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A gaseousmixtureconsistsof16gofheliumand16gofoxygenthe ratioof twospecificheatsof themixtureis |
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Answer» Solution :Number of MOLES of helium `(n)=16/4=4,` Number of moles of OXYGEN `(n^(.))=16/32=1/2` For For the monoatomic helium gas, degrees of freedom, f=3 So `C_(V)=f/2R=3/2R` `C_(V)("MIXTURE")=(nC_(V)+n^(.)C_(V))/(n+n^(.))=(4xx3/2R+1/2xx5/2R)/((4+1/2))=29/18R therefore C_(P)-C_(V)=R` `GAMMA("mixture")=C_(P)/C_(V)=1+R/(C_(V)("mixture"))=1+R/((29/18R))=1+18/29=47/29` `gamma("mixture")=1.62` |
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