Question

If the temperature is raised by 1k from 300k, the percentage change in the speed of sound in the gaseous mixture is $\left( {R = 8.31\dfrac{J}{{mol.k}}} \right)$
(A) $0.167\% $
(B) $0.334\% $
(C) $1\% $
(D) $2\% $

Answer 1

Hint: To find the percentage change in the speed of sound, we use the formula of speed of sound in terms of temperature i.e., ${V_S} = \sqrt {\dfrac{{\tau RT}}{M}} $
Where $R = $ gas constant
$T = $ absolute temperature
$M = $ Molecular mass
$\tau = $ Adiabatic constant

Complete step by step solution:
We know that according to replace Laplace correction formula of velocity of sound in terms of temperature is given as ${V_S} = \sqrt {\dfrac{{\tau RT}}{M}} $
Here we use an error method to calculate the percentage change.

So, $\dfrac{{\Delta {V_S}}}{{{V_S}}} = \dfrac{1}{2}\dfrac{{\Delta T}}{T}$
Given that change in temperature $\Delta T = 1k$
& initial temperature $T = 100k$
$\dfrac{{\Delta {V_S}}}{{{V_S}}} \times 100\% = \dfrac{1}{2} \times \dfrac{1}{{300}} \times 100\% $
$\% $ change in ${V_S} = \dfrac{{\Delta {V_S}}}{{{V_S}}} \times 100\% = \dfrac{1}{6}$
$ = 0.1666\% $

Hence option A is correct answer.

Note: Error method gives an approximate answer. If we want to calculate exact answer then we have to put the values of $\tau, $R, H and T in the given formula
${V_S} = \sqrt {\dfrac{{\tau RT}}{M}} $
Now, ${T_i} = 300k$
So, ${V_S} = \sqrt {\dfrac{{\tau R(300)}}{M}} $
Finally T becomes $ = 300 + 1 = 301k$
So, $V_S^1 = \sqrt {\dfrac{{\tau R(301)}}{M}} $
Now for $\% $ change in ${V_S}$
$\left( {\dfrac{{V_S^1 - {V_S}}}{{{V_S}}}} \right) \times 100\% = \dfrac{{\sqrt {\dfrac{{\tau R}}{M}} \left[ {\sqrt {301} - \sqrt {300} } \right]}}{{\sqrt {\dfrac{{\tau R}}{M}} \sqrt {300} }} \times 100$
$\% $ change in ${V_S} = \dfrac{{\left( {\sqrt {301} - \sqrt {300} } \right)}}{{\sqrt {300} }} \times 100\% $
$ = \dfrac{{(17.349 - 17.320)}}{{17.320}} \times 100$
$ = 0.0016743 \times 100\% $
$ = 0.1674\% $