Question

The root mean square velocity of gas molecules at $0^{\circ }C$ will be if at N.T.P. its density is $1.43Kg/m^3$

• A. 461 m/s
• B. 164 m/s
• C. 461 cm/s
• D. 164 cm/s

Answer 1

Answer: (C)

According to formula
$\Rightarrow P=\frac{1}{3}\rho {\bar{C}}^2$
Where $P$ is pressure of gas
$\bar{C}$ is rms velocity
$\rho$ is density of gas
Given,
$\Rightarrow P=1atm=1.013\times 10^5$ pascal
$\Rightarrow \rho =1.43kg/m^3$
So, $\bar{C}=\sqrt{\frac{3P}{\rho }}=\sqrt{\frac{3\times 1.013\times 10^5}{1.43}}$
$\Rightarrow \overline{C}\cong 461m/s$
Hence, the answer is $461m/s$.