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Question: The speed of sound in air at \(15^\circ C\) and 76 cm of Hg is \(340\dfrac{m}{s}\). The speed of sou...

The speed of sound in air at 15C15^\circ C and 76 cm of Hg is 340ms340\dfrac{m}{s}. The speed of sound in air at 30C30^\circ C and 75 cm of Hg will be (in m/s).
A) 340303288340\sqrt {\dfrac{{303}}{{288}}}
B) 340288303340\sqrt {\dfrac{{288}}{{303}}}
C) 3402340\sqrt 2
D) 3402×7576340\sqrt {\dfrac{{2 \times 75}}{{76}}}

Explanation

Solution

The speed of sound changes with change in temperature. The speed of sound increases with increase in temperature as the molecules will vibrate at a faster rate and pass the sound energy to the next molecule.

Formula used:
The speed of sound in adiabatic process is equal to,
v=γRTM\Rightarrow v = \sqrt {\dfrac{{\gamma RT}}{M}}
Where T is the temperature, M is the mass, RR is universal gas constant.

Complete step by step solution:
It is given in the problem that the speed of sound in air at 15C15^\circ C and 76 cm of Hg is 340ms340\dfrac{m}{s} and we need to find the speed of sound in air at 30C30^\circ C and 75 cm of Hg will be (in m/s).
The speed of sound in adiabatic process is equal to,
v=γRTM\Rightarrow v = \sqrt {\dfrac{{\gamma RT}}{M}}
The ratio of the speed is equal to,
v1v2=γRT1MγRT2M\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {\dfrac{{\gamma R{T_1}}}{M}} }}{{\sqrt {\dfrac{{\gamma R{T_2}}}{M}} }}
v1v2=T1T2\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }}
The temperatureT1{T_1},
T1=273+15\Rightarrow {T_1} = 273 + 15
T1=288K\Rightarrow {T_1} = 288K
The temperatureT2{T_2},
T2=273+30\Rightarrow {T_2} = 273 + 30
T2=303K\Rightarrow {T_2} = 303K
v1v2=288303\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {288} }}{{\sqrt {303} }}
The velocityv1=340ms{v_1} = 340\dfrac{m}{s}.
340v2=288303\Rightarrow \dfrac{{340}}{{{v_2}}} = \dfrac{{\sqrt {288} }}{{\sqrt {303} }}
v2=340303288ms\Rightarrow {v_2} = 340\sqrt {\dfrac{{303}}{{288}}} \dfrac{m}{s}
The speed of sound at 30C30^\circ C is equal to v2=340303288ms{v_2} = 340\sqrt {\dfrac{{303}}{{288}}} \dfrac{m}{s}.

The correct answer for this problem is option A.

Additional information: The sound needs a medium to travel unlike the light which can also travel in the space as the sunlight reaches the earth crossing the space. The speed of sound depends upon the inverse of the density of the medium if the density is more than the speed is less and if the density of medium is less than the speed of the sound is more.

Note: The students are advised to understand and remember the formula of the speed of sound for adiabatic processes as it is very useful for solving these types of problems. Sound is a longitudinal form of wave as the wave propagates parallel to the direction of motion of sound.