Question

Find the ratio of speed of sound in hydrogen to the speed of sound in oxygen at 1270K.

Answer 1

Hint: The speed of sound is defined as the distance travelled by the sound wave per unit time. This speed depends on the temperature of the medium and as well as on the medium. The propagation of sound is an adiabatic process according to Laplace. Think of what can be applied to find the speed of sound. Find the speed of sound in both the gases, hydrogen and oxygen. Then find the ratio of the speed.

Complete step by step solution:
According to Laplace, the speed of sound is $v = \sqrt {\dfrac{{\gamma RT}}{M}} $. Here, $\gamma $ is the specific heat ratio, $R$ is the universal gas constant, $T$ is the temperature in kelvins and $M$ is the molecular mass of the gas.
For hydrogen,
${v_h} = \sqrt {\dfrac{{{\gamma _h}RT}}{{{M_h}}}} $
Hydrogen is a diatomic gas, $\gamma $ for diatomic gas is given by ${\gamma _h} = \dfrac{7}{5}$
$
  {v_h} = \sqrt {\dfrac{{7RT}}{{5{M_h}}}} \\
  \therefore {v_h} = \sqrt {\dfrac{1}{{{M_h}}}} \sqrt {\dfrac{{7RT}}{5}} \\
 $
For oxygen,
${v_o} = \sqrt {\dfrac{{{\gamma _o}RT}}{{{M_o}}}} $

Oxygen is a diatomic gas, $\gamma $for oxygen will be ${\gamma _h} = \dfrac{7}{5}$
$
  {v_o} = \sqrt {\dfrac{{{\gamma _o}RT}}{{{M_o}}}} \\
\implies {v_o} = \sqrt {\dfrac{{7RT}}{{5{M_o}}}} \\
  \therefore {v_o} = \sqrt {\dfrac{1}{{{M_o}}}} \sqrt {\dfrac{{7RT}}{5}} \\
 $
Now, the ratio will be given by $\dfrac{{{v_h}}}{{{v_o}}} = \sqrt {\dfrac{{{M_o}}}{{{M_h}}}} $
$
  {M_h} = 2 \\
  {M_o} = 16 \\
 $
$
  \dfrac{{{v_h}}}{{{v_o}}} = \sqrt {\dfrac{{32}}{2}} \\
\implies \dfrac{{{v_h}}}{{{v_o}}} = \sqrt {16} \\
  \therefore \dfrac{{{v_h}}}{{{v_o}}} = 4 \\
 $
Therefore, the ratio of speed of sound in hydrogen to the speed of sound in oxygen at $1270K$ is $4$.

Additional Information:
The rate by which the sound wave changes the position is referred to as the speed of sound. It depends on the temperature and medium. The speed of sound decreases as one goes from solid-liquid-gas, that is, the sound travels slower in gases, comparatively faster in liquids and fastest in solids.
If sound is travelling in a mixture of gases having molecular masses as ${M_1},{M_2},{M_3},......{M_k}$, having number of moles ${n_1},{n_2},{n_3},.....{n_k}$ and with molar specific heat at constant volume as ${C_{{v_1}}},{C_{{v_2}}},{C_{{v_3}}},....{C_{{v_k}}}$
Then the speed of sound is given by $v = \sqrt {\dfrac{{{\gamma _{mix}}RT}}{{{M_{mix}}}}} $, where ${\gamma _{mix}}$is the ratio of specific heats at constant pressure and volume of the gas mixture, and ${M_{mix}}$ is the molecular mass of the mixture.
${M_{mix}}$ is given by ${M_{mix}} = \dfrac{{\sum\limits_{i = 1}^k {{n_i}{M_i}} }}{{\sum\limits_{i = 1}^k {{n_i}} }}$
${C_{{v_{mix}}}}$ is given by ${C_{{v_{mix}}}} = \dfrac{{\sum\limits_{i = 1}^k {{n_i}{C_{{v_i}}}} }}{{\sum\limits_{i = 1}^k {{n_i}} }}$
Now,
$
  {C_{{p_{mix}}}} - {C_{{v_{mix}}}} = R \\
\implies {C_{{p_{mix}}}} = {C_{{v_{mix}}}} + R \\
 $
Now ${\gamma _{mix}}$ can be calculated as follows
$
  {\gamma _{mix}} = \dfrac{{{C_{{p_{mix}}}}}}{{{C_{{v_{mix}}}}}} \\
\implies {\gamma _{mix}} = \dfrac{{{C_{{v_{mix}}}} + R}}{{{C_{{v_{mix}}}}}} \\
\implies {\gamma _{mix}} = 1 + \dfrac{R}{{{C_{{v_{mix}}}}}} \\
\implies {\gamma _{mix}} = 1 + \dfrac{R}{{\left( {\dfrac{{\sum\limits_{i = 1}^k {{n_i}{C_{{v_i}}}} }}{{\sum\limits_{i = 1}^k {{n_i}} }}} \right)}} \\
 $
Now, the speed of sound can be calculated using the above expressions.

Note:
While solving, keep in mind the temperature of the gas/gas mixture should be in kelvins. Also remember that the speed of the gas is directly proportional to the square root of $\gamma $ and temperature, and inversely proportional to the square root of the molecular mass. This information can be used for questions based on comparison.