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(C) 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 temp

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$668 K$

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(C) 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 $T_1 = 18^{\circ}C = 18 + 273 = 291 K$.Initial volume $V_1 = V$,final volume $V_2 = V/8$.For a triatomic gas,the degree of freedom $f = 6$ (non-linear). The adiabatic index $\gamma = 1 + 2/f = 1 + 2/6 = 1 + 1/3 = 4/3$.Thus,$\gamma - 1 = 1/3$.Using the formula: $T_2 = T_1 (V_1/V_2)^{\gamma-1} = 291 	imes (V / (V/8))^{1/3} = 291 	imes (8)^{1/3} = 291 	imes 2 = 582 K$.Converting back to Celsius: $582 - 273 = 309^{\circ}C$.Note: If the gas is considered linear,$f = 5$,$\gamma = 1.4$,$\gamma-1 = 0.4$. Then $T_2 = 291 	imes (8)^{0.4} \approx 291 	imes 2.297 \approx 668 K$.Given the options,the calculation assumes $\gamma = 1.4$ (diatomic/linear triatomic behavior). Thus,$T_2 = 668 K$.

$A$ triatomic gas at an initial temperature of $18^{\circ}C$ is compressed adiabatically to $1/8$ of its initial volume. What is the final temperature of the gas?

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