Question

The temperature at which R.M.S velocity of $ S{O_2} $ molecules is half that of helium molecules as $ 300{\text{ K}} $ ?
A. $ {\text{150 K}} $
B. $ 600{\text{ K}} $
C. $ 900{\text{ K}} $
D. $ 1200{\text{ K}} $

Answer 1

Hint: Root means square (R.M.S) velocity of a molecule is calculated by using the formula:
 $ {v_{r.m.s}}{\text{ = }}\sqrt {\dfrac{{3RT}}{M}} $
We will find the dependencies on which the root mean square velocity of a molecule depends. Then with the help of the relation between the R.M.S velocity of helium and $ S{O_2} $ molecules we can find the temperature at which R.M.S velocity of $ S{O_2} $ molecules is half that of helium molecules.
 $ {v_{r.m.s}}{\text{ = }}\sqrt {\dfrac{{3RT}}{M}} $
where,
 $ R{\text{ = }} $ Universal Gas constant
 $ {\text{T = }} $ Temperature of gas
 $ {\text{M = }} $ Molecular mass of gas.

Complete Step By Step Answer:
The root mean square (R.M.S) velocity is the square root of the mean of the all velocities of a molecule. It is represented by $ {v_{r.m.s}} $ and it can be calculated by using the formula:
 $ \Rightarrow {\text{ }}{v_{r.m.s}}{\text{ = }}\sqrt {\dfrac{{3RT}}{M}} $
We can observe that the only variables on which it depends is temperature of gas and molecular mass of gas molecules. Since all other terms are constant this can be deduced as,
 $ \Rightarrow {\text{ }}{v_{r.m.s}}{\text{ = }}\sqrt {\dfrac{T}{M}} $
For $ S{O_2} $ molecules it can be written as,
 $ {v_{r.m.s\left( {S{O_2}} \right)}}{\text{ = }}\sqrt {\dfrac{{{T_{S{O_2}}}}}{{{M_{S{O_2}}}}}} $ ___________ $ {\text{(1)}} $
Similarly for Helium molecules it can be written as,
 $ {v_{r.m.s\left( {H{e_2}} \right)}}{\text{ = }}\sqrt {\dfrac{{{T_{H{e_2}}}}}{{{M_{H{e_2}}}}}} $ _____________ $ {\text{(2)}} $
According to question $ {v_{r.m.s(S{O_2})}} $ is half of $ {v_{r.m.s(H{e_2})}} $ . It can be represented as:
 $ \dfrac{{{v_{r.m.s(S{O_2})}}}}{{{v_{r.m.s(H{e_2})}}}}{\text{ = }}\dfrac{1}{2} $ ______________ $ (3) $
Also $ {T_{He}} $ is given as $ 300{\text{ K}} $ and we can calculate molecular mass of both gases as,
Molecular mass of $ S{O_2} $ molecules $ {M_{S{O_2}}}{\text{ = }}\left( {32{\text{ + 16 }} \times {\text{ 2}}} \right){\text{ = 64 g}} $
Molecular mass of $ H{e_2} $ molecules $ {M_{H{e_2}}}{\text{ = }}\left( {2{\text{ }} \times {\text{ 2}}} \right){\text{ = 4 g}} $
Now dividing equation $ {\text{(1)}} $ and $ {\text{(2)}} $ we get the result as,
 $ \Rightarrow {\text{ }}\dfrac{{{v_{r.m.s(S{O_2})}}}}{{{v_{r.m.s(H{e_2})}}}}{\text{ = }}\sqrt {\dfrac{{{T_{S{O_2}}}{\text{ }} \times {\text{ }}{M_{H{e_2}}}}}{{{T_{H{e_2}}}{\text{ }} \times {\text{ }}{M_{S{O_2}}}}}} $
On substituting the result from equation $ (3) $ , we get the result as
 $ \Rightarrow {\text{ }}\dfrac{1}{2}{\text{ = }}\sqrt {\dfrac{{{T_{S{O_2}}}{\text{ }} \times {\text{ }}{M_{H{e_2}}}}}{{{T_{H{e_2}}}{\text{ }} \times {\text{ }}{M_{S{O_2}}}}}} $
On substituting the given values we get the result as,
 $ \Rightarrow {\text{ }}\dfrac{1}{2}{\text{ = }}\sqrt {\dfrac{{{T_{S{O_2}}}{\text{ }} \times {\text{ 4}}}}{{{\text{300 }} \times {\text{ 64}}}}} $
On squaring both sides and rearrange the terms for finding $ {T_{S{O_2}}} $ ,
 $ \Rightarrow {\text{ }}{{\text{T}}_{S{O_2}}}{\text{ = }}\dfrac{{300{\text{ }} \times {\text{ 64}}}}{{16}} $
 $ \Rightarrow {\text{ }}{{\text{T}}_{S{O_2}}}{\text{ = 1200 K}} $
Hence we get the temperature as $ {\text{1200 K}} $ .
Thus the correct option is D.

Note:
The value of universal gas constant changes only when the units of other parameters change. Thus we can treat it as constant. Helium in molecular form is present as $ H{e_2} $ not as $ He $ . The temperature of the gas should be in kelvin. If it is in degree celsius then do convert it into Kelvin scale.