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

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

Explanation

Solution

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=γRTMv = \sqrt {\dfrac{{\gamma RT}}{M}} . Here, γ\gamma is the specific heat ratio, RR is the universal gas constant, TT is the temperature in kelvins and MM is the molecular mass of the gas.
For hydrogen,
vh=γhRTMh{v_h} = \sqrt {\dfrac{{{\gamma _h}RT}}{{{M_h}}}}
Hydrogen is a diatomic gas, γ\gamma for diatomic gas is given by γh=75{\gamma _h} = \dfrac{7}{5}
vh=7RT5Mh vh=1Mh7RT5  {v_h} = \sqrt {\dfrac{{7RT}}{{5{M_h}}}} \\\ \therefore {v_h} = \sqrt {\dfrac{1}{{{M_h}}}} \sqrt {\dfrac{{7RT}}{5}} \\\
For oxygen,
vo=γoRTMo{v_o} = \sqrt {\dfrac{{{\gamma _o}RT}}{{{M_o}}}}

Oxygen is a diatomic gas, γ\gamma for oxygen will be γh=75{\gamma _h} = \dfrac{7}{5}
vo=γoRTMo     vo=7RT5Mo vo=1Mo7RT5  {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 vhvo=MoMh\dfrac{{{v_h}}}{{{v_o}}} = \sqrt {\dfrac{{{M_o}}}{{{M_h}}}}
Mh=2 Mo=16  {M_h} = 2 \\\ {M_o} = 16 \\\
vhvo=322     vhvo=16 vhvo=4  \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 1270K1270K is 44.

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 M1,M2,M3,......Mk{M_1},{M_2},{M_3},......{M_k}, having number of moles n1,n2,n3,.....nk{n_1},{n_2},{n_3},.....{n_k} and with molar specific heat at constant volume as Cv1,Cv2,Cv3,....Cvk{C_{{v_1}}},{C_{{v_2}}},{C_{{v_3}}},....{C_{{v_k}}}
Then the speed of sound is given by v=γmixRTMmixv = \sqrt {\dfrac{{{\gamma _{mix}}RT}}{{{M_{mix}}}}} , where γmix{\gamma _{mix}}is the ratio of specific heats at constant pressure and volume of the gas mixture, and Mmix{M_{mix}} is the molecular mass of the mixture.
Mmix{M_{mix}} is given by Mmix=i=1kniMii=1kni{M_{mix}} = \dfrac{{\sum\limits_{i = 1}^k {{n_i}{M_i}} }}{{\sum\limits_{i = 1}^k {{n_i}} }}
Cvmix{C_{{v_{mix}}}} is given by Cvmix=i=1kniCvii=1kni{C_{{v_{mix}}}} = \dfrac{{\sum\limits_{i = 1}^k {{n_i}{C_{{v_i}}}} }}{{\sum\limits_{i = 1}^k {{n_i}} }}
Now,
CpmixCvmix=R     Cpmix=Cvmix+R  {C_{{p_{mix}}}} - {C_{{v_{mix}}}} = R \\\ \implies {C_{{p_{mix}}}} = {C_{{v_{mix}}}} + R \\\
Now γmix{\gamma _{mix}} can be calculated as follows
γmix=CpmixCvmix     γmix=Cvmix+RCvmix     γmix=1+RCvmix     γmix=1+R(i=1kniCvii=1kni)  {\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 γ\gamma and temperature, and inversely proportional to the square root of the molecular mass. This information can be used for questions based on comparison.