Question

Root mean square speed of $ {N_2} $ in air is $ 515\dfrac{m}{s} $ . What is the average speed of $ He $ gas in the same air?

Answer 1

Hint: The root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas. Another point to be noted is that if the medium of the two gases remains the same, then the temperature and other constants must also be taken the same.

Complete answer:
Another way to define the Root mean square speed is that it can be said to be the value of the square root of the sum of the squares of the stacking velocity values divided by the number of values.
Some important points to remember are:
(1) All gas particles move with random speed and direction.
(2) Solving for the average velocity of gas particles gives us the average velocity of zero, assuming that all particles are moving equally in different directions.
(3) You can acquire the average speed of gaseous particles by taking the root of the square of the average velocity.
(4) The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect a material’s kinetic energy.
The formula for calculating the root mean square speed is $ {V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} ........(1) $
Where, $ {V_{rms}} $ is the root mean square speed.
 $ R $ is the gas constant.
 $ T $ is the temperature.
 $ M $ is the molar mass.
Now, we know that the molar mass of nitrogen is $ 24gm $
By putting the molar mass in equation (1), we get,
 $ {({V_{rms}})_{nitrogen}} = \sqrt {\dfrac{{3RT}}{{24}}} ........(2) $ $ {({V_{rms}})_{nitrogen}} = \sqrt {\dfrac{{3RT}}{{24}}} ........(2) $
Also, we know that the molar mass of helium is $ 4gm $
 Again, By putting the molar mass in equation (1), we get,
 $ {({V_{rms}})_{helium}} = \sqrt {\dfrac{{3RT}}{4}} ........(3) $
 As the medium remains same then R and T of both the gases are also same,
On dividing equation (2) and (3); 3, R and T gets cancelled and we get,
 $ \dfrac{{{{({V_{rms}})}_{helium}}}}{{{{({V_{rms}})}_{nitrogen}}}} = \sqrt {\dfrac{{24}}{4}} $
 $ \dfrac{{{{({V_{rms}})}_{helium}}}}{{{{({V_{rms}})}_{nitrogen}}}} = \sqrt 6 $
On further solving, we get,
 $ \dfrac{{{{({V_{rms}})}_{helium}}}}{{{{({V_{rms}})}_{nitrogen}}}} = 2.6 $
Now, we are given that $ {({V_{rms}})_{nitrogen}} = 515\dfrac{m}{s} $
So, on putting this value in the above equation, we get,
 $ {({V_{rms}})_{helium}} = (2.6 \times 515)\dfrac{m}{s} $
 $ {({V_{rms}})_{helium}} = (1339)\dfrac{m}{s} $
So, the final answer is $ {({V_{rms}})_{helium}} = (1339)\dfrac{m}{s} $ .

Note:
Often average speed is used for calculations but when average speed does not work the root mean square speed is used. The major difference between average velocity and root mean square speed is that the average speed is a vector quantity whereas the root mean square speed is a scalar quantity.