1.

Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.

Answer»

Solution :Helium =-2 mole, OXYGEN =- 4mol
He and `O_(2)` are mixed, hence molecular weight of the mixture of gases
`M_("mix") =(n_(1)M_(1) + n_(2)M_(2))/(n_(1)+n_(2)) =(2 xx 4+ 4 xx 32)/(2+4)`
`=(8+128)/6 = 136/8 = 22.6 xx 10^(-3) kg mol^(-1)`
Futher as He is monoatomic
`Cv_(1)=3R//2`
and `O_(2)` is diatomic `Cv_(2) = 5R//2`, hence for the mixture
`(Cp)_("mix") =(CV)_("mix") +R = (13R)/6 +R= (19R)/6`
`therefore lambda_("mix") =(Cp)/(Cv) =(19R//6)/(13R//6) = 19/13`
Acceleration to Laplace, be speed of sound
`v=SQRT((lambda RT)/(M))`
`=sqrt(19/13 xx (8.31 xx 300 xx 6)/(1.36 xx 10^(-3))) = sqrt((284202)/1768 xx 10^(3))`
`=sqrt((28420)/1768 xx 10^(4))`
`v= 400.9 ms^(-1)`


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