Question

The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300K are $6.21\times {{10}^{-21}}J$and 484 m/s respectively. The corresponding values at 600K are nearly (assume ideal gas behaviours)
(A) $12.42\times {{10}^{-21}}J.968\,m/s$
(B) $8.78\times {{10}^{-21}}J.684\,m/s$
(C) $6.21\times {{10}^{-21}}J.968\,m/s$
(D) $12.42\times {{10}^{-21}}J.684m/s$

Answer 1

We know that translational motion is the motion by which a body shifts from one point in space to another. One example of translational motion is the motion of a bullet fired from a gun. An object has a rectilinear motion when it moves along a straight line. Train moving on a track, any object freely falling due to gravity, driving a car on the road, motion of bullets fired from a gun and expanding of galaxies are some examples of translational kinetic energy. Temperature is a measure of the average kinetic energy of all the molecules in a gas. As the temperature and, therefore, kinetic energy, of a gas changes, the RMS speed of the gas molecules also changes. The RMS speed of the molecules is the square root of the average of each individual velocity squared. Using the above concept, we have to solve this question.

Complete step-by step answer
We should know that translational energy relates to the displacement of molecules in a space as a function of the normal thermal motions of matter. The kinetic energy of the translational motion of an ideal gas depends on its temperature. The formula for the kinetic energy of a gas defines the average kinetic energy per molecule. The average translational kinetic energy of a single molecule of an ideal gas is (Joules).
Let us mention the values that are mentioned in the question at first:
E = average energy = $6.21\times {{10}^{-21}}j$
${{V}_{rms}}=rms\,speed=484m/s$
${{T}_{1}}=300K$
As $E\times T,\,if\,T=600K$, temperature is doubled so energy (E) will also be doubled.
So, the expression for the energy is formed as:
$\therefore E=2\times 6.21\times {{10}^{-21}}j=12.42\times {{10}^{-21}}j$
Now we have to find the value of rms of velocity as:
The root mean square velocity is the square root of the average of the square of the velocity. As such, it has units of velocity. The reason we use the rms velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions.The root mean square velocity is the square root of the average of the square of the velocity.The reason we use the rms velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions.
$Also\,{{V}_{rms}}=\sqrt{3K\frac{bT}{m}\,,{{V}_{rms}}^{2}\times T}$
The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass.Average velocity is a concept that measures the displacement of an object over time.The average speed of molecules is the mean of all magnitudes of velocity at which molecules of the given gas are moving. The root-mean-square speed of molecules is the speed at which all the molecules have the same total kinetic energy as in case of their actual speed.
$\frac{{{V}^{2}}_{rms}(600K)}{v={{V}^{2}}_{rms}(300K)}=\left( \frac{600}{300} \right)=2\Rightarrow 2\times {{484}^{2}}$
The answer is obtained as:
$\Rightarrow 684.7m/s$

Hence, the correct answer is Option D.

Note We should know that generally, a gas behaves more like an ideal gas at higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them. The particles are so small that their volume is negligible compared with the volume occupied by the gas. The particles don't interact. There are no attractive or repulsive forces between them. The average kinetic energy of the gas particles is proportional to temperature.
It should be known to us that ideal gases are defined as having molecules of negligible size with an average molar kinetic energy dependent only on temperature. At a low temperature, most gases behave enough like ideal gases that the ideal gas law can be applied to them. An ideal gas is also known as a perfect gas.