Q 11-12-016JEE MainMedium
A gas has density $1.2$ kg m$^{-3}$ at a pressure of $10^5$ Pa. Find the rms speed of its molecules in m s$^{-1}$.
Numerical value type. Enter your answer.
Answer: 500
Concept. Kinetic theory: $p = \dfrac13\rho v_{\text{rms}}^2$.
Step 1. Solve.
$$v_{\text{rms}} = \sqrt{\frac{3p}{\rho}} = \sqrt{\frac{3 \times 10^5}{1.2}} = \sqrt{2.5 \times 10^5} = 500\ \text{m s}^{-1}$$
Answer. $500$.
Solution by Sreeraj P, M.Sc Physics