Compare the theoretical speeds of sound in hydrogen (`\M_H` = 2.0 g/mol, `\Ɣ_H` =1.4) with helium (`\M_{He}` = 4.0 g mol⁻¹, `\Ɣ_{He}` = 1.66 & R = 8334 j k⁻¹ mol⁻¹) at 0°C .
( `\v_He` = 0.77 `\v_H`)


Given:

`\M_H` = 2.0 g/mol
`\Ɣ_H` =1.4
`\M_{He}` = 4.0 g mol⁻¹
`\Ɣ_{He}` = 1.66 
R = 8334 j k⁻¹ mol⁻¹
T = 0°C = 273 K

To Find:

Comparison of the theoretical speeds of sound in hydrogen with helium = `\frac{v_{He}}{v_H}` = ?

Solution:

To find the speed v of sound in a gas is given by:

v = `\sqrt frac {ƔRT}{M}`

Speed `\v_{He} of sound in Helium (He) gas

`\v_{He}` = `\sqrt frac {Ɣ_{He}ｘRｘT}{M}`

`\v_{He}`  = `\sqrt frac {1.66 ｘ 8334 j k⁻¹ mol⁻¹ ｘ 273 K}{4 kg}`

`\v_{He}`  = `\sqrt {944,200.53m² s⁻²}`

`\v_{He}` = 972 m s⁻¹

Speed vH of sound in Hydrogen (H) gas

`\v_H` = `\sqrt frac {Ɣ_HｘRｘT}{M}`

`\v_H` = `\sqrt frac {1.4 ｘ 8334 j k⁻¹ mol⁻¹ ｘ 273 K}{2 kg}`

`\v_H` = `\sqrt {1,592627.4 m² s⁻²}`

`\v_H` = 1262 m s⁻¹


Now 

`\frac{v_{He}}{v_H}` = `\frac{972 m s⁻¹}{1262 m s⁻¹}`

`\frac{v_{He}}{v_H}` = 0.77

or

`\v_{He}` = 0.77 `\v_H`



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