Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

what is the ratio of the mean speed of $\text{ }{{\text{O}}_{\text{3}}}\text{ }$molecules to the RMS speed of a $\text{ }{{\text{O}}_{2}}\text{ }$molecule at the same T?
A) $\text{ }{{\left( \dfrac{3\text{ }\!\!\pi\!\!\text{ }}{7} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\text{ }$
B) $\text{ }{{\left( \dfrac{16}{9\text{ }\!\!\pi\!\!\text{ }} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\text{ }$
C) $\text{ }{{\left( 3\text{ }\!\!\pi\!\!\text{ } \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\text{ }$
D) $\text{ }{{\left( \dfrac{\text{4 }\!\!\pi\!\!\text{ }}{9} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\text{ }$

seo-qna
Last updated date: 20th Jun 2024
Total views: 395.7k
Views today: 9.95k
Answer
VerifiedVerified
395.7k+ views
Hint: The mean square speed of the gas is given as, $\text{ mean speed = }\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ M}}}\text{ }$and the root mean square (RMS) speed of the gas molecules is equal to the average speed of particles. It is given as, $\text{ rms = }\sqrt{\dfrac{\text{3RT}}{\text{M}}}\text{ }$.where, R is gas constant, T is the absolute temperature and M is the molar mass of molecules.

Complete step by step solution:
We know that the mean speed is an average of the speed particles. It is a square root of the average velocity of the molecules in a gas.
The root mean square speed is given as,
$\text{ mean speed = }\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ M}}}\text{ }$
Where R is gas constant, T is the absolute temperature and M is the molar mass of molecules.
The root mean square (RMS) speed is used to measure the average speed of particles in a gas. Mathematically it is represented as follows
$\text{ rms = }\sqrt{\dfrac{\text{3RT}}{\text{M}}}\text{ }$
Where R is gas constant, T is the absolute temperature and M is the molar mass of molecules.
Let’s first calculate the molecular weight of ozone $\text{ }{{\text{O}}_{\text{3}}}\text{ }$ and oxygen gas $\text{ }{{\text{O}}_{2}}\text{ }$.molecular weight $\text{ }{{\text{O}}_{\text{3}}}\text{ }$is:
$\text{ MW of }{{\text{O}}_{\text{3}}}\text{ }=\text{ 3}\times \left( 16 \right)\text{ = 48 }$
The molecular weight of oxygen gas $\text{ }{{\text{O}}_{2}}\text{ }$is:
$\text{ MW of }{{\text{O}}_{2}}\text{ }=\text{ 2}\times \left( 16 \right)\text{ = 32 }$
The mean speed ( $\text{ }{{\text{V}}_{\text{mean}}}\text{ }$) for the ozone $\text{ }{{\text{O}}_{\text{3}}}\text{ }$molecule is written as,
$\text{ mean speed of }{{\text{O}}_{\text{3}}}\left( {{\text{V}}_{\text{mean}}} \right)\text{= }\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ M}}}\text{ = }\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ }\times \text{48}}}\text{ }$ (1)
Let’s this as an equation (1) .now the root mean square or RMS speed ($\text{ }{{\text{V}}_{\text{rms}}}\text{ }$) for oxygen gas is written as,
$\text{ rms speed of }{{\text{O}}_{\text{2}}}\text{ molecule }\left( {{\text{V}}_{\text{rms}}} \right)\text{= }\sqrt{\dfrac{\text{3RT}}{\text{M}}}\text{ =}\sqrt{\dfrac{\text{3RT}}{32}\text{ }}\text{ }$ (2)
We are interested to determine the ratio of the mean speed of a $\text{ }{{\text{O}}_{\text{3}}}\text{ }$molecule to the RMS speed of the oxygen gas. Let’s divide equation (1) by equation (2).On dividing we have,
$\begin{align}
& \text{ }\dfrac{\text{mean speed of }{{\text{O}}_{\text{3}}}}{\text{rms speed of }{{\text{O}}_{\text{2}}}}\text{ = }\dfrac{\text{ }{{\text{V}}_{\text{mean}}}\text{ }}{\text{ }{{\text{V}}_{\text{rms}}}\text{ }}\text{ = }\dfrac{\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 48}}}}{\sqrt{\dfrac{\text{3RT}}{\text{32}}\text{ }}} \\
& \Rightarrow \dfrac{\text{ }{{\text{V}}_{\text{mean}}}\text{ }}{\text{ }{{\text{V}}_{\text{rms}}}\text{ }}\text{ = }\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 48}}}\text{ }\times \text{ }\sqrt{\dfrac{32}{\text{3RT}}} \\
& \Rightarrow \dfrac{\text{ }{{\text{V}}_{\text{mean}}}\text{ }}{\text{ }{{\text{V}}_{\text{rms}}}\text{ }}\text{ = }\sqrt{\dfrac{16}{9\text{ }\!\!\pi\!\!\text{ }}}\text{ } \\
\end{align}$
Thus the ratio of the mean speed of ozone gas and to the root mean square speed of oxygen gas is equal to, $\sqrt{\dfrac{16}{9\text{ }\!\!\pi\!\!\text{ }}}\text{ }$or $\text{ }{{\left( \dfrac{16}{9\text{ }\!\!\pi\!\!\text{ }} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\text{ }$.

Hence, (B) is the correct option.

Note: Note that, for a particular gas the ratio of the speed of rms to the average speed is equal to,
$\text{ }\dfrac{{{\text{V}}_{\text{rms}}}}{{{\text{V}}_{\text{mean}}}}\text{= }\sqrt{\dfrac{\text{3RT}}{\text{M}}}\text{:}\sqrt{\dfrac{\text{8RT}}{\text{ }\!\!\pi\!\!\text{ M}}}\text{ = }\sqrt{\text{3}}\text{:}\sqrt{\dfrac{\text{8}}{\text{ }\!\!\pi\!\!\text{ }}}\text{ = 1}\text{.181 : 1 }$
This relation is applicable for the same gas only.