Question

The average velocity of an ideal gas molecule at \[27^\circ {\rm{C}}\] is \[0.3{\rm{m/sec}}\] . The average velocity at \[927^\circ {\rm{C}}\] will be
A. \[{\text{0}}{\text{.6 m/sec}}\]
B. \[{\text{0}}{\text{.3 m/sec}}\]
C. \[{\text{0}}{\text{.9 m/sec}}\]
D. \[{\text{3}}{\text{.0 m/sec}}\]

Answer 1

Hint: Write the expression for the average velocity of an ideal gas molecule
 \[u = \sqrt {\dfrac{{8RT}}{{\pi M}}} \]
Thus, the average velocity of an ideal gas molecule is directly proportional to the square root of absolute temperature.

Complete step by step answer:
Write the expression for the average velocity of an ideal gas molecule
\[u = \sqrt {\dfrac{{8RT}}{{\pi M}}} \]
Here, u is the average velocity of an ideal gas molecule, R is the ideal gas constant, T is absolute temperature and M is the molecular weight. The value of \[\pi \] is 3.1416.
For a given gas molecule, M is constant. Also R and \[\pi \] are constant. So for a given gas molecule, \[\sqrt {\dfrac{{8R}}{{\pi M}}} \] is constant. Hence, \[u \propto \sqrt T \].
Thus, the average velocity of an ideal gas molecule is directly proportional to the square root of absolute temperature.
For two different temperatures, the ratio of the average velocities is
\[\dfrac{{{u_2}}}{{{u_1}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}} \]
Rearrange above expression
\[{u_2} = {u_1} \times \sqrt {\dfrac{{{T_2}}}{{{T_1}}}} \] … …(1)
The initial temperature is \[27^\circ {\rm{C}}\] . To convert the unit of temperature from degree Celsius to kelvin, add 273.
${T_1} = 27^\circ {\rm{C}} \\
{T_1}{\rm{ = }}\left( {27 + 273} \right){\rm{K}} \\
{T_1}{\rm{ = 300K}} \\$
The final temperature is \[927^\circ {\rm{C}}\]. To convert the unit of temperature from degree Celsius to kelvin, add 273.
${T_1} = 927^\circ {\rm{C}} \\
{T_1}{\rm{ = }}\left( {927 + 273} \right){\rm{K}} \\
{T_1}{\rm{ = 1200K}} \\$
The initial speed is \[0.3{\rm{m/sec}}\] .
Substitute values in equation (1) and calculate the final speed.
${u_2} = {u_1} \times \sqrt {\dfrac{{{T_2}}}{{{T_1}}}} \\
{u_2} = 0.3{\rm{m/sec}} \times \sqrt {\dfrac{{1200}}{{300}}} \\
{u_2} = 0.3{\rm{m/sec}} \times \sqrt 4 \\
{u_2} = 0.3{\rm{m/sec}} \times 2 \\
{u_2} = 0.6{\rm{m/sec}} \\$
Hence, the final speed of the ideal gas molecule is 0.6 m/sec .

Hence, the option A ) is the correct option.

Note: To avoid calculation error, it is necessary to convert the unit of temperature from degree celsius to kelvin by adding 273. Also do not forget that the formula has a square root.