
At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecules have at ${35^ \circ }C$ ?
Answer
484.8k+ views
Hint: The root mean square (RMS or rms) can be defined as the square root of the mean square, i.e., the arithmetic mean of the squares of a given set of numbers. RMS or Root Mean Square value can be calculated by taking the square root of the arithmetic mean of squared observations.
Formula used:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
${V_{rms}} = $ Speed of the molecules
$R = $Universal gas constant
$T = $Temperature in Kelvin
$M = $Molar mass
Complete answer:
Given:
Temperature$ = {35^ \circ }C = (35 + 273)K$
$ = 308K$
Molar mass of nitrogen $ = 28g$
Room mean square velocity of a gas is given by the formula:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
${V_{{N_2}}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $
Let us assume that the root mean square velocity of hydrogen is same as that of nitrogen at ${35^ \circ }C$.
Molar mass of hydrogen$ = 2g$
${V_{{R_2}}} = {V_{{N_2}}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $ $ - - - - 1$
From the formula,
${V_{{H_2}}} = \sqrt {\dfrac{{3RT}}{2}} $ $ - - - - 2$
From equation $1$ and $2$ we have,
$\sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $
Squaring both the sides, we have
$\dfrac{{3RT}}{2} = \dfrac{{3R \times 308}}{{28}}$
$T = \dfrac{{308}}{{28}}$
$T = 22K$
Hence, at $22K$ hydrogen molecules will have the same root mean square speed as nitrogen molecules have at ${35^ \circ }C$ .
Note:
Gaseous particles, according to Kinetic Molecular Theory, are in a state of constant random motion; individual particles move at varying speeds, colliding and changing directions frequently. The movement of gas particles is described using velocity, which takes into account both speed and direction. By squaring the velocities and calculating the square root, we can eliminate the "directional" component of velocity while also obtaining the average velocity of the particles. We now refer to the value as the average speed because it does not include the direction of the particles.
Formula used:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
${V_{rms}} = $ Speed of the molecules
$R = $Universal gas constant
$T = $Temperature in Kelvin
$M = $Molar mass
Complete answer:
Given:
Temperature$ = {35^ \circ }C = (35 + 273)K$
$ = 308K$
Molar mass of nitrogen $ = 28g$
Room mean square velocity of a gas is given by the formula:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
${V_{{N_2}}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $
Let us assume that the root mean square velocity of hydrogen is same as that of nitrogen at ${35^ \circ }C$.
Molar mass of hydrogen$ = 2g$
${V_{{R_2}}} = {V_{{N_2}}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $ $ - - - - 1$
From the formula,
${V_{{H_2}}} = \sqrt {\dfrac{{3RT}}{2}} $ $ - - - - 2$
From equation $1$ and $2$ we have,
$\sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} $
Squaring both the sides, we have
$\dfrac{{3RT}}{2} = \dfrac{{3R \times 308}}{{28}}$
$T = \dfrac{{308}}{{28}}$
$T = 22K$
Hence, at $22K$ hydrogen molecules will have the same root mean square speed as nitrogen molecules have at ${35^ \circ }C$ .
Note:
Gaseous particles, according to Kinetic Molecular Theory, are in a state of constant random motion; individual particles move at varying speeds, colliding and changing directions frequently. The movement of gas particles is described using velocity, which takes into account both speed and direction. By squaring the velocities and calculating the square root, we can eliminate the "directional" component of velocity while also obtaining the average velocity of the particles. We now refer to the value as the average speed because it does not include the direction of the particles.
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