Question

The root mean square velocity of an ideal gas in a closed container of the fixed volume is increased from \[5 \times {10^4}{\rm{cm}}\,{{\rm{s}}^1}\] to \[10 \times {10^4}\,{\rm{cm}}\,{{\rm{s}}^{ - 1}}\] . Which of the following statements correctly explains how the change is accomplished?
A. By heating the gas the temperature is doubled
B. By heating the gas, the pressure is quadrupled (i.e. made four times)
C. By heating the gas, the temperature is quadrupled
D. By heating the gas the pressure is doubled

Answer 1

Hint: From the formula, it is clear that RMS is dependent on the temperature and molar mass of a gas. Here, we have to understand the RMS velocity of an ideal gas. So, we have to use the equation of an ideal gas, that is, PV=nRT.

Complete Step by Step Solution:
The rms velocity can be find out by the following formula,

 \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
\[{v^2}_{rms} = \dfrac{{3RT}}{M}\]
On rearranging,
\[T = \dfrac{{{v^2}_{rms}M}}{{3R}}\]
So, \[{T_1} = \dfrac{{{v^2}_{rms(1)}M}}{{3R}}\]and \[{T_2} = \dfrac{{{v^2}_{rms(2)}M}}{{3R}}\]
Given, \[{v_{rms(1)}} = 5 \times {10^4}\,{\rm{cm}}\,{{\rm{s}}^{ - 1}}\] and \[{v_{rms(2)}} = 10 \times {10^4}\,{\rm{cm}}\,{{\rm{s}}^{ - 1}}\]
\[\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{v^2}_{rms(1)}}}{{{v^2}_{rms(2)}}}\]
\[\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{{\left( {5 \times {{10}^4}} \right)}^2}}}{{{{\left( {10 \times {{10}^4}} \right)}^2}}}\]
\[\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{25}}{{100}} = \dfrac{1}{4}\]
 \[{T_2} = 4{T_1}\]
Therefore, on transferring heat, temperature increases to four times.
Also given that, the gas is an ideal gas and ideal gas equation is \[PV = nRT\] , where P, V , R, n and T stands for pressure, volume, Gas constant, , mole of electrons and temperature respectively. Now, we will understand effect of pressure on rms velocity.
\[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Also, \[PV = RT\]
\[{v^2}_{rms} = \dfrac{{3PV}}{M}\]
On rearranging,
\[P = \dfrac{{{v^2}_{rms}M}}{{3V}}\]
So, \[{P_1} = \dfrac{{{v^2}_{rms(1)}M}}{{3V}}\]and \[{P_2} = \dfrac{{{v^2}_{rms(2)}M}}{{3V}}\]
Given that, pressure is constant.
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{v^2}_{rms(1)}}}{{{v^2}_{rms(2)}}}\]
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{{\left( {5 \times {{10}^4}} \right)}^2}}}{{{{\left( {10 \times {{10}^4}} \right)}^2}}} = \dfrac{1}{4}\]
\[{P_2} = 4{P_1}\]
Therefore, the pressure also increased to four times on heating the gas.
Hence, options B and C are correct.

Note: It is to be noted that average velocity is different from rms velocity. Average velocity defines the arithmetic mean calculation of velocities of different gaseous molecules at a particular temperature.