For an adiabatic process with gamma = 2.5,if | vedclass

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(C) For an adiabatic process,the relation between pressure and volume is given by $PV^\gamma = 	ext{constant}$.This implies $P_1 V_1^\gamma = P_2 V_2

Vedclass · Accepted Answer

$P' = P 	imes 2^{15/2}$

Vedclass · Answer

(C) For an adiabatic process,the relation between pressure and volume is given by $PV^\gamma = 	ext{constant}$.This implies $P_1 V_1^\gamma = P_2 V_2^\gamma$,or $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} ight)^\gamma$.Given $\gamma = 2.5 = 5/2$ and the new volume $V_2 = \frac{V_1}{8}$,we have $\frac{V_1}{V_2} = 8$.Substituting these values: $\frac{P'}{P} = (8)^{5/2}$.Since $8 = 2^3$,we have $(2^3)^{5/2} = 2^{15/2}$.Therefore,$P' = P 	imes 2^{15/2}$.

For an adiabatic process with $\gamma = 2.5$,if the volume of a gas is reduced to $1/8$ of its initial volume,what will be the new pressure $P'$ in terms of the initial pressure $P$?

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