The speed of sound in hydrogen at NTP 1270 m/s. Then, the speed in a mixture of hydrogen and oxygen in the ratio \[4:1\] by volume will be:
A. 317 m/s
B. 635 m/s
C. 830 m/s
D. 950 m/s
Answer
643.2k+ views
Hint: First of all, consider the volume and density of hydrogen as variables and then find the volume and density of oxygen by using the given ratio. Then calculate the density of the mixture of gases which leads to finding the required speed in the mixture of hydrogen and oxygen.
Complete step-by-step solution -
Formula used: \[m = v \times d\], \[V = \sqrt {\dfrac{{\gamma p}}{d}} \]
Given the ratio of hydrogen and oxygen by volume is \[4:1\]
Let the volume of oxygen be V then the volume of the hydrogen will be 4V.
So, total volume of mixture of hydrogen and oxygen = 4V + V = 5V
Let d be the density of the hydrogen.
We know that density is directly proportional to molar mass of element.
So, density of oxygen = 16 \[ \times \] d = 16d
Let D be the density of the mixture of gases.
Now, mass of mixture = mass of hydrogen + mass of oxygen
We know that \[m = v \times d\] where \[m\] is mass of the gas, \[v\] is the velocity of the gas and \[d\] is the density of the gas.
So, we have
\[
\Rightarrow {\text{Volume of mixture}} \times {\text{density of mixture}} = {\text{volume of hydrogen }} \times {\text{density of hydrogen}}\; \\
{\text{ + volume of oxygen}} \times {\text{density of oxygen}}\; \\
\Rightarrow 5{\text{V}} \times {\text{D}} = 4{\text{V}} \times {\text{d}} + {\text{V}} \times {\text{16d}} \\
\Rightarrow {\text{V}}\left( {5{\text{D}}} \right) = {\text{V}}\left( {4{\text{d}} + {\text{16d}}} \right) \\
\Rightarrow 5{\text{D}} = 20{\text{d}} \\
\therefore {\text{D}} = 4{\text{d}} \\
\]
We know that the speed of sound in the gas is given by \[V = \sqrt {\dfrac{{\gamma p}}{d}} \]
As the speed of the sound is inversely proportional to the square root of density i.e., \[V \propto \sqrt {\dfrac{1}{d}} \]
So, the ratio of speed of mixture and speed of hydrogen is given by
\[
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{d}{D}} \\
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{d}{{4d}}} {\text{ }}\left[ {{\text{D}} = 4{\text{d}}} \right] \\
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2} \\
\]
Given speed of sound in hydrogen is 1270 m/s i.e., \[{V_{{{\text{H}}_2}}} = 1270\]
So, we have
\[
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{1270}} = \dfrac{1}{2} \\
\Rightarrow {V_{{\text{mix}}}} = \dfrac{{1270}}{2} \\
\therefore {V_{{\text{mix}}}} = 635{\text{ m/s}} \\
\]
Thus, the correct answer is B. 635 m/s
Note: The density of a gas depends directly on the molar mass of the gas. The mass of a substance is equal to the product of the volume and density of the substance. And the speed of the sound is inversely proportional to the square root of density at NTP (stands for normal temperature and pressure).
Complete step-by-step solution -
Formula used: \[m = v \times d\], \[V = \sqrt {\dfrac{{\gamma p}}{d}} \]
Given the ratio of hydrogen and oxygen by volume is \[4:1\]
Let the volume of oxygen be V then the volume of the hydrogen will be 4V.
So, total volume of mixture of hydrogen and oxygen = 4V + V = 5V
Let d be the density of the hydrogen.
We know that density is directly proportional to molar mass of element.
So, density of oxygen = 16 \[ \times \] d = 16d
Let D be the density of the mixture of gases.
Now, mass of mixture = mass of hydrogen + mass of oxygen
We know that \[m = v \times d\] where \[m\] is mass of the gas, \[v\] is the velocity of the gas and \[d\] is the density of the gas.
So, we have
\[
\Rightarrow {\text{Volume of mixture}} \times {\text{density of mixture}} = {\text{volume of hydrogen }} \times {\text{density of hydrogen}}\; \\
{\text{ + volume of oxygen}} \times {\text{density of oxygen}}\; \\
\Rightarrow 5{\text{V}} \times {\text{D}} = 4{\text{V}} \times {\text{d}} + {\text{V}} \times {\text{16d}} \\
\Rightarrow {\text{V}}\left( {5{\text{D}}} \right) = {\text{V}}\left( {4{\text{d}} + {\text{16d}}} \right) \\
\Rightarrow 5{\text{D}} = 20{\text{d}} \\
\therefore {\text{D}} = 4{\text{d}} \\
\]
We know that the speed of sound in the gas is given by \[V = \sqrt {\dfrac{{\gamma p}}{d}} \]
As the speed of the sound is inversely proportional to the square root of density i.e., \[V \propto \sqrt {\dfrac{1}{d}} \]
So, the ratio of speed of mixture and speed of hydrogen is given by
\[
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{d}{D}} \\
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{d}{{4d}}} {\text{ }}\left[ {{\text{D}} = 4{\text{d}}} \right] \\
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{{V_{{{\text{H}}_2}}}}} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2} \\
\]
Given speed of sound in hydrogen is 1270 m/s i.e., \[{V_{{{\text{H}}_2}}} = 1270\]
So, we have
\[
\Rightarrow \dfrac{{{V_{{\text{mix}}}}}}{{1270}} = \dfrac{1}{2} \\
\Rightarrow {V_{{\text{mix}}}} = \dfrac{{1270}}{2} \\
\therefore {V_{{\text{mix}}}} = 635{\text{ m/s}} \\
\]
Thus, the correct answer is B. 635 m/s
Note: The density of a gas depends directly on the molar mass of the gas. The mass of a substance is equal to the product of the volume and density of the substance. And the speed of the sound is inversely proportional to the square root of density at NTP (stands for normal temperature and pressure).
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

