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(C) Using the ideal gas law, the total number of moles $n_{total} = n_1 + n_2 = 3.0 + 4.0 = 7.0 \, 	ext{moles}$.The total volume $V_{total} = V_1 + V

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Using the ideal gas law, the total number of moles $n_{total} = n_1 + n_2 = 3.0 + 4.0 = 7.0 \, 	ext{moles}$.The total volume $V_{total} = V_1 + V_2 = 4.5 + 5.5 = 10.0 \, L$.Assuming the temperature $T$ remains constant (as the problem implies a simple mixing process), we use the relation $P_{final} V_{total} = (n_1 + n_2) RT$.From the initial states:$P_1 V_1 = n_1 RT \Rightarrow (2.0)(4.5) = 3.0 RT \Rightarrow 9.0 = 3.0 RT \Rightarrow RT = 3.0$.$P_2 V_2 = n_2 RT \Rightarrow (3.0)(5.5) = 4.0 RT \Rightarrow 16.5 = 4.0 RT \Rightarrow RT = 4.125$.Since the temperatures are different, we use the conservation of internal energy: $U_1 + U_2 = U_{mix}$.$U = \frac{f}{2} nRT = \frac{f}{2} PV$.$\frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2 = \frac{f}{2} P_{final} (V_1 + V_2)$.$(2.0)(4.5) + (3.0)(5.5) = P_{final} (4.5 + 5.5)$.$9.0 + 16.5 = P_{final} (10.0)$.$25.5 = 10.0 P_{final} \Rightarrow P_{final} = 2.55 \, 	ext{atm}$.Given $P_{final} = x 	imes 10^{-1} \, 	ext{atm}$, we have $2.55 = x 	imes 10^{-1} \Rightarrow x = 25.5$.

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