Answer
405k+ views
Hint: As we know that due to collision of molecules the speeds and kinetic energies of individual molecules keeps on changing and to calculate these speeds like RMS i.e the root mean square velocity as the name implies we first take square of the individual speeds and then their mean.
Complete Step by step answer:
> As we know that due to collision of molecules the speeds and kinetic energies of individual molecules keeps on changing. However, at a given temperature the average kinetic energy of a gas molecule remains constant.
> Let us assume that at a given temperature we have to calculate these speeds like RMS i.e the root mean square velocity as the name implies we first take the square of the individual speeds which may be assumed as ${V_1},{V_2}..$and ${n_1},{n_2}..$be the number of molecules then we will take mean and finally the square root of mean. Mathematically it can be represented as:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $where R is the gas constant, T is the temperature and M is the molecular mass of the gas.
Using this formula we can identify the RMS velocity of the gas. But in the question we are given with density and pressure of the gas and no information regarding which gas so to calculate the RMS speed we need to change the above formula in terms of pressure and density.
As we know that, $PV = nRT$ and $density = \dfrac{{mass}}{{volume}}$ using these two,
$
PV = nRT \\
\Rightarrow PM = dRT \\
\Rightarrow M = \dfrac{{dRT}}{P} \\
$
Putting the value of M in the RMS speed formula we get,
$
{V_{rms}} = \sqrt {\dfrac{{3RTP}}{{dRT}}} \\
\Rightarrow {V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\
$
Now, we know $d = 1.43 \times {10^{ - 2}}g\;c{c^{ - 1}}$and , to calculate the speed in $m{s^{ - 1}}$we will change $d = 1.43\;Kg{m^{ - 3}}$$P = 1atm = 1.013 \times {10^5}N{m^{ - 2}}$thus putting these values in the formula:
$
{V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\
\Rightarrow {V_{rms}} = \sqrt {\dfrac{{3 \times 1.013 \times {{10}^5}}}{{1.43}}} \\
\Rightarrow {V_{rms}} = 461m{s^{ - 1}} \\
$
Therefore, the correct answer is, ${V_{rms}} = 461m{s^{ - 1}}$.
Note: Remember that the RMS speed is directly proportional to the square root of the temperature and inversely proportional to the square root of mass, thus when the temperature is increased its RMS speed will increase.
Complete Step by step answer:
> As we know that due to collision of molecules the speeds and kinetic energies of individual molecules keeps on changing. However, at a given temperature the average kinetic energy of a gas molecule remains constant.
> Let us assume that at a given temperature we have to calculate these speeds like RMS i.e the root mean square velocity as the name implies we first take the square of the individual speeds which may be assumed as ${V_1},{V_2}..$and ${n_1},{n_2}..$be the number of molecules then we will take mean and finally the square root of mean. Mathematically it can be represented as:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $where R is the gas constant, T is the temperature and M is the molecular mass of the gas.
Using this formula we can identify the RMS velocity of the gas. But in the question we are given with density and pressure of the gas and no information regarding which gas so to calculate the RMS speed we need to change the above formula in terms of pressure and density.
As we know that, $PV = nRT$ and $density = \dfrac{{mass}}{{volume}}$ using these two,
$
PV = nRT \\
\Rightarrow PM = dRT \\
\Rightarrow M = \dfrac{{dRT}}{P} \\
$
Putting the value of M in the RMS speed formula we get,
$
{V_{rms}} = \sqrt {\dfrac{{3RTP}}{{dRT}}} \\
\Rightarrow {V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\
$
Now, we know $d = 1.43 \times {10^{ - 2}}g\;c{c^{ - 1}}$and , to calculate the speed in $m{s^{ - 1}}$we will change $d = 1.43\;Kg{m^{ - 3}}$$P = 1atm = 1.013 \times {10^5}N{m^{ - 2}}$thus putting these values in the formula:
$
{V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\
\Rightarrow {V_{rms}} = \sqrt {\dfrac{{3 \times 1.013 \times {{10}^5}}}{{1.43}}} \\
\Rightarrow {V_{rms}} = 461m{s^{ - 1}} \\
$
Therefore, the correct answer is, ${V_{rms}} = 461m{s^{ - 1}}$.
Note: Remember that the RMS speed is directly proportional to the square root of the temperature and inversely proportional to the square root of mass, thus when the temperature is increased its RMS speed will increase.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)