}{{{\rho _1}}}} $
In the above question we are given the ratio of vapor densities for hydrogen and oxygen to be $\dfrac{1}{{16}}$ .
Substituting this is the above relation, we get:
$\dfrac{{{{\left( {{v_{rms}}} \right)}_{{H_2}}}}}{{{{\left( {{v_{rms}}} \right)}_{{O_2}}}}} = \sqrt {\dfrac{{16}}{1}} = 4$
Thus, the correct option is A.



Note: In terms of density and pressure, the root mean square velocity of individual gas molecules is given by ${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $ . With pressure as a constant value, you can clearly deduce from this relation that the RMS velocity of a gas is inversely proportional to the square root of its density