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(D) Given: Initial volume $V_1 = 800 \ cm^3$,Final volume $V_2 = 200 \ cm^3$,Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$,and adiabat

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$300$

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(D) Given: Initial volume $V_1 = 800 \ cm^3$,Final volume $V_2 = 200 \ cm^3$,Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$,and adiabatic index $\gamma = 1.5$.For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $300 	imes (800)^{1.5-1} = T_2 	imes (200)^{1.5-1}$.$300 	imes (800)^{0.5} = T_2 	imes (200)^{0.5}$.$T_2 = 300 	imes \left( \frac{800}{200} ight)^{0.5} = 300 	imes (4)^{0.5} = 300 	imes 2 = 600 \ K$.The change in temperature is $\Delta T = T_2 - T_1 = 600 \ K - 300 \ K = 300 \ K$.

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