An ideal gas undergoes an adiabatic process. | vedclass

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(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $P V^\gamma = 	ext{constant}$.Taking the natural logarithm

Vedclass · Accepted Answer

$0.06$

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(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $P V^\gamma = 	ext{constant}$.Taking the natural logarithm on both sides: $\ln P + \gamma \ln V = 	ext{constant}$.Differentiating both sides: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.This implies $\frac{dV}{V} = -\frac{1}{\gamma} \frac{dP}{P}$.Given that the pressure is reduced by $0.1 \%$, we have $\frac{dP}{P} = -0.001$.Given $\gamma = \frac{5}{3}$, we substitute these values into the equation:$\frac{dV}{V} = -\frac{1}{5/3} 	imes (-0.001) = \frac{3}{5} 	imes 0.001 = 0.6 	imes 0.001 = 0.0006$.Converting this to percentage: $0.0006 	imes 100 \% = 0.06 \%$.Thus, the volume increases by $0.06 \%$.

An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$, then the volume is changed by (Given $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$) (in $\%$)

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