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(D) Since the system is isolated,the total change in internal energy is zero: $\Delta E_{	ext{int}} = 0$.This implies: $(\Delta E_{	ext{int}})_{N_2}

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$T_f = \frac{3}{2} T_o$

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(D) Since the system is isolated,the total change in internal energy is zero: $\Delta E_{	ext{int}} = 0$.This implies: $(\Delta E_{	ext{int}})_{N_2} + (\Delta E_{	ext{int}})_{He} = 0$.For nitrogen ($N_2$,diatomic),$C_V = \frac{5}{2}R$. For helium ($He$,monatomic),$C_V = \frac{3}{2}R$.Using the formula $\Delta E_{	ext{int}} = \mu C_V \Delta T$:$1 	imes (\frac{5}{2}R)(T_f - T_o) + 1 	imes (\frac{3}{2}R)(T_f - \frac{7}{3}T_o) = 0$.Dividing by $R/2$:$5(T_f - T_o) + 3(T_f - \frac{7}{3}T_o) = 0$.$5T_f - 5T_o + 3T_f - 7T_o = 0$.$8T_f = 12T_o$.$T_f = \frac{12}{8}T_o = \frac{3}{2}T_o$.

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