For an adiabatic process,if gamma = 2.5 and t | vedclass

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

Vedclass · Accepted Answer

$P' = P 	imes (2)^{7.5}$

Vedclass · Answer

(C) For an adiabatic process,the relationship between pressure and volume is given by $PV^\gamma = 	ext{constant}$.Thus,$P_1 V_1^\gamma = P_2 V_2^\gamma$.Given: $P_1 = P$,$V_2 = \frac{V_1}{8}$,and $\gamma = 2.5 = \frac{5}{2}$.Substituting these values:$P 	imes V_1^{5/2} = P' 	imes \left(\frac{V_1}{8}ight)^{5/2}$.$P' = P 	imes \left(\frac{V_1}{V_1/8}ight)^{5/2} = P 	imes (8)^{5/2}$.Since $8 = 2^3$,we have $P' = P 	imes (2^3)^{5/2} = P 	imes 2^{15/2} = P 	imes 2^{7.5}$.Therefore,$P' = P 	imes (2)^{7.5}$.

For an adiabatic process,if $\gamma = 2.5$ and the volume becomes $1/8$ of its initial volume,then the new pressure $P'$ is (initial pressure $= P$):

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