Question

The molecules of a given mass of gas have an rms velocity $200m{s^{ - 1}}$ at $27^\circ C$ and ${10^5}N{m^{ - 2}}$ pressure. When the absolute temperature is doubled and the pressure is halved, the rms speed of the molecules of the same gas is:
(A) $200m{s^{ - 1}}$
(B) $400m{s^{ - 1}}$
(C) $200\sqrt 2 m{s^{ - 1}}$
(D) $400\sqrt 2 m{s^{ - 1}}$

Answer 1

From the formula for the rms velocity of the gas, we see that the velocity is directly proportional to the square root of temperature and is independent of its pressure. So by taking the ratio of the velocities in the two situations, we can find the rms velocity in case 2 in terms of the rms velocity in the first case.

Formula Used: In this question, we will be using the following formula,
$\Rightarrow {v_{rms}} = \sqrt {\dfrac{{3KT}}{m}}$
where ${v_{rms}}$ is the rms velocity,
$K$ is the Boltzmann constant
$T$ is the given temperature and
$m$ is the mass of the given gas.

Complete step by step answer
Here we are given the rms velocity of a given mass of gas at a certain temperature and pressure. The rms velocity is the square root of the average of the squares of the velocities. The rms velocity of a certain mass of gas at a particular temperature is given by,
$\Rightarrow {v_{rms}} = \sqrt {\dfrac{{3KT}}{m}}$
From the formula, we can see that the rms velocity is only dependent on the temperature of certain gas and not the pressure of the surroundings. This is because, when the pressure varies then the density of the gas also changes.
Therefore from here, we can write,
$\Rightarrow {v_{rms}} \propto \sqrt T$. This is because for a certain gas all other given variables are constant.
Now, in the question, we are given that the rms velocity in the first case is given by,
$\Rightarrow {v_{rms}}_1 \propto \sqrt {{T_1}}$
and for the second case, we have,
$\Rightarrow {v_{rms}}_2 \propto \sqrt {{T_2}}$
Therefore if we take the ratio of the velocities in the 2 cases we have,
$\Rightarrow \dfrac{{{v_{rms}}_2}}{{{v_{rms}}_1}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}}$
Now we are given that the absolute temperature gets doubled. Therefore we can write, $\Rightarrow {T_2} = 2{T_1}$
By substituting these values we get,
$\Rightarrow \dfrac{{{v_{rms}}_2}}{{{v_{rms}}_1}} = \sqrt {\dfrac{{2{T_1}}}{{{T_1}}}}$
We can therefore cancel out the ${T_1}$ from both the numerator and the denominator, and we get
$\Rightarrow \dfrac{{{v_{rms}}_2}}{{{v_{rms}}_1}} = \sqrt 2$
Now the rms velocity in the first case is given as, ${v_{rms}}_1 = 200m{s^{ - 1}}$
So by taking the ${v_{rms}}_1$ from the L.H.S to the R.H.S we have,
$\Rightarrow {v_{rms}}_2 = \sqrt 2 {v_{rms}}_1$
$\Rightarrow {v_{rms}}_2 = \sqrt 2 \times 200m{s^{ - 1}}$
Hence, we get the rms velocity in the second case as,
$\Rightarrow {v_{rms}}_2 = 200\sqrt 2 m{s^{ - 1}}$
So the correct answer is $200\sqrt 2 m{s^{ - 1}}$ and the correct option is (C).

Note
In this case though it is said in the question that the pressure changes in the second case, but still that won’t have any effect on the rms velocity. This is because as from the formula we see that the rms velocity is independent of the pressure on the gas molecules.