Home
Class 11
PHYSICS
The velocity of sound in a gas depends o...

The velocity of sound in a gas depends on its pressure and density . Obtain the relation between velocity , pressure and density.

Text Solution

AI Generated Solution

The correct Answer is:
To find the relationship between the velocity of sound (v), pressure (P), and density (ρ) in a gas, we can start by expressing the dependencies in terms of dimensional analysis. ### Step 1: Identify the variables We know that: - Velocity of sound (v) - Pressure (P) - Density (ρ) ### Step 2: Write the dimensional formulas 1. **Velocity (v)** has dimensions of length per time: \[ [v] = L^1 T^{-1} \] 2. **Pressure (P)** is defined as force per unit area. The dimensional formula for force (F) is mass times acceleration: \[ [F] = M^1 L^1 T^{-2} \] Therefore, pressure can be expressed as: \[ [P] = \frac{[F]}{[A]} = \frac{M^1 L^1 T^{-2}}{L^2} = M^1 L^{-1} T^{-2} \] 3. **Density (ρ)** is mass per unit volume: \[ [ρ] = \frac{M^1}{L^3} = M^1 L^{-3} T^0 \] ### Step 3: Establish the relationship Assuming the velocity of sound (v) is proportional to the pressure (P) and inversely proportional to the square root of density (ρ), we can write: \[ v \propto \frac{P^x}{ρ^y} \] where x and y are constants to be determined. ### Step 4: Write the dimensional equation Substituting the dimensions into the equation gives: \[ [L^1 T^{-1}] = \frac{[M^1 L^{-1} T^{-2}]^x}{[M^1 L^{-3}]^y} \] ### Step 5: Simplify the right-hand side This expands to: \[ [L^1 T^{-1}] = \frac{M^{x} L^{-x} T^{-2x}}{M^{y} L^{-3y}} \] This can be simplified to: \[ [L^1 T^{-1}] = M^{x-y} L^{-x+3y} T^{-2x} \] ### Step 6: Equate dimensions Now we equate the dimensions on both sides: 1. For mass (M): \[ x - y = 0 \quad \text{(1)} \] 2. For length (L): \[ -x + 3y = 1 \quad \text{(2)} \] 3. For time (T): \[ -2x = -1 \quad \Rightarrow \quad x = \frac{1}{2} \quad \text{(3)} \] ### Step 7: Solve the equations From equation (3), substituting \(x = \frac{1}{2}\) into equation (1): \[ \frac{1}{2} - y = 0 \quad \Rightarrow \quad y = \frac{1}{2} \] Now substituting \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\) into equation (2): \[ -\frac{1}{2} + 3 \left(\frac{1}{2}\right) = 1 \quad \Rightarrow \quad 1 = 1 \quad \text{(True)} \] ### Conclusion Thus, the final relationship between the velocity of sound (v), pressure (P), and density (ρ) is: \[ v = k \sqrt{\frac{P}{ρ}} \] where \(k\) is a proportionality constant.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • UNITS AND DIMENSION

    NN GHOSH|Exercise EXERCISE|26 Videos
  • THERMOMETRY

    NN GHOSH|Exercise All Questions|9 Videos
  • VECTOR AND SCALARS

    NN GHOSH|Exercise All Questions|48 Videos

Similar Questions

Explore conceptually related problems

Write the relation between pressure and density of gas.

Give the relationship between pressure and density of gas.

Knowledge Check

  • The velocity of sound in any gas depends upon

    A
    wavelength of sound only
    B
    density and elasticity of gas
    C
    intensity of sound waves only,
    D
    amplitude and frequency of sound
  • The velocity of sound in a gas at pressure P and density P

    A
    `V = sqrt(Yp)/(rho)`
    B
    `V = sqrt(P/(Y_rho))`
    C
    `V = y sqrt(p/rho)`
    D
    `V = sqrt((2P)/(rho))`
  • Velocity of sound in a gas….. With increase of pressure

    A
    increase
    B
    does not vary
    C
    decrease
    D
    either 'a' or 'c' depending on the gas
  • Similar Questions

    Explore conceptually related problems

    At a constant temperature, what is the relation between pressure P and density rho of gas?

    Velocity of sound in solids is affected by their density .

    The velocity of sound is ___ to the density of the medium

    The velocity of sound in air is taken as 332 m/s.If the pressure of gas is taken double ,then what will be the velocity of sound in air?

    At constant pressure density of a gas is