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Question: At what temperature is the velocity of sound in oxygen equal to half its velocity in chlorine at 42\...

At what temperature is the velocity of sound in oxygen equal to half its velocity in chlorine at 42C{}^{\circ }C? The ratio of molecular weights of oxygen and chlorine is 16 : 35.

Explanation

Solution

Use the formula for the speed of sound waves travelling in a given gas. Consider the speed of sound in oxygen be v and that of chlorine be v2\dfrac{v}{2}. Then write the equations for both the speeds. Then divide the two equations to find the temperature of oxygen.
Formula used:
v=γRTMv=\sqrt{\dfrac{\gamma RT}{M}}

Complete answer:
We know that the speed of sound waves in air is approximately equal to 340ms1m{{s}^{-1}}. However, it is not constant in all the gases. The speed of the sound waves changes with change of the gas in which it travels.
The speed of sound traveling in a gas is given as v=γRTMv=\sqrt{\dfrac{\gamma RT}{M}} …. (i).
Here, γ\gamma is the adiabatic index, R is the universal gas constant, T is the temperature of the gas and M is the molecular mass of the gas.
It is given that velocity of sound in oxygen is equal to half its velocity in chlorine at a temperature 42C{}^{\circ }C.
Let the speed of sound in oxygen be v. Then the speed of sound in chlorine is v2\dfrac{v}{2}.
Let the temperature of oxygen at which we can get this relation be T. And the temperature of chlorine is 42C=315K42{}^{\circ }C=315K
Let the molecular weights of oxygen and chlorine be Mo{{M}_{o}} and Mc{{M}_{c}}.
The value of γ\gamma for both will be the same.
Therefore, from (i) we get,
v=γRTMov=\sqrt{\dfrac{\gamma RT}{{{M}_{o}}}} …. (ii).
And v2=γR(315)Mc\dfrac{v}{2}=\sqrt{\dfrac{\gamma R(315)}{{{M}_{c}}}} …. (iii).
Divide (ii) and (iii).
vv2=γRTMoγRMc\Rightarrow \dfrac{v}{\dfrac{v}{2}}=\dfrac{\sqrt{\dfrac{\gamma RT}{{{M}_{o}}}}}{\sqrt{\dfrac{\gamma R}{{{M}_{c}}}}}
2=McT315Mo\Rightarrow 2=\sqrt{\dfrac{{{M}_{c}}T}{315{{M}_{o}}}}.
It is given that McMo=3516\dfrac{{{M}_{c}}}{{{M}_{o}}}=\dfrac{35}{16}.
2=35T315(16)\Rightarrow 2=\sqrt{\dfrac{35T}{315(16)}}
4=35T315(16)\Rightarrow 4=\dfrac{35T}{315(16)}
T=4×315×1635=576K\Rightarrow T=\dfrac{4\times 315\times 16}{35}=576K.
Therefore, the temperature of oxygen must be 576K.

Note:
Some students may make a mistake by substituting the values of the temperature of chlorine in degree Celsius. The temperature in the formula for speed of sound is in Kelvin.
Therefore, always remember to substitute the temperature in Kelvin in any problem that requires temperature. Otherwise, the answer may go wrong.