
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
515.1k+ 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 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
