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  • Need explanation for: The speed of sound in oxygen (O2) at a certain temperature is 460 ms-1 The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal)

The speed of sound in oxygen  (O2) at a certain temperature is  460 ms-1 The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal)

  • Option 1)

    330 ms^{-1}

  • Option 2)

    460 ms^{-1}

  • Option 3)

    500 ms^{-1}

  • Option 4)

    650 ms^{-1}

 

Answers (2)

\nu = \sqrt{\frac{\gamma P}{\rho }}= \sqrt{\frac{\gamma RT}{M}}

\gamma \: for\: O_{2}= 1+2/5= 1.4

\gamma \: for\: H\! e= 1+2/3=5/3

\frac{\nu _{2}}{\nu _{1}}= \left ( \sqrt{\frac{\gamma _{He}}{4} \times \frac{32}{\gamma _{O_{2}}}}\right )\times 460

=460\times \sqrt{\frac{5}{3}\times \frac{1}{4}\times \frac{32\times 5}{7}}=1420m/s

The value of the speed of sound in He should have been 965 m/s and that of O2 about 320 m/s. The value of the velocity given fo O2 is  quite high.

Option not given.

V\:=\sqrt{\frac{\gamma P}{P}}\:=\:\sqrt{\frac{\gamma RT}{M}}

\gamma for O_{2}\:=\:1+\frac{2}{5}=\frac{7}{5}

\gamma for H_{e}=1+\frac{2}{3}=\frac{5}{3}

\frac{v_{2}}{v_{1}} = \sqrt{\frac{\gamma H_{e}}{4}\times \frac{32}{\gamma O_{2}}}\:\times460\:\:\:\:=\sqrt{\frac{5}{3}\times\frac{1}{4}\times\frac{32\times 5}{7}}\:\times460\:=\:1420\:ms^{-1}


Option 1)

330 ms^{-1}

Option 2)

460 ms^{-1}

Option 3)

500 ms^{-1}

Option 4)

650 ms^{-1}

Posted by

Sabhrant Ambastha

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Laplace formula (correct value )

= \sqrt{\frac{\gamma P}{\rho }}

-

 

 

\nu = \sqrt{\frac{\gamma P}{\rho }}= \sqrt{\frac{\gamma RT}{M}}

\gamma \: for\: O_{2}= 1+\frac{2}{5}=\frac{7}{5}

\gamma \: for\: H\! e= 1+\frac{2}{3}=\frac{5}{3}

\frac{\nu _{2}}{\nu _{1}}= \sqrt{\frac{\gamma _{He}}{4} \times \frac{32}{\gamma _{O_{2}}}}\times 460= \sqrt{\frac{5}{3}\times \frac{1}{4}\times \frac{32\times 5}{7}} \times 460=1420m/s

Posted by

Safeer PP

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