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A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture isA. `1.59`B. `1.62`C. `1.4`D. `1.54` |
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Answer» Correct Answer - B `C_(v) =(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))` For helium, `n_(1) = (16)/(4) = 4` and `gamma_(1) = (5)/(3)` For oxygen `n_(2) = (16)/(32) = (1)/(6)` and `gamma_(2) = (7)/(5)` `C_(v_(1)) = (R )/(gamma_(1)-1) = (R )/((5)/(3)-1) -(3)/(2)R` `C_(v_(2)) = (R )/(gamma_(2)-1) = (R)/((7)/(5)-1)=(5)/(2)` `:. C_(v) = (4xx(3)/(2)R+(1)/(2),(5)/(2)R)/(4+(1)/(2)) = (6R+(5)/(4)R)/((9)/(2))` `= (29 R xx 2)/(9xx4) = (29R)/(18), C_(P) = C_(V) +R` `rArr (C_(P))/(C_(V)) = (C_(V)+R)/(C_(V)) = 1+(R)/(C_(V))` `rArr (C_(P))/(C_(V)) = (R)/((29)/(18)R) +1 rArr (C_(P))/(C_(V)) = (18)/(19) +1` `= (18 +19)/(29) = 1.62` |
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