
The root means square speed of gas molecules is $C_{rms}$. The root mean square in a specific direction will be:
A. $C_{rms}$.
B. $\dfrac{C_{rms}}{3}$.
C. $\dfrac{C_{rms}}{\sqrt 3}$.
D. $\sqrt 3 C_{rms}$
Answer
543.6k+ views
Hint: We know that the gases molecules move in random direction, when not confined to a definite boundary. We also know that velocity of the gases is a vector quantity. Using this understanding of gases we can answer this question.
Complete answer:
Consider some gas molecules in a definite box, whose boundaries are fixed. Since the nature of the gas molecules is to move in random directions, they will interact with the boundaries of the box and will bounce back at different time intervals.
Let $C_{rms}$ be the rms of the gas molecules, then $C_{rms}=\sqrt{c^{2}}$, where $c$ is the speed of the gases.
Now to account for the direction of these gases let us consider three dimensional spaces, we live in and let us limit the direction of velocity of the gases to the three coordinate system. Since velocity is a vector, we can say that the velocity of the gases is nothing but the vector sum of the velocity in the three directions.
Here, it is given that $C_{rms}$is the root mean square velocity of the gas molecules, then
$c^2=X^2+Y^2+Z^2$ where $X,\; Y\;,Z$ are the rms velocity of the gas in the x,y, z coordinate axis respectively.
Since gases move in a random fashion, let us assume that $X=Y=Z$
Then the $c^{2}=3X^2$
Then the velocity of the gases in any one direction is given as $\dfrac{1}{3}c^2$
Then the rms in any direction is given as $U_{rms}=\sqrt{\dfrac{1}{3}c^{2}}$
$\implies U_{rms}=\sqrt{\dfrac{1}{3}}C_{rms}$
Thus the correct answer is option C. $\dfrac{C_{rms}}{\sqrt 3}$.
Note:
As each gas molecule is moving with a different velocity, and in a different direction, it is very difficult to understand the nature of the gases as a whole .Hence to avoid this we use a new velocity called the root mean square velocity. We also know that velocity is a vector quantity which has both direction and magnitude. To avoid the randomness in the magnitude, we use the root mean square velocity.
Complete answer:
Consider some gas molecules in a definite box, whose boundaries are fixed. Since the nature of the gas molecules is to move in random directions, they will interact with the boundaries of the box and will bounce back at different time intervals.
Let $C_{rms}$ be the rms of the gas molecules, then $C_{rms}=\sqrt{c^{2}}$, where $c$ is the speed of the gases.
Now to account for the direction of these gases let us consider three dimensional spaces, we live in and let us limit the direction of velocity of the gases to the three coordinate system. Since velocity is a vector, we can say that the velocity of the gases is nothing but the vector sum of the velocity in the three directions.
Here, it is given that $C_{rms}$is the root mean square velocity of the gas molecules, then
$c^2=X^2+Y^2+Z^2$ where $X,\; Y\;,Z$ are the rms velocity of the gas in the x,y, z coordinate axis respectively.
Since gases move in a random fashion, let us assume that $X=Y=Z$
Then the $c^{2}=3X^2$
Then the velocity of the gases in any one direction is given as $\dfrac{1}{3}c^2$
Then the rms in any direction is given as $U_{rms}=\sqrt{\dfrac{1}{3}c^{2}}$
$\implies U_{rms}=\sqrt{\dfrac{1}{3}}C_{rms}$
Thus the correct answer is option C. $\dfrac{C_{rms}}{\sqrt 3}$.
Note:
As each gas molecule is moving with a different velocity, and in a different direction, it is very difficult to understand the nature of the gases as a whole .Hence to avoid this we use a new velocity called the root mean square velocity. We also know that velocity is a vector quantity which has both direction and magnitude. To avoid the randomness in the magnitude, we use the root mean square velocity.
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