Question

The density of a gas is $6 \times {10^{ - 2}}kg{m^{ - 3}}$ and the root mean square velocity of the gas molecules is $500m{s^{ - 1}}$. The pressure exerted by the gas on the walls of the vessel is
A. $5 \times {10^3}N{m^{ - 2}}$
B. $1.2 \times {10^{ - 4}}N{m^{ - 2}}$
C. $0.83 \times {10^{ - 4}}N{m^{ - 2}}$
D. $30N{m^{ - 2}}$

Answer 1

Hint:According to the question, we need to find the pressure exerted by the gas on the vessel having certain density and root mean square velocity. We can use the formula of pressure $P = \dfrac{1}{3}\rho {v^2}$ where we had used all the standard symbols for density and velocity respectively.

Complete step by step answer:
Let us first understand the meaning of root mean square velocity. Root mean square velocity is the average speed of particles in gas. It is represented as,
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
where $M$ is the molecular weight of the gas and $T$ is the temperature of the gas.
Coming to the question, applying the formula for pressure.
Pressure exerted by the gas is $P = \dfrac{1}{3}\rho {v^2}$ where $\rho $ is the density of the gas and $v$ is the root mean square velocity of the gas.
From the question we had $\rho = 6 \times {10^{ - 2}}kg{m^{ - 3}}$ and $v = 500m{s^{ - 1}}$
Substituting the values
$P = \dfrac{1}{3}\left( {6 \times {{10}^{ - 2}}} \right){\left( {500} \right)^2}$
$\therefore P = 5 \times {10^3}N{m^{ - 2}}$

Hence, the correct option is A.

Note:The question was related to the topic Kinetic theory of gases. This theory assumes that molecules are identical, collisions between molecules to be elastic, no intermolecular attraction between the molecules and volume of molecule is negligible as compared to the volume of the container.