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(C) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monatomic gas,the adiabati

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$900$

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(C) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monatomic gas,the adiabatic index $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.Given $V_2 = 8 V_1$ and $T_1 = 100 K$.$T_2 = T_1 \left(\frac{V_1}{V_2}ight)^{\gamma-1} = 100 	imes \left(\frac{1}{8}ight)^{2/3} = 100 	imes \left(\frac{1}{4}ight) = 25 K$.The change in internal energy is given by $\Delta U = n C_v \Delta T$.For a monatomic gas,$C_v = \frac{3}{2} R$.$\Delta U = 1 	imes \frac{3}{2} 	imes 8.0 	imes (25 - 100) = 12 	imes (-75) = -900 J$.The decrease in internal energy is $|\Delta U| = 900 J$.

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is $100 K$ and the universal gas constant $R = 8.0 J mol^{-1} K^{-1}$,the decrease in its internal energy,in Joule,is. . . . .

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