Question

temperature at which velocity of sound in air is twice its velocity at 15°C​

Answer 1

Answer:

Velocity of sound is directly proportional to square root of absolute temperature.

(v2/v1) = √(T2 / T1)

Where v2 is final velocity, v1 is initial velocity, T2 is final absolute temperature, T1 is initial temperature.

Absolute temperature is equal to degree Celsius plus 273.

Given v2 = 2 × v1, temperature in degree Celsius = 10

Temperature in Kelvin = 10 + 273 = 283

Putting values

(2 × v1)/v1 = √(T2/283)

2 = √ (T2 /283)

Squaring both sides

4 = (T2/283)

T2 = 4 × 283

T2 = 1132 kelvin

Temperature in degrees Celsius = 1132 - 273 = 859 degree Celsius.

At 859 degree Celsius or at 1132 Kelvin temperature velocity will be double the velocity at 10 degree Celsius.

Answer 2

Concept:

The equation of velocity of sound in air is represented by the equation, v ∝ √T

Given:

Temperature, T = 15°C = 273 + 15 = 288 K

Find:

We need to determine the temperature at which the velocity of sound in air is twice its velocity at 15°C

Solution:

We know that the velocity of sound in air is represented by the equation,

v ∝ √T where v is the velocity and T is the temperature.

Temperature has an impact on the sound wave speed as well. High temperatures give molecules more energy, which allows them to vibrate more quickly. Sound waves can move more swiftly because the molecules vibrate more quickly. Thus, a drop in temperature results in a decrease in the sound velocity in air.

It can also be represented as, v² ∝ T

Therefore, v1²/v2² = T1 / T2

v1²/(2v1)² = 288 / T2

Therefore, T2 = 288 × 4

T2 = 1152 K

T2 = 1152 - 273 = 879°C

Thus, the temperature at which velocity of sound in air is twice its velocity at 15°C​ is 879°C.

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