Ans: We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\] …...(i)
Where,
Density, \[\rho =\frac{Mass}{Volume}=\frac{M}{V}\]
M = Molecular weight of gas
V = Volume of gas
Hence, equation (i) becomes:
\[v=\sqrt{\frac{\gamma PV}{M}}\] ……(ii)
Now ideal gas equation for \[n=1\] is:
\[PV=RT\]
For constant T, \[PV=Constant\]
Both \[M\] and \[\gamma \] are constants, \[v=Constant\]
Hence, the speed of sound is independent of the change in the pressure of the gas at a constant temperature.
Ans: We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\] ……(i)
Now ideal gas equation for \[n=1\] is:
\[PV=RT\]
\[P=\frac{RT}{V}\] ……(ii)
Substituting (ii) in (i), we get:
\[v=\sqrt{\frac{\gamma RT}{V\rho }}=\sqrt{\frac{\gamma RT}{M}}\] ……(iii)
Where,
Mass, \[M=\rho V\]is a constant
\[\gamma \]and \[R\]are also constants.
We get from equation (iii),
\[v\propto \sqrt{T}\].
Hence, the sound speed in a gas is directly proportional to the square root of the gaseous medium’s temperature, i.e., the sound speed increases with rise in the gaseous medium’s temperature and vice versa.
Ans: Let \[{{v}_{m}}\] and \[{{v}_{d}}\] are the sound speed in moist air and dry air respectively and \[{{\rho }_{m}}\] and \[{{\rho }_{d}}\] are the densities of moist air and dry air respectively.
We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\]
The speed of sound in moist air is:
\[{{v}_{m}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}}\] ……(i)
The speed of sound in dry air is:
\[{{v}_{d}}=\sqrt{\frac{\gamma P}{{{\rho }_{d}}}}\] ……(ii)
On dividing equations (i) and (ii), we get:
\[\frac{{{v}_{m}}}{{{v}_{d}}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}\times \frac{{{\rho }_{d}}}{\gamma P}}=\sqrt{\frac{{{\rho }_{d}}}{{{\rho }_{m}}}}\] ……(iii)
However, the presence of water vapour decreases the density of air, i.e.,
\[{{\rho }_{d}}<{{\rho }_{m}}\]
\[\Rightarrow {{v}_{m}}>{{v}_{d}}\]
Hence, the speed of sound in moist air is higher than it is in dry air. Thus, in a gaseous medium, the sound speed increases with humidity.