
The rms speed of the molecules of a gas in the vessel is $400m{{s}^{-1}}$ . If the temperature of the gas is reduced to half then the rms speed of the remaining molecules will be ……….
Answer
521.4k+ views
Hint: Temperature of the system is directly proportional to the rms velocity. The rms speed at a particular temperature is given to us. The ratio between the rms speed at temperature T and rms speed at temperature T/2 gives us the required answer.
Formula used: ${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
Complete step by step answer:
Measuring the velocities of particles at a given time results in a large distribution of values. Some particles may move very slowly, others very quickly, and because they are constantly moving in different directions, the velocity could equal zero. (Velocity is a vector quantity, equal to the speed and direction of a particle). To properly estimate the average velocity, average the squares of the velocities and take the square root of that value. This is known as the root-mean-square (RMS) velocity. We can represent the root mean square velocity by:
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
Where, R is the gas constant,
T is absolute temperature,
and M is the molar mass of the gas particles in $kgmo{{l}^{-1}}$ .
So any change in any of the variable parameters may lead to a change in ${{v}_{rms}}$ .
Let us only consider the temperature in the equation for this particular question.
From the equation, we can see that,
${{v}_{rms}}\propto \sqrt{T}$ .
Now, let us assume that the temperature at which molecules having rms velocity equal to $400m{{s}^{-1}}$ as T.
Then, we need to find the rms velocity of the molecules when this temperature is made half.
i.e. $T\to \dfrac{T}{2}$ .
We know, ${{v}_{rms}}\propto \sqrt{T}$ and ${{v}_{rms}}$ at temperature T is $400m{{s}^{-1}}$.
Then, ${{v}_{rms}}$ at temperature $\dfrac{T}{2}$ could be found by,
$\begin{align}
& \dfrac{400}{{{v}_{rms}}}=\dfrac{\sqrt{T}}{\sqrt{\dfrac{T}{2}}} \\
& \Rightarrow {{v}_{rms}}=\dfrac{400}{\sqrt{2}}=282.84m/s \\
\end{align}$
Therefore, the rms velocity of the molecules if temperature is reduced by half of the initial temperature is found to be $282.84m/s$.
Note: In the similar manner, we can find the root mean square velocity of any sample gas at a particular temperature. We can square the root mean square velocity and use it to obtain the kinetic energy of the particles in the system.
Formula used: ${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
Complete step by step answer:
Measuring the velocities of particles at a given time results in a large distribution of values. Some particles may move very slowly, others very quickly, and because they are constantly moving in different directions, the velocity could equal zero. (Velocity is a vector quantity, equal to the speed and direction of a particle). To properly estimate the average velocity, average the squares of the velocities and take the square root of that value. This is known as the root-mean-square (RMS) velocity. We can represent the root mean square velocity by:
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
Where, R is the gas constant,
T is absolute temperature,
and M is the molar mass of the gas particles in $kgmo{{l}^{-1}}$ .
So any change in any of the variable parameters may lead to a change in ${{v}_{rms}}$ .
Let us only consider the temperature in the equation for this particular question.
From the equation, we can see that,
${{v}_{rms}}\propto \sqrt{T}$ .
Now, let us assume that the temperature at which molecules having rms velocity equal to $400m{{s}^{-1}}$ as T.
Then, we need to find the rms velocity of the molecules when this temperature is made half.
i.e. $T\to \dfrac{T}{2}$ .
We know, ${{v}_{rms}}\propto \sqrt{T}$ and ${{v}_{rms}}$ at temperature T is $400m{{s}^{-1}}$.
Then, ${{v}_{rms}}$ at temperature $\dfrac{T}{2}$ could be found by,
$\begin{align}
& \dfrac{400}{{{v}_{rms}}}=\dfrac{\sqrt{T}}{\sqrt{\dfrac{T}{2}}} \\
& \Rightarrow {{v}_{rms}}=\dfrac{400}{\sqrt{2}}=282.84m/s \\
\end{align}$
Therefore, the rms velocity of the molecules if temperature is reduced by half of the initial temperature is found to be $282.84m/s$.
Note: In the similar manner, we can find the root mean square velocity of any sample gas at a particular temperature. We can square the root mean square velocity and use it to obtain the kinetic energy of the particles in the system.
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