Question

Calculate the temperature values at which the molecules of the first two members of the homologous series,${C_n}{H_{2n + 2}}$ will have the same rms speed as $C{O_2}$ gas at $770K$. The normal b.p. of n − butane is $273K$ .
This question has multiple answers:
A. $280K$
B. $525K$
C. ${280^ \circ }C$
D. ${280^ \circ }C$

Answer 1

${v_{rms}}$, which is the root mean square of velocity, relates the velocity of a gas with its molecular mass at different temperatures. If the value of ${v_{rms}}$ is the same for two components, they can be equated and hence, if we know one temperature value, we can find out the other.

Formula used: ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
Where, $R =$ Gas constant
$T =$ Temperature
$M =$ Mass
${v_{rms}}$= root mean square velocity

Complete step by step answer:
Let us first understand a few terms which we will be using today.
${v_{rms}}$ is the root mean square of the velocity of individual gas molecules. We use this value of ${v_{rms}}$ as a general value for the velocity of the gas.
The formula we will be using is:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
$R =$ Gas constant
$T =$ Temperature
$M =$ Mass
In the question, we have been told, the about the of first two member ${C_n}{H_{2n + 2}}$
These values will be
$n = 1,C{H_4}$
$n = 2,{C_2}{H_6}$

Let us take case 1
Where $n = 1,C{H_4}$
Mass of $C{H_4}$ is $C(12) + H \times 4(1 \times 4) \Rightarrow 16amu$
Mass for $C{O_2}$ is $C(12) + O \times 2(16 \times 2) \Rightarrow 44amu$
We have been given in the question that, ${v_{rms}}$ is the same for $C{O_2}$ and $C{H_4}$ .
${v_{rms}}_{(C{H_4})} = \sqrt {\dfrac{{3R{T_{C{H_4}}}}}{{{M_{C{H_4}}}}}}$ Let this be equation one
And, ${v_{rms(C{O_2})}} = \sqrt {\dfrac{{3R{T_{(C{O_2})}}}}{{{M_{(C{O_2})}}}}}$ let this be equation two
from the question we know that
${v_{rms(C{H_4})}} = {v_{rms(C{O_2})}}$
Hence, equation one and two we get,
$\sqrt {\dfrac{{{T_{C{H_4}}}}}{{{M_{C{H_4}}}}}} = \sqrt {\dfrac{{{T_{C{O_2}}}}}{{{M_{CO2}}}}}$
where,
${T_{C{H_4}}}$temperature for methane
${T_{C{O_2}}}$temperature for carbon dioxide
${M_{C{H_4}}}$molecular mass of methane
${M_{C{O_2}}}$molecular mass of carbon dioxide.
We know, ${T_{C{O_2}}} = 773K$
Substituting the values in the above equation, we get
${T_{C{H_4}}} = \dfrac{{773 \times 16}}{{44}} \Rightarrow 280K$

Case 2,
$n = 2,{C_2}{H_6}$
Mass of $C{H_4}$ is $C \times 2(12 \times 2) + H \times 6(1 \times 6) \Rightarrow 30amu$
Mass for $C{O_2}$ is $C(12) + O \times 2(16 \times 2) \Rightarrow 44amu$
We have been given the question that ${v_{rms}}$ is the same for $C{O_2}$ and ${C_2}{H_6}$ .
Hence, we get
${v_{rms({C_2}{H_6})}} = {v_{rms(C{O_2})}}$
So, further substituting the formula we, get
$\sqrt {\dfrac{{{T_{{C_2}{H_6}}}}}{{{M_{{C_2}{H_6}}}}}} = \sqrt {\dfrac{{{T_{C{O_2}}}}}{{{M_{CO2}}}}}$
${T_{{C_2}{H_6}}}$temperature for ethane
${T_{C{O_2}}}$temperature for carbon dioxide
${M_{{C_2}{H_6}}}$molecular mass of ethane
${M_{C{O_2}}}$molecular mass of carbon dioxide.

We know, ${T_{C{O_2}}} = 773K$
Substituting the values, we get
${T_{{C_2}{H_6}}} = \dfrac{{773 \times 30}}{{44}} \Rightarrow 525K$

So, the correct answer is Option B.

Note: The reason we need to use ${v_{rms}}$ instead of average velocity is because for any molecule, average velocity will be zero. Since all the gas molecules are in constant random motion, they cancel each other out and hence there will be no net value for velocity. ${v_{rms}}$ helps in determining diffusion rates also