Question

Find the rms speed of an argon molecule at $27^0C$ (Molecular weight of argon=$40$ gm/mol)
A.$234.2{\text{ }}m/s$
B.$342.2{\text{ }}m/s$
C.$432.2{\text{ }}m/s$
D.$243.2{\text{ }}m/s$

Answer 1

Hint: -First convert the temperature in Kelvin by adding $273.15$ to the given value. Then convert the molecular mass from gm to kg. Then use the relation between rms speed, temperature and molecular mass, which is given as-
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ Where ${v_{rms}}$ is rms speed of the gas, T is the temperature, M is the molecular mass of the gas and R is universal gas constant. The value of R=$8.314$ J/molK. Now put the given values in the equation and solve it to get the answer.

Complete step by step answer:
Given temperature=$27^0C$
We know that to convert degree Celsius into Kelvin we add $273.15$ to the value of temperature.
So temperature T=$273.15 + 27 = 300.15$ K
Also given the molecular mass of argon=$40$ gm/mol
And we know that $1kg = 1000gm$ so –
$ \Rightarrow 1gm = {10^{ - 3}}kg$
Then the molecular mass of argon M=$40 \times {10^{ - 3}}$kg/mol
Now we have to find the rms speed of argon.
We know that the relation between rms speed, temperature and molecular mass is given as-
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ Where ${v_{rms}}$ is rms speed of the gas, T is the temperature, M is the molecular mass of the gas and R is universal gas constant. The value of R=$8.314$ J/molK
On putting the given values in the formula, we get-
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{3 \times 8.314 \times 300.15}}{{40 \times {{10}^{ - 3}}}}} $
On solving, we get-
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{7486.3413}}{{40 \times {{10}^{ - 3}}}}} $
On division, we get-
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{187.1585325}}{{{{10}^{ - 3}}}}} $
On simplifying, we get-
\[ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{187158.5325 \times {{10}^{ - 3}}}}{{{{10}^{ - 3}}}}} = \sqrt {187158.5325} \]
$ \Rightarrow {v_{rms}} = 432.2{\text{ m/s}}$

Hence the correct answer is C.

Note:
Argon is a noble gas which is colorless and odorless. It is an inert gas. It is used for-
-It is used in fluorescent tubes and low-energy light bulbs.
-It is used in the double-glazed windows to fill the space between panes.
-It is used in the tyres of luxury cars to reduce road noise and to protect the rubber.
-It is used in processing of titanium and in arc welding.
-The atmosphere created using this gas is used to grow crystals of silicon and germanium.