The enclosed air inside a closed box of rigid walls has pressure P 1 . The density of air inside this box is constant. On heating the gas, the pressure of the enclosed air is increased from P 1 to

Question

The enclosed air inside a closed box of rigid walls has pressure P 1 ​ . The density of air inside this box is constant. On heating the gas, the pressure of the enclosed air is increased from P 1 ​ to P 2 ​ . Now it is observed that the sound travels 1.4 times faster than at pressure P 1 ​ . The ratio P 2 ​ by P 1 ​ is

TestHub · Accepted Answer

The speed of sound in a gas is given by $v = \frac{γ P}{ρ}$ .

Since the box has rigid walls, the volume is constant. The mass of the enclosed air is also constant, which means the density $ρ$ remains constant.

Thus, the speed of sound is directly proportional to the square root of pressure, $v \propto P$ .

$\frac{v _{2}}{v _{1}} = \frac{P _{2}}{P _{1}}$

Given that $v_{2} = 1.4 v_{1}$ , we have:

$1.4 = \frac{P _{2}}{P _{1}}$

Squaring both sides gives:

$\frac{P _{2}}{P _{1}} = (1.4)^{2} = 1.96$

Answer: $1.96$

Physics - Sound Wave Question with Solution | TestHub

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