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(B) For an adiabatic process, the relationship between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given: Initial

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$-182$

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(B) For an adiabatic process, the relationship between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given: Initial temperature $T_1 = 273 \ K$, final volume $V_2 = 81 V_1$, and adiabatic index $\gamma = 1.25$.Rearranging the formula for the final temperature $T_2$:$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma - 1}$Substituting the values:$T_2 = 273 	imes \left( \frac{V_1}{81 V_1} ight)^{1.25 - 1}$$T_2 = 273 	imes \left( \frac{1}{81} ight)^{0.25}$Since $0.25 = \frac{1}{4}$, we have:$T_2 = 273 	imes \left( \frac{1}{81} ight)^{1/4}$$T_2 = 273 	imes \frac{1}{(3^4)^{1/4}} = \frac{273}{3} = 91 \ K$To convert the temperature from Kelvin to Celsius:$T(^\circ C) = T(K) - 273$$T(^\circ C) = 91 - 273 = -182 ^\circ C$.

$A$ certain mass of gas at $273 \ K$ is expanded to $81$ times its volume under adiabatic condition. If $\gamma = 1.25$ for the gas, then its final temperature is ..... $^\circ C$

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