Question

The rms velocity of a gas molecule of a given amount of a gas at $27^{\circ}C$ and $1.0 \times 10^{5}Nm^{-2}$ pressure is $200m/s$. If temperature and pressure are respectively $127^{\circ}C$ and $0.5\times10^{5}Nm^{-2}$ the rms velocity will be:
A. $\dfrac{400}{\sqrt3}ms^{-1}$
B. $100\sqrt2ms^{-1}$
C. $100\dfrac{\sqrt2}{3}ms^{-1}$
D. $50\sqrt(\dfrac{2}{3})ms^{-1}$

Answer 1

Hint: We know from gases law, that the $v_{rms}\propto\sqrt T$, here we need to compare the two rms velocity with respect to temperature, to find the unknown rms velocity.

Formula used: We know from gases law, that the $v_{rms}\propto\sqrt T$

Complete step-by-step solution -
Root-mean-square velocity of gases is the root of the mean of the squares of velocity of all the gas particles in the system; this is taken into calculation , because of the random motion and velocities of the gas particles.
It is represented as $v_{rms}=\sqrt{(\dfrac{3RT}{M_{m}})}$ where $R$ is the gas constant, $T$ is the absolute temperature and $M_{m}$ is the molar mass of the gas particles.
We know from gases law, that the $v_{rms}\propto\sqrt T$
It is given that ,
$(v_{rms})_{1}=200m/s$
$(v_{rms})_{2}=?$
Taking temperature in kelvin, $T_{1}= 273+27=300K$
 $T_{2}= 273+127=400K$
Then we can write,
$\dfrac{( v_{rms})_{1}}{ (v_{rms})_{2}}=\sqrt{(\dfrac{ T_{1}}{ T_{2}})}$
$\dfrac{200}{ (v_{rms})_{2}}=\sqrt{(\dfrac{ 300}{ 400})}$
$(v_{rms})_{2}=\sqrt{(\dfrac{ 4}{ 3})}\times 200$
$(v_{rms})_{2}=(\dfrac{ 400}{ \sqrt 3})$

Hence , the new rms is A.$\dfrac{400}{\sqrt3}ms^{-1}$

Additional Information:
The mean speed, most probable speed and root-mean-square speed are properties of the Maxwell-Boltzmann distribution, which studies the molecular collision of the gas molecules, on the basis of statistical thermodynamics. Maxwell–Boltzmann statistics gives the average number of particles found in a given single-particle microstate.It is assumed that the particles don’t interact, and exist as independent particles.

Note: Rms velocity is taken instead of normal velocity because of the random motion and velocities of the gas particles. From the equation it is clear that $v_{rms}\propto\sqrt T$, $v_{rms}\propto\dfrac{1}{\sqrt M}$. Here, it is assumed that the particles don’t interact, and exist as independent particles.