Answer to Question #154499 in Physics for Abhinaba Banerjee

Question

Calculate the root mean square velocity of air molecules at N.T.P.(Density of air = 1.293 Kg/m³).

Answer 1

The root mean square velocity of air molecules can be found from the formula:

$$v_{rms}=\sqrt{\dfrac{3RT}{M}}.$$

Let's express $\dfrac{RT}{M}$ in terms of $P$ and $\rho$. From the ideal gas law, we have:

$$PV=nRT,$$

here, $n = \dfrac{m}{M}.$

Let's divide both sides of the equation by $V$:

$$P=\dfrac{m}{V}\dfrac{RT}{M}.$$

Since $\rho=\dfrac{m}{V}$ we have:

$$P=\rho\dfrac{RT}{M}$$

Finally, we get:

$$\dfrac{RT}{M}=\dfrac{P}{\rho}.$$

Then, we can calculate the root mean square velocity of air molecules at normal temperature and pressure:

$$v_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}},$$ $$v_{rms}=\sqrt{\dfrac{3\cdot 1.013\cdot10^5\ Pa}{1.293\ \dfrac{kg}{m^3}}}=485\ \dfrac{m}{s}.$$

Answer:

$v_{rms}=485\ \dfrac{m}{s}.$