Question

The r.m.s speed of the molecules of a gas at \[100{}^\circ C\] is ‘v’. The temperature at which the r.m.s speed will be \[\sqrt{3}v\] is
A.) \[546{}^\circ C\]
B.) \[646{}^\circ C\]
C.) \[746{}^\circ C\]
D.) \[846{}^\circ C\]

Answer 1

Hint: We have given two r.m.s speed of the same gas. As we know, the temperature can make a great impact on the kinetic energy of the gases. Here, we can simply say that the temperature of gas increases since the r.m.s speed increases. We can calculate the temperature from the equation of r.m.s speed of an ideal gas.

Formula used:
\[{{v}_{r.m.s}}=\sqrt{\dfrac{3RT}{M}}\], where R is the universal gas constant, T is the temperature and M is the molar mass.

Complete step by step answer:
According to the kinetic theory of gases, the root mean square (r.m.s) speed of an ideal gas can be written as,

\[{{v}_{r.m.s}}=\sqrt{\dfrac{3RT}{M}}\], where R is the universal gas constant, T is the temperature and M is the molar mass.

Since we are dealing with only the temperature and r.m.s speed, we don’t have to consider other terms. According to the equation of root mean square speed, the speed is in a direct relationship with the temperature.

\[{{v}_{r.m.s}}\propto T\]

We can consider two cases in this scenario. At \[100{}^\circ C\] the r.m.s speed is v is the first case. The second case is at a particular temperature (T) the r.m.s speed is \[\sqrt{3}v\]. To find the temperature we can find the ratio between them.

\[\dfrac{{{v}^{2}}}{3{{v}^{2}}}=\dfrac{100{}^\circ C}{T}\]
\[T=\dfrac{100{}^\circ C\times 3{{v}^{2}}}{{{v}^{2}}}\]

We have to convert the temperature into the Kelvin scale.

\[T=\dfrac{373\times 3{{v}^{2}}}{{{v}^{2}}}\]
\[T=1119K\]

But the options are on the Celsius scale. So, we have to convert the Kelvin scale into the Celsius scale.

\[T=1119-273=846{}^\circ C\]

Therefore, the correct option is D.

Additional information:
We can discuss the postulates of the kinetic theory of gases;
a.) Each gas is composed of several particles called atoms and molecules and they behave like small spheres.
b.) These particles will undergo random motion. They will try to move in straight lines, but if they hit with another particle or the walls of the container, the direction will change.
c.) There is no force of attraction or repulsion under normal conditions.
d.) The collisions between the gas particles and walls of the container are purely elastic collisions. Thus, no transfer of energy occurs.
e.) The kinetic energy of the gas particles depends only upon the temperature of the gas.

Note: If we are doing with the Celsius scale we won’t get the right answer. Because the r.m.s speed is calculating with the Kelvin scale. Therefore, first, we have to convert the Celsius scale to the Kelvin scale. After the calculation, we have to convert again the Kelvin scale to the Celsius scale. Since we are doing the ratio, we don’t have to take the other terms that are involved in the r.m.s speed.