
The velocity of sound in air at $27{}^\circ C$ is 340m/s. Calculate the velocity of sound in air at $127{}^\circ C$.
Answer
521.1k+ views
Hint: As a first step, you could recall the expression of velocity in terms of Temperature. Assuming all the other quantities in that expression are constant, we could find the proportionality relation between velocity and temperature. Thus we could find the answer easily from it.
Formula used:
Velocity,
$v=\sqrt{\dfrac{\gamma RT}{M}}$
Complete answer:
In the question we are given the velocity of sound in air at room temperature that is $27{}^\circ C$ to be 340m/s and we are supposed to find the velocity of sound in air at a greater temperature, that is, $127{}^\circ C$.
All we have to do here is to find the relation between the velocity of sound in air with temperature. We know that this relation could be given by,
$v=\sqrt{\dfrac{\gamma RT}{M}}$
Where, $\gamma $ is the adiabatic index, R is the universal gas constant, T is the absolute temperature and M is the molecular mass.
Since the medium remains the same (air) in both the given cases we could say that $\gamma $ and M remains constant and also, R is already a constant. So, we could make this conclusion that,
$v\alpha \sqrt{T}$
$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{{{T}_{1}}}{{{T}_{2}}}}$ …………………………………………. (1)
Before substituting, let us convert the temperatures into their SI units.
${{T}_{1}}=27{}^\circ C=300K$
${{T}_{2}}=127{}^\circ C=400K$
Now, we could substitute these in (1) to get,
${{v}_{2}}={{v}_{1}}\sqrt{\dfrac{{{T}_{2}}}{{{T}_{1}}}}=340\sqrt{\dfrac{400}{300}}$
$\therefore {{v}_{2}}=392.6m/s$
Therefore, we found the velocity of sound in air at $127{}^\circ C$ to be 392.6m/s.
Note:
In problems involving the calculation with temperatures, make sure that you convert the given temperatures into their respective SI units. If you forget to do this step, you will get the answer wrong. We have taken molecular mass M and adiabatic index to be constant as they are dependent on the medium and the medium remains the same in both cases.
Formula used:
Velocity,
$v=\sqrt{\dfrac{\gamma RT}{M}}$
Complete answer:
In the question we are given the velocity of sound in air at room temperature that is $27{}^\circ C$ to be 340m/s and we are supposed to find the velocity of sound in air at a greater temperature, that is, $127{}^\circ C$.
All we have to do here is to find the relation between the velocity of sound in air with temperature. We know that this relation could be given by,
$v=\sqrt{\dfrac{\gamma RT}{M}}$
Where, $\gamma $ is the adiabatic index, R is the universal gas constant, T is the absolute temperature and M is the molecular mass.
Since the medium remains the same (air) in both the given cases we could say that $\gamma $ and M remains constant and also, R is already a constant. So, we could make this conclusion that,
$v\alpha \sqrt{T}$
$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{{{T}_{1}}}{{{T}_{2}}}}$ …………………………………………. (1)
Before substituting, let us convert the temperatures into their SI units.
${{T}_{1}}=27{}^\circ C=300K$
${{T}_{2}}=127{}^\circ C=400K$
Now, we could substitute these in (1) to get,
${{v}_{2}}={{v}_{1}}\sqrt{\dfrac{{{T}_{2}}}{{{T}_{1}}}}=340\sqrt{\dfrac{400}{300}}$
$\therefore {{v}_{2}}=392.6m/s$
Therefore, we found the velocity of sound in air at $127{}^\circ C$ to be 392.6m/s.
Note:
In problems involving the calculation with temperatures, make sure that you convert the given temperatures into their respective SI units. If you forget to do this step, you will get the answer wrong. We have taken molecular mass M and adiabatic index to be constant as they are dependent on the medium and the medium remains the same in both cases.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

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

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

