As per Newton’s formula, what is the velocity of sound at N.T.P. ?
Answer
543.6k+ views
Hint: To calculate the velocity of sound at Normal temperature and pressure (NTP), we need to first find the condition of NTP. The parameters at Normal temperature and pressure (NTP) are ${{20}^{\circ }}C$ and 1 atm pressure. We will use this value of pressure and temperature to calculate the speed of sound using Newton’s formula:
Complete answer:
Let us assume that the speed of sound at Normal temperature and pressure (NTP) is denoted by ‘v’. Then, the Newton’s formula for speed of sound is given by:
$\Rightarrow v=\sqrt{\dfrac{P}{\rho }}$
Where,
P is the pressure of the medium in which the sound wave is travelling. And,
$\rho $ is the density of the medium (in this case, air)
Now, to calculate the density of air, we know that at NTP, 1 mole of any gas occupies 24.04Ltr of volume.
Thus, the density of air can be calculated as:
$\begin{align}
& \Rightarrow \rho =\dfrac{29\times {{10}^{-3}}kg}{24.04\times {{10}^{-3}}{{m}^{3}}} \\
& \Rightarrow \rho =1.206kg{{m}^{-3}} \\
\end{align}$
Also, 1 atm pressure when converted into standard unit comes out to be:
$\begin{align}
& \Rightarrow P=1.01325\times {{10}^{5}}Pa \\
& \Rightarrow P=101325Pa \\
\end{align}$
Now, putting the value of pressure and density in Newton's equation for speed of sound. We get:
$\begin{align}
& \Rightarrow v=\sqrt{\dfrac{101325Pa}{1.206kg{{m}^{-3}}}} \\
& \Rightarrow v\approx 290m{{s}^{-1}} \\
\end{align}$
Hence, as per Newton's formula the velocity of sound at N.T.P. comes out to be $290m{{s}^{-1}}$ .
Note:
While deriving the above speed of sound formula, Newton assumed that compression and rare-fraction in the air is an isothermal process. But, this assumption was later on corrected by Laplace who said that the compression and rare-fraction of air is an adiabatic process. This is known as Laplace correction, and the new formula is: $v=\sqrt{\dfrac{\gamma P}{\rho }}$ , where ‘$\gamma $’ is the adiabatic index.
Complete answer:
Let us assume that the speed of sound at Normal temperature and pressure (NTP) is denoted by ‘v’. Then, the Newton’s formula for speed of sound is given by:
$\Rightarrow v=\sqrt{\dfrac{P}{\rho }}$
Where,
P is the pressure of the medium in which the sound wave is travelling. And,
$\rho $ is the density of the medium (in this case, air)
Now, to calculate the density of air, we know that at NTP, 1 mole of any gas occupies 24.04Ltr of volume.
Thus, the density of air can be calculated as:
$\begin{align}
& \Rightarrow \rho =\dfrac{29\times {{10}^{-3}}kg}{24.04\times {{10}^{-3}}{{m}^{3}}} \\
& \Rightarrow \rho =1.206kg{{m}^{-3}} \\
\end{align}$
Also, 1 atm pressure when converted into standard unit comes out to be:
$\begin{align}
& \Rightarrow P=1.01325\times {{10}^{5}}Pa \\
& \Rightarrow P=101325Pa \\
\end{align}$
Now, putting the value of pressure and density in Newton's equation for speed of sound. We get:
$\begin{align}
& \Rightarrow v=\sqrt{\dfrac{101325Pa}{1.206kg{{m}^{-3}}}} \\
& \Rightarrow v\approx 290m{{s}^{-1}} \\
\end{align}$
Hence, as per Newton's formula the velocity of sound at N.T.P. comes out to be $290m{{s}^{-1}}$ .
Note:
While deriving the above speed of sound formula, Newton assumed that compression and rare-fraction in the air is an isothermal process. But, this assumption was later on corrected by Laplace who said that the compression and rare-fraction of air is an adiabatic process. This is known as Laplace correction, and the new formula is: $v=\sqrt{\dfrac{\gamma P}{\rho }}$ , where ‘$\gamma $’ is the adiabatic index.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

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

Which among the following are examples of coming together class 11 social science CBSE

Discuss the various forms of bacteria class 11 biology CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram showing structure of LS of class 11 biology CBSE

