he temperature at which the r.m.s. speed of $ {{\text{O}}_{\text{2}}} $ is equal to that of neon at $ \text{300 K} $ is:
A. $ \text{280 K} $
B. $ \text{480 K} $
C. $ \text{680 K} $
D. $ \text{180 K} $
VerifiedHint: The r.m.s. Speed or the root mean square speed of any gas is defined as the square root of the average of the squares of the speeds of the molecules. It is a measure of the speed of the molecules.
$ {{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3RT}}{\text{M}}} $ ,
where R is the universal gas constant, M is the molecular weight of the gas, T is the temperature in absolute scale.
Complete Answer:
To find out the rms speed of neon at $ \text{300 K} $ we need to know the molecular weights of the gases. For oxygen, the molecular weight is equal to 32 and for neon it is atomic weight since the noble gases do not form molecules and the atomic weight of neon is 10. Putting the values of the molecular weights in the earlier equation we get,
$ \left( {{\text{V}}_{\text{rms}}} \right)\text{oxygen =}\sqrt{\dfrac{\text{3RT}}{32}} $
And that for neon is, $ \left( {{\text{V}}_{\text{rms}}} \right)\text{neon =}\sqrt{\dfrac{\text{3RT}}{20}} $ , the value of T = $ \text{300 K} $ , $ \left( {{\text{V}}_{\text{rms}}} \right)\text{neon =}\sqrt{\dfrac{\text{3R300}}{20}}=\sqrt{\text{45R}} $
Now the temperature at which the temperature $ \left( {{\text{V}}_{\text{rms}}} \right)\text{oxygen =}\sqrt{\text{45R}} $ is, $ \sqrt{\text{45R}}\text{ =}\sqrt{\dfrac{\text{3RT}}{32}} $
Or, $ \text{ }\sqrt{\text{T}}=\sqrt{\dfrac{45\times 32}{3}} $ , hence the temperature is equal to $ \text{480 K} $ .
So, the correct answer is option B.
Note: The noble gases like neon, argon, krypton, etc. do not have the tendency to accept electrons in their valence shells because they are already filled up with the maximum number of electrons. They either have two electrons, valid only for helium or eight electrons in their valence shell which is the maximum number of electrons allowed in a valence shell. Hence they do not form molecules.
