Question

What is the approximate root means speed of line smoke particles of mass ${{10}^{-17}}kg$ at ${{27}^{o}}C$?
$\left( k=1.38\times {{10}^{-23}}J{{k}^{-1}} \right)$
$\begin{align}
  & A.\text{ }2.0\times {{10}^{-2}}m{{s}^{-1}} \\
 & B.\text{ }2.7\times {{10}^{-2}}m{{s}^{-1}} \\
 & C.\text{ }3.5\times {{10}^{-2}}m{{s}^{-1}} \\
 & D.\text{ }4.4\times {{10}^{-2}}m{{s}^{-1}} \\
\end{align}$

Answer 1

Hint: Evaluate the kinetic energy in terms of molar mass m and then find kinetic energy in terms of temperature according to the kinetic theory of gases. After that, compare two relations of kinetic energy and find out the root mean speed.
Formula used: $K.E=\dfrac{1}{2}mv_{rms}^{2}$
$K.E=\dfrac{3}{2}KT$

Complete step-by-step solution
We know that the mass of a particle is ${{10}^{-17}}kg$ and temperature${{27}^{o}}C$according to the kinetic theory of gases.
$K.E=\dfrac{3}{2}KT......\left( 1 \right)$
Where ‘k’ is Boltzmann constant having a value of $1.38\times {{10}^{-23}}$ and ‘T’ is the temperature which is given in the question. We also know that,
$K.E=\dfrac{1}{2}mv_{rms}^{2}.......\left( 2 \right)$
Where ‘m’ is the mass of a particle and ${{v}_{rms}}$ is the root mean speed.
On comparing (1) and (2) equations of kinetic energy we get,
$\dfrac{1}{2}mv_{rms}^{2}=\dfrac{3}{2}KT$
After simplifying this equation, find the value of ${{v}_{rms}}$ that is the root mean speed.
${{v}_{rms}}=\sqrt{\dfrac{3KT}{m}}$
Now substitute the values of K,T and m in the equation, we get,
${{v}_{rms}}=\sqrt{\dfrac{3\times 1.38\times {{10}^{-23}}\times 300}{{{10}^{-17}}}}$
After calculating this we get the value of root mean speed.
${{v}_{rms}}=3.5\times {{10}^{-2}}m{{s}^{-1}}$
Hence the correct option is (c).

Additional Information:
The square root of mean square speed is called root-mean-square speed or rms speed. It is denoted by the symbol ${{v}_{rms}}$. And for a given sample of gas higher temperature means a higher value of ${{v}_{rms}}$ and similarly, lower temperature means a lower value of ${{v}_{rms}}$.

Note: If any object having mass m and velocity of ${{v}_{rms}}$ then it is kinetic energy becomes $\dfrac{1}{2}mv_{rms}^{2}$ because as we know that kinetic energy is the energy possessed by the motion of the object. Potential energy is another form of energy, which is possessed by an object by virtue of its position in a system.