An air bag on adiabatic expansion undergoes 5 | vedclass

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(C) For an adiabatic process, the relation between pressure and volume is $Pv^{\gamma} = K$.Differentiating both sides, we get $v^{\gamma} dP + P \gam

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) For an adiabatic process, the relation between pressure and volume is $Pv^{\gamma} = K$.Differentiating both sides, we get $v^{\gamma} dP + P \gamma v^{\gamma-1} dv = 0$.Rearranging the terms, we get $\frac{dP}{P} = -\gamma \frac{dv}{v}$.To find the percentage change, multiply both sides by $100$:$\frac{dP}{P} 	imes 100 = -\gamma \left( \frac{dv}{v} 	imes 100 ight)$.Given $\gamma = 1.4$ and $\frac{dv}{v} 	imes 100 = 5 \%$, we substitute these values:$	ext{Percentage change in pressure} = -1.4 	imes 5 = -7 \%$.The negative sign indicates a decrease in pressure. Thus, the magnitude of the percentage change is $7 \%$.

An air bag on adiabatic expansion undergoes $5 \%$ increase in its volume. The percentage change in pressure is $\left[\gamma_{air}=1.4\right]$ (in $\%$)

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