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Question: At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecul...

At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecules have at 35∘C{35^ \circ }C ?

Explanation

Solution

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:
Vrms=3RTM{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}}
Vrms={V_{rms}} = Speed of the molecules
R=R = Universal gas constant
T=T = Temperature in Kelvin
M=M = Molar mass

Complete answer:
Given:
Temperature=35∘C=(35+273)K = {35^ \circ }C = (35 + 273)K
=308K= 308K
Molar mass of nitrogen =28g = 28g
Room mean square velocity of a gas is given by the formula:
Vrms=3RTM{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}}
VN2=3R×30828{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∘C{35^ \circ }C.
Molar mass of hydrogen=2g = 2g
VR2=VN2=3R×30828{V_{{R_2}}} = {V_{{N_2}}} = \sqrt {\dfrac{{3R \times 308}}{{28}}} −−−−1 - - - - 1
From the formula,
VH2=3RT2{V_{{H_2}}} = \sqrt {\dfrac{{3RT}}{2}} −−−−2 - - - - 2
From equation 11 and 22 we have,
3RT2=3R×30828\sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R \times 308}}{{28}}}
Squaring both the sides, we have
3RT2=3R×30828\dfrac{{3RT}}{2} = \dfrac{{3R \times 308}}{{28}}
T=30828T = \dfrac{{308}}{{28}}
T=22KT = 22K
Hence, at 22K22K hydrogen molecules will have the same root mean square speed as nitrogen molecules have at 35∘C{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.