Question

Determine the gas temperature at which
(a) The root mean square velocity of hydrogen molecules exceeds their most probable velocity by $ \Delta v = 400m/s $ .
(b) The velocity distribution function $ F(v) $ for the oxygen molecules will have the maximum value at the velocity $ v = 420m/s $ .

Answer 1

Hint: The key to solve this question is to remember the formulae of the root mean square velocity and the most probable velocity with their definition. Also, we have to be careful while solving the question as there are two types of molecules-hydrogen and oxygen having different molecular weights. So, their molecular mass should be known to be put in the formula.

Complete Step By Step Answer:
(a) The square root of the mean of squares of the speed of different molecules is called the root mean square speed. Mathematically, it is given by:
 $ {v_{rms}} = \sqrt {\dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + .....}}{N}} $
Where,
 $ {v_1},{v_2},{v_3} $ are the velocity of each particle.
 $ N $ is the total number of particles.
For, ideal gas, the root mean square velocity is given by:
 $ {v_{rms}} = \sqrt {\dfrac{{3kT}}{m}} .......(1) $
Where,
 $ {v_{rms}} $ is the root mean square velocity.
 $ k $ is Boltzmann's constant.
 $ T $ is the temperature.
 $ m $ is the molecular mass.
The speed which is possessed by the maximum fraction of the total number of molecules of the gas is called the most probable speed. Mathematically it is given as:
 $ {v_{mp}} = \sqrt {\dfrac{{2kT}}{m}} ......(2) $
Where,
 $ {v_{mp}} $ is the most probable velocity.
  $ k $ is Boltzmann's constant.
 $ T $ is the temperature.
 $ m $ is the molecular mass.
According to the question, $ \Delta v = 400m/s $ we get,
 $ {v_{rms}} - {v_{mp}} = \Delta v......(3) $
Substituting the values in equation (3) from equation (1) & (2) we get,
 $ \sqrt {\dfrac{{3kT}}{m}} - \sqrt {\dfrac{{2kT}}{m}} = 400 $
 $ \Rightarrow (\sqrt 3 - \sqrt 2 )\sqrt {\dfrac{{kT}}{m}} = 400 $
Squaring both sides, we get,
 $ {(\sqrt 3 - \sqrt 2 )^2}\dfrac{{kT}}{m} = {400^2} $


So, the value of temperature is given by the following formula:
 $ T = {\left( {\dfrac{{400}}{{\sqrt 3 - \sqrt 2 }}} \right)^2}\dfrac{m}{k} $
Now, the mass of the Hydrogen molecule, $ m \approx 2\,amu $ . We know that,
 $ 1\,amu = 1.661 \times {10^{ - 27}}kg $
So, the mass of hydrogen molecule in kg is $ 2 \times 1.661 \times {10^{ - 27}}kg = 3.322 \times {10^{ - 27}}kg $ and the value of Boltzmann’s constant is $ 1.38 \times {10^{ - 23}}J/K $ .
Now, substitute these values in equation (4), we get, $ T = {\left( {\dfrac{{400}}{{\sqrt 3 - \sqrt 2 }}} \right)^2}\dfrac{{3.322 \times {{10}^{ - 27}}}}{{1.38 \times {{10}^{ - 23}}}} $
 $ \Rightarrow T = 381.3K $

(b) At the most probable temperature, the speed of any gas is maximum. Therefore we use the formula of most probable velocity here. In the question, the value of velocity $ {v_{mp}} = 420m/s $ is given.
 $ {v_{mp}} = \sqrt {\dfrac{{2kT}}{m}} $
Squaring both sides, we get,
 $ {v_{mp}}^2 = \dfrac{{2kT}}{m} $
 $ \Rightarrow T = \dfrac{{{v_{mp}}^2m}}{{2k}}......(5) $
Now, the mass of the oxygen molecule, $ m \approx 32\,am{u_{}} $ . We know that,
 $ 1\,amu = 1.661 \times {10^{ - 27}}kg $
So, the mass of oxygen molecule in kg is $ 32 \times 1.661 \times {10^{ - 27}}kg = 53.152 \times {10^{ - 27}}kg $ and the value of Boltzmann’s constant is $ 1.38 \times {10^{ - 23}}J/K $ .
Now, substitute these values in equation (5), we get,
 $ T = \dfrac{{{{(420)}^2} \times 53.152 \times {{10}^{ - 27}}}}{{2 \times 1.38 \times {{10}^{ - 23}}}} $
Solving this we get,
 $ T = 339.7K $ .

Note:
The root means square velocity is directly proportional to the square root of the temperature. This implies that as the temperature increases, the RMS speed of the gas molecules also increases.
The moon has no atmosphere as the root mean square velocity of gas molecules is more than the escape velocity. The atmosphere is possible only when the root mean square velocity of gas molecules is less than the escape velocity of the planet or the satellite.