
As per Newton’s formula, what is the velocity of sound at N.T.P. ?
Answer
512.4k+ 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
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

