Q. If $r.m.s.$ velocity of a gas is $V_{rms} = 1840\ m/s$ and its density $\rho = 8.99 \times 10^{- 2}kg/m^{3},$ the pressure of the gas will be
A) $1.01\ N/m^{2}$
B) $1.01 \times 10^{3}N/m^{2}$
C) $1.01 \times 10^{5}N/m^{2}$
D) $1.01 \times 10^{7}N/m^{2}$
Explanation
(c) $V_{rms} = \sqrt{\frac{3P}{\rho}}\ or\ P = \frac{\rho V_{rms}^{2}}{3}$
$= \frac{8.99 \times 10^{- 2} \times 1840 \times 1840}{3} = 1.01 \times 10^{5}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.