A container has two chambers of volumes V_1=2 | vedclass

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(B) For an ideal gas,the internal energy is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the partition is a thermal insulator,the total inte

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$1.6$

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(B) For an ideal gas,the internal energy is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the partition is a thermal insulator,the total internal energy of the system is conserved when the partition is removed.$U_{	ext{initial}} = U_{	ext{final}}$$\frac{f_1}{2} P_1 V_1 + \frac{f_2}{2} P_2 V_2 = \frac{f_{	ext{mix}}}{2} P_{	ext{mix}} (V_1 + V_2)$Assuming the gas is the same in both chambers,$f_1 = f_2 = f_{	ext{mix}} = f$.$P_1 V_1 + P_2 V_2 = P_{	ext{mix}} (V_1 + V_2)$Substituting the given values:$(1 \ atm 	imes 2 \ L) + (2 \ atm 	imes 3 \ L) = P_{	ext{mix}} (2 \ L + 3 \ L)$$2 + 6 = P_{	ext{mix}} (5)$$8 = 5 P_{	ext{mix}}$$P_{	ext{mix}} = \frac{8}{5} \ atm = 1.6 \ atm$.

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