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Chapter 9: Problem 74

If the temperature is raised by \(1 \mathrm{~K}\) from \(300 \mathrm{~K}\) the percentage change in the speed of sound in a gaseous mixture is \((R=8.31 \mathrm{~J} /\) mole \(-\mathrm{K})\) (A) \(0.167 \%\) (B) \(2 \%\) (C) \(1 \%\) (D) \(0.334 \%\)

Short Answer

Expert verified
The percentage change in the speed of sound when the temperature is raised by 1 K from 300 K can be found using the expression \[\% \text{ change} = \frac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}} \times 100\]. Substituting the given values of T1 = 300 K and T2 = 301 K, we get \[\% \text{ change} = \frac{\sqrt{301} - \sqrt{300}}{\sqrt{300}} \times 100 \approx 0.167\%\]. Thus, the correct answer is (A) \(0.167\%\).

Step by step solution

01

Write down the formula for the speed of sound in a gas

The formula for the speed of sound in an ideal gas is given by: \[v = \sqrt{\frac{\gamma R T}{M}}\] Where \(v\) is the speed of sound, \(\gamma\) is the adiabatic index, R is the gas constant, T is the temperature, and M is the molar mass of the gas.
02

Calculate the initial speed of sound

We will use the given initial temperature T1 = 300 K and the gas constant in our formula. To make the calculation simple, we will not use any specific value for \(\gamma\) and M. \[v_1 = \sqrt{\frac{\gamma R T_1}{M}}\]
03

Calculate the final speed of sound

The problem states that the temperature is raised by 1 K, so T2 = 301 K. We will use this temperature in our formula. \[v_2 = \sqrt{\frac{\gamma R T_2}{M}}\]
04

Calculate the percentage change in the velocity

The percentage change in the speed of sound is defined as: \[\% \text{ change} = \frac{v_2 - v_1}{v_1} \times 100\] Substitute the expressions for \(v_1\) and \(v_2\): \[\% \text{ change} = \frac{\sqrt{\frac{\gamma R T_2}{M}} - \sqrt{\frac{\gamma R T_1}{M}}}{\sqrt{\frac{\gamma R T_1}{M}}} \times 100\]
05

Simplify the expression

We can simplify the expression by cancelling out the common terms: \[\% \text{ change} = \frac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}} \times 100\]
06

Plug in the given values and solve

Substitute the given values of T1 = 300 K and T2 = 301 K into the expression and solve: \[\% \text{ change} = \frac{\sqrt{301} - \sqrt{300}}{\sqrt{300}} \times 100 \approx 0.167\%\] The percentage change in the speed of sound is approximately 0.167% when the temperature is raised by 1 K from 300 K. Therefore, the correct answer is (A) \(0.167\%\).

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