Question

If temperature is raised by $ 1{\text{K}} $ from $ 1{\text{K}} $ , find the percentage change in the speed of sound in the gaseous mixture.

Answer 1

Hint:
To solve this question, we need to find the expression of the speed of the sound in terms of the temperature. Then, applying the concept of relative errors on that expression will give the answer.
Formula Used: The formula used to solve this question is
 $ v = \sqrt {\dfrac{{\gamma P}}{\rho }} $ , where $ v $ is the velocity of sound in a gaseous medium having pressure $ P $ density $ \rho $ and the ratio of the specific heats at constant pressure and at constant volume as $ \gamma $ .

Complete step by step answer:
The original temperature of the gaseous mixture is given as
 $ T = 300{\text{K}} $ …………………..(1)
And the change in the temperature is given as
 $ \Delta T = 1{\text{K}} $ …………………..(2)
We know that the velocity of sound in a gaseous medium is given by the Laplace correction formula as
 $ v = \sqrt {\dfrac{{\gamma P}}{\rho }} $ …………………..(3)
But we need to find the velocity in terms of the temperature of the gaseous mixture, as we have information about the temperature only.
From the ideal gas equation we have
 $ PV = nRT $
Dividing both sides by $ V $ we get
 $ P = \dfrac{{nRT}}{V} $
We know that the number of moles is given by
 $ n = \dfrac{M}{{{M_0}}} $
Putting this we get
 $ P = \dfrac{{MRT}}{{{M_0}V}} $
Substituting $ \rho = \dfrac{M}{V} $ , we have
  $ P = \dfrac{{\rho RT}}{{{M_0}}} $
Substituting this value of the pressure in (3)
 $ v = \sqrt {\dfrac{\gamma }{\rho }\dfrac{{\rho RT}}{{{M_0}}}} $
Cancelling $ \rho $ we get
 $ v = \sqrt {\dfrac{{\gamma RT}}{{{M_0}}}} $
As $ \gamma $ , $ R $ , and $ {M_0} $ are constants for a given mixture, so combining all these constants into a single constant $ k $ , we get
 $ v = k\sqrt T $
Taking relative errors on both the sides of this equation
 $ \dfrac{{\Delta v}}{v} = \dfrac{1}{2}\dfrac{{\Delta T}}{T} $
Substituting the values from (1) and (2), we get
 $ \dfrac{{\Delta v}}{v} = \dfrac{1}{2} \times \dfrac{1}{{300}} $
Now, the percentage change will be equal to
 $ \dfrac{{\Delta v}}{v} \times 100 $
 $ \dfrac{1}{2} \times \dfrac{1}{{300}} \times 100 $
 $ = 0.167\% $
Thus, the percentage change in the value of the speed of the sound in the given gaseous mixture is equal to $ 0.167\% $ .

Note:
Do not understand the relative error concept as simply the multiplication of the exponents of the quantities with their corresponding relative errors. Instead, the modulus of the exponents is multiplied. This is done to ensure that the relative errors of all the quantities are being added. Always remember, the errors are always added, never subtracted.