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In adiabatic process, the pressure is increased by 2//3%. Ifgamma=3//2then the volume decreases by nearly – |
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Answer» `(4)/(9)%` `(DeltaP)/(DeltaV)=-(gammaP)/(V)` `(DeltaP)/(P) xx 100 = -gamma (DeltaV)/(V)xx 100 =(2)/(3)` `(DeltaV)/(V) xx 100 = -(1)/(gamma) xx (2)/(3)` `= -(2)/(3) xx (2)/(3) = -(4)/(9)%` |
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