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

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.