The volume of air increases by 5% in its adia | vedclass

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(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = K$,where $\gamma$ is the adiabatic index.Taki

Vedclass · Accepted Answer

$7$

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(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = K$,where $\gamma$ is the adiabatic index.Taking the logarithmic derivative of both sides: $\ln P + \gamma \ln V = \ln K$.Differentiating both sides,we get: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.This implies $\frac{dP}{P} = -\gamma \frac{dV}{V}$.For air (a diatomic gas),the adiabatic index $\gamma = 1.4$.Given the percentage increase in volume $\frac{dV}{V} 	imes 100 = 5\%$.The percentage change in pressure is $\frac{dP}{P} 	imes 100 = -\gamma \left( \frac{dV}{V} 	imes 100 ight)$.Substituting the values: $\frac{dP}{P} 	imes 100 = -1.4 	imes 5 = -7\%$.The negative sign indicates a decrease in pressure. Therefore,the percentage decrease in pressure is $7\%$.

The volume of air increases by $5\%$ in its adiabatic expansion. The percentage decrease in its pressure will be ...... $\%$

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