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
From $p = \dfrac{1}{3}\rho v_{\text{rms}}^2$:
$$v_{\text{rms}} = \sqrt{\frac{3p}{\rho}} = \sqrt{\frac{3 \times 10^5}{1.2}} = \sqrt{2.5 \times 10^5} = 500\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics