Question

The density of air at a pressure of ${{10}^{5}}N{{m}^{-2}}$ is $1.2kg{{m}^{-3}}$. Under these conditions, the root mean square velocity of the air molecules in $m{{s}^{-1}}$ is:
$\begin{align}
  & (A)500 \\
 & (B)1000 \\
 & (C)1500 \\
 & (D)3000 \\
\end{align}$

Answer 1

Hint: For a gas consisting of vey high but finite number of molecules within itself, the root mean square velocity or the RMS velocity of the gas is defined as the root of mean of sum of the squares of all the velocities of all the different particles at any fixed point. We will use the standard formula of root mean square velocity to get the answer of our problem.

Complete step-by-step answer:
Let the molar mass of air be given by ${{M}_{0}}$ and the temperature at which we are calculating the root mean square velocity be T.
Then, the formula for root mean square velocity is given by:
$\Rightarrow {{U}_{rms}}=\sqrt{\dfrac{3RT}{{{M}_{0}}}}$ [Let this expression be equation number (1)]
Now, we need to convert this equation in terms of pressure and density.
Therefore, on using ideal gas equation we have:
$\Rightarrow PV=nRT$ [Let this expression be equation number (2)]
Here,
$\Rightarrow V=\dfrac{M}{\rho }$ , where M is the mass of the gas and $\rho $ its density.
Also, the number of moles can be written as:
$\Rightarrow n=\dfrac{M}{{{M}_{0}}}$
Putting these two values in equation number (2), we get:
$\begin{align}
  & \Rightarrow P\left( \dfrac{M}{\rho } \right)=\left( \dfrac{M}{{{M}_{0}}} \right)RT \\
 & \Rightarrow \dfrac{P}{\rho }=\dfrac{RT}{{{M}_{0}}} \\
\end{align}$
Multiplying both sides by 3 and then taking square roots both side, we have:
$\Rightarrow \sqrt{\dfrac{3P}{\rho }}=\sqrt{\dfrac{3RT}{{{M}_{0}}}}$
Now, using equation number (1). We get:
$\Rightarrow {{U}_{rms}}=\sqrt{\dfrac{3P}{\rho }}$
Putting the values of pressure and density as:
 $\begin{align}
  & \Rightarrow P={{10}^{5}}N{{m}^{-2}} \\
 & \Rightarrow \rho =1.2kg{{m}^{-3}} \\
\end{align}$
We get the root mean square velocity as:
$\begin{align}
  & \Rightarrow {{U}_{rms}}=\sqrt{\dfrac{3\times {{10}^{5}}}{1.2}} \\
 & \therefore {{U}_{rms}}\approx 500m{{s}^{-1}} \\
\end{align}$
Hence, the root mean square velocity comes out to be $500m{{s}^{-1}}$ .

So, the correct answer is “Option A”.

Note: For a gas under a certain pressure, there are three types of velocities. The average speed of molecules, the root mean square speed and the most probable speed of the molecules. Their formulas are almost similar, so one should not confuse with another while solving a question. Their order of magnitude being RMS speed, then average speed and then most probable speed.