Question

If the density of the gas is $4kg{\text{ }}{m^{ - 3}}$ and its pressure is $1.2 \times {10^5}N{\text{ }}{m^{ - 2}}$ , how do I calculate the root-mean-square speed?

Answer 1

As, we can see in the question that it is not given any identity of gas so we will either assume or the variable cancel so we do not need to remember the molar mass, firstly, we have found the value of $RT$ with the help of ideal gas law and then put in the formula of root-mean-square speed and hence we will get the required solution.

Formula used:
${v_{RMS}} = \sqrt {\dfrac{{3RT}}{{{M_m}}}}$
Where, $R$ is the universal gas constant, $T$ is the temperature in kelvin and ${M_m}$ is the molar mass of the gas.

Complete step by step answer:
Let us consider ideal gas law,
i.e., $PV = nRT$
We can rewrite the above formula as,
$\dfrac{{PV}}{n} = RT$
Now substituting the above obtained formula in RMS equation,
${v_{RMS}} = \sqrt {\dfrac{{3RT}}{{{M_m}}}} \\\ \Rightarrow {v_{RMS}} = \sqrt {\dfrac{{3PV}}{{n{M_m}}}} \\\$
And we know that density is mass per unit volume so putting the value in denominator because there it is written in reciprocal,
${v_{RMS}} = \sqrt {\dfrac{{3P}}{\rho }} \\\ \Rightarrow {v_{RMS}} = \sqrt {\dfrac{{3 \times 1.2 \times {{10}^5}}}{4}} \\\ \Rightarrow {v_{RMS}} = \sqrt {0.9 \times {{10}^5}} \\\ \therefore {v_{RMS}} = 300\,m{\text{ }}{s^{ - 1}} \\\$
Hence, the root-mean-square speed is $300\,m{\text{ }}{s^{ - 1}}$.

Note: Note that in the question it is given in the same unit so we don’t have to worry about that but take care to convert all the given units in SI units for easy calculations while solving the problem. The topic of the question was kinetic theory of gases. This hypothesis posits that molecules are identical, that collisions between them are elastic, that there is no intermolecular attraction between them, and that the volume of each molecule is insignificant in comparison to the volume of the container.