Question

The rms velocities of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ at a temperature T (in Kelvin) is x cm/sec. At what temperature (in Kelvin) the rms velocity of nitrous oxide would be $4x\,cm/\sec$:

A.16 T
B.2T
C.4T
D.32T

Answer 1

We know that, root mean square velocity is the square root of means of square of speeds of the different molecules in a gas at a given temperature. Here, we have to use the formula of root mean speed, ${U_{RMS}} = \sqrt {\dfrac{{3RT}}{M}}$ to calculate temperature of nitrous oxide.

Complete step by step answer: Here, rms velocities of carbon dioxide and nitrous oxide are given. Also, the temperature of carbon dioxide is given as T and the temperature of nitrous oxide needs to be calculated. So, if we use the root mean square speed equation of the two gases, then we can easily calculate the temperature of nitrous oxide using the rms speed, temperature and molar masses of the gases.

We know that, formula of the root mean square speed is,

${U_{RMS}} = \sqrt {\dfrac{{3RT}}{M}}$

Where, ${U_{RMS}}$ is the root mean square speed, R is gas constant, T is temperature and M is molar mass of the gas.

As the value of 3R is constant in the root mean square speed formula, we can say that,

${U_{RMS}}\alpha \sqrt {\dfrac{T}{M}}$

For the mixture of two gases namely, carbon dioxide $\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)$and nitrous oxide $\left( {{{\rm{N}}_{\rm{2}}}{\rm{O}}} \right)$ , root mean square equation is,

$\dfrac{{{U_{RMS\,({\rm{C}}{{\rm{O}}_{\rm{2}}})}}}}{{{U_{RMS\left( {{{\rm{N}}_{\rm{2}}}{\rm{O}}} \right)}}}} = \sqrt {\dfrac{{{T_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}{{{M_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}} \times \dfrac{{{M_{{{\rm{N}}_2}{\rm{O}}}}}}{{{T_{{{\rm{N}}_2}{\rm{O}}}}}}}$
…… (1)

Given that RMS of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ at T (in Kelvin) is x cm/sec and RMS of ${{\rm{N}}_2}{\rm{O}}$ at ${T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}}$ (in Kelvin) is 4x cm/sec.

Now, we have to calculate the molar mass of both the gases namely, nitrous oxide and carbon dioxide.

Molar mass of ${\rm{C}}{{\rm{O}}_{\rm{2}}} = 12 + 16 \times 2 = 12 + 32 = 44\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$

Molar mass of ${{\rm{N}}_{\rm{2}}}{\rm{O}} = {\rm{14}} \times {\rm{2 + 16 = 28 + 16 = 44}}\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$

Now, we have to put the temperature and molar masses of both the gases in equation (1).

$\Rightarrow \dfrac{x}{{4x}} = \sqrt {\dfrac{T}{{{T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}}}} \times \dfrac{{44}}{{44}}}$

$\Rightarrow \dfrac{1}{4} = \sqrt {\dfrac{T}{{{T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}}}} \times \dfrac{{44}}{{44}}}$

$\Rightarrow \dfrac{1}{4} = \sqrt {\dfrac{T}{{{T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}}}}}$

Squaring both sides,

$\Rightarrow \dfrac{1}{{16}} = \dfrac{T}{{{T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}}}}$

$\Rightarrow {T_{{{\rm{N}}_{\rm{2}}}{\rm{O}}}} = 16T$

Therefore, the temperature of nitrous oxide is 16T.

Hence, correct answer is option A.

Additional information:
Very small particles (molecules) present in a gaseous mixture. These molecules are not in a stationary state rather they collide with the container’s walls and among themselves. Due to these collisions, there is an alteration in the speed of the molecules. Maxwell and Boltzmann showed that all molecules in a certain gas may not have the same speed but it is also not necessary that all may possess different speeds.

Note: Different types of speed possessed by gaseous molecules are average speed, most probable speed, root mean square speed etc. The speed which is possessed by the maximum fraction of gas molecules at a given temperature is the most probable speed. At a given temperature, RMS speed of any gas is maximum while probable speed is minimum.