If gamma = 2.5 and the final volume is frac{1 | vedclass

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

If $\gamma = 2.5$ and the final volume is $\frac{1}{8}$ times the initial volume, then the final pressure $P'$ is equal to (Initial pressure $= P$).

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