
The r.m.s speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the r.m.s speed becomes:
A.\[v\]
B. $\sqrt 2 v$
C. $2v$
D. $4v$
Answer
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Hint:To calculate the r.m.s. (root mean square) velocity of gas, get the formula of r.m.s speed and put the values in the formula .The formula is $\sqrt {\dfrac{{3RT}}{M}} $where, you have to put value of $T$ i.e. new temperature and $M$ i.e. new Molecular weight and you will get the desired result.
Complete step by step answer:Root mean square velocity of a gas is the measure of velocity of gas particles.
rms speed of any gas is given by:$\sqrt {\dfrac{{3RT}}{M}} $
where $T$ is the temperature and
$M$ is the molecular weight of the gas
Here, we can see that rms speed is directly proportional to square-root of temperature and inversely proportional to square root of molecule.
i.e.
rms speed $\alpha \sqrt T $ and,
rms speed $\alpha \dfrac{1}{{\sqrt M }}$
So, in question it would directly suggest the change is with the reference of temperature and molecular weight individually also. But to see the combined effect we have to use the formula of rms speed.
Now, in question it is given that, temperature is doubled:
That is, let us take new temperature to be \[T'\]
So, $T' = 2T - - - (1)$
And, in question it is given that oxygen molecule, dissociate into oxygen atoms:
Let us take new molecular weight be $M'$
So, $M' = \dfrac{M}{2} - - - (2)$
To calculate the new rms speed
Put, the values in formula, in $\sqrt {\dfrac{{3RT'}}{{M'}}} - - - (3)$
Put the values from $(1)$ and $(2)$in $(3)$
We get, $\sqrt {\dfrac{{3R(2T)}}{{M/2}}} $
Solving we get: \[\sqrt {\dfrac{{4(3RT)}}{M}} \]
From here, we can see that new velocity is double of old velocity
Let’s take new velocity be $v'$
$v' = 2v$
Therefore, option c is the correct answer.
Note: remember in the formula of rms ,R used is a constant, it’s value is same everywhere i.e. $R = 8.315J/mol.K$.and while solving questions always take care of units, all the units must be the same to get the desired results.
Complete step by step answer:Root mean square velocity of a gas is the measure of velocity of gas particles.
rms speed of any gas is given by:$\sqrt {\dfrac{{3RT}}{M}} $
where $T$ is the temperature and
$M$ is the molecular weight of the gas
Here, we can see that rms speed is directly proportional to square-root of temperature and inversely proportional to square root of molecule.
i.e.
rms speed $\alpha \sqrt T $ and,
rms speed $\alpha \dfrac{1}{{\sqrt M }}$
So, in question it would directly suggest the change is with the reference of temperature and molecular weight individually also. But to see the combined effect we have to use the formula of rms speed.
Now, in question it is given that, temperature is doubled:
That is, let us take new temperature to be \[T'\]
So, $T' = 2T - - - (1)$
And, in question it is given that oxygen molecule, dissociate into oxygen atoms:
Let us take new molecular weight be $M'$
So, $M' = \dfrac{M}{2} - - - (2)$
To calculate the new rms speed
Put, the values in formula, in $\sqrt {\dfrac{{3RT'}}{{M'}}} - - - (3)$
Put the values from $(1)$ and $(2)$in $(3)$
We get, $\sqrt {\dfrac{{3R(2T)}}{{M/2}}} $
Solving we get: \[\sqrt {\dfrac{{4(3RT)}}{M}} \]
From here, we can see that new velocity is double of old velocity
Let’s take new velocity be $v'$
$v' = 2v$
Therefore, option c is the correct answer.
Note: remember in the formula of rms ,R used is a constant, it’s value is same everywhere i.e. $R = 8.315J/mol.K$.and while solving questions always take care of units, all the units must be the same to get the desired results.
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