Question

At what temperature, the average speed of gas molecules will be double that at $27{}^\circ C$ ?
A.$27{}^\circ C$
B.$327{}^\circ C$
C.$527{}^\circ C$
D.$927{}^\circ C$

Answer 1

We know the kinetic energy of molecules so we know that the average kinetic energy and total kinetic energy is a function of temperature. In average kinetic energy we used translational degrees of freedom while in the calculation of total kinetic energy we use total degrees of freedom.

Complete answer:
As we know that in the kinetic theory of gases there are several types of velocities like the average speed or the mean speed, the most probable velocity or the root mean square velocity. Each velocity has its own importance and thus none of them can be neglected. Mean speed is the average speed of all the molecules irrespective of their direction of motion. Whereas the most probable speed is the speed whose particle in number is maximum which means that in the category of velocity we will find the maximum number of particles. The root mean square is a very important term in the energy calculation for the gases. It directly corresponds to the energy of the sample unlike the mean speed and the most probable speed. The chances of mistakes anyone could make is that they get confused while writing the formula.
Here we know, Average speed is given by ${{V}_{av}}\alpha {{\left( T \right)}^{\dfrac{1}{2}}}$ at $27{}^\circ C;$
$\dfrac{{{\left( {{V}_{av}} \right)}_{2}}}{{{\left( {{V}_{av}} \right)}_{1}}}=\sqrt{\dfrac{{{T}_{2}}}{{{T}_{1}}}}$ here as we know that $\dfrac{{{\left( {{V}_{av}} \right)}_{2}}}{{{\left( {{V}_{av}} \right)}_{1}}}=2.$
$2=\sqrt{\dfrac{{{T}_{2}}}{{{T}_{1}}}}$ On squaring both the sides we get; ${{\left( 2 \right)}^{2}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow {{T}_{2}}=4{{T}_{1}}$
From here we get; ${{T}_{1}}=273+27=300K$
Thus, ${{T}_{2}}=4\times 300=1200-273=927{}^\circ C.$
Therefore, the correct answer is option D.

Note:
Remember that kinetic energy of molecules so we know that the average kinetic energy and total kinetic energy is a function of temperature. In average kinetic energy we used translational degrees of freedom while in the calculation of total kinetic energy we use total degrees of freedom.