
The RMS velocity of smoke particle of mass \[2.5\times {{10}^{-17}}kg\] in their Brownian motion NTP is (\[k=1.38\times {{10}^{-23}}J/mol\])?
Answer
473.1k+ views
Hint: We need to find the relation of the mass of the particle under consideration and the root mean square velocity of the particle at the Normal temperature and pressure conditions as per the problem to find the solution for it.
Complete answer:
We know that the smoke particles in the air move about in the random directions due to the collisions with the air molecules and the smoke particles itself. The random motion of the molecular sized particles in a medium due to the continuous inelastic collision is called the Brownian movement.
We know that for a particle which collides and moves throughout the medium will not possess a constant velocity. The magnitude and direction of the velocity vector changes as a result of the Brownian movement.
According to the Kinetic theory of gases, any gas molecule of certain mass at a particular temperature possesses a kinetic energy that remains constant for the given pressure conditions. We can equate the average kinetic energy of a particle using the relation as –
\[\dfrac{1}{2}m{{v}_{rms}}^{2}=\dfrac{3}{2}kT\]
Where, k is the Boltzmann constant,
T is the temperature of the surrounding.
We can apply this in the situation given to us. The smoke particle is moving in the normal temperature and pressure conditions, which is 293K temperature and 1 atm pressure.
\[\begin{align}
& \dfrac{1}{2}m{{v}_{rms}}^{2}=\dfrac{3}{2}kT \\
& \Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3kT}{m}} \\
& \Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3\times 1.38\times {{10}^{-23}}\times 293}{2.5\times {{10}^{-17}}}} \\
& \therefore {{v}_{rms}}=2.2\times {{10}^{-2}}m{{s}^{-1}} \\
\end{align}\]
The RMS velocity of the smoke particles is \[2.2cm{{s}^{-1}}\]in normal temperature and pressure conditions. This is the required solution.
Note:
We should be careful when considering the temperature of the Brownian movement. Here, it is clearly mentioned that the temperature is to be considered at normal temperature and pressure condition which is different from the standard (STP) condition.
Complete answer:
We know that the smoke particles in the air move about in the random directions due to the collisions with the air molecules and the smoke particles itself. The random motion of the molecular sized particles in a medium due to the continuous inelastic collision is called the Brownian movement.
We know that for a particle which collides and moves throughout the medium will not possess a constant velocity. The magnitude and direction of the velocity vector changes as a result of the Brownian movement.
According to the Kinetic theory of gases, any gas molecule of certain mass at a particular temperature possesses a kinetic energy that remains constant for the given pressure conditions. We can equate the average kinetic energy of a particle using the relation as –
\[\dfrac{1}{2}m{{v}_{rms}}^{2}=\dfrac{3}{2}kT\]
Where, k is the Boltzmann constant,
T is the temperature of the surrounding.
We can apply this in the situation given to us. The smoke particle is moving in the normal temperature and pressure conditions, which is 293K temperature and 1 atm pressure.
\[\begin{align}
& \dfrac{1}{2}m{{v}_{rms}}^{2}=\dfrac{3}{2}kT \\
& \Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3kT}{m}} \\
& \Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3\times 1.38\times {{10}^{-23}}\times 293}{2.5\times {{10}^{-17}}}} \\
& \therefore {{v}_{rms}}=2.2\times {{10}^{-2}}m{{s}^{-1}} \\
\end{align}\]
The RMS velocity of the smoke particles is \[2.2cm{{s}^{-1}}\]in normal temperature and pressure conditions. This is the required solution.
Note:
We should be careful when considering the temperature of the Brownian movement. Here, it is clearly mentioned that the temperature is to be considered at normal temperature and pressure condition which is different from the standard (STP) condition.
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