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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. |
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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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