The container shown in the figure has two chambers,separated by a partition,with volumes $V_1 = 2.0 \, L$ and $V_2 = 3.0 \, L$. The chambers contain $\mu_1 = 4.0 \, mol$ and $\mu_2 = 5.0 \, mol$ of a gas at pressures $P_1 = 1.00 \, atm$ and $P_2 = 2.00 \, atm$. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

(1.60 ATM) Given:
$V_1 = 2.0 \, L, V_2 = 3.0 \, L$
$\mu_1 = 4.0 \, mol, \mu_2 = 5.0 \, mol$
$P_1 = 1.00 \, atm, P_2 = 2.00 \, atm$
For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} PV$,where $f$ is the degrees of freedom.
Since the gases mix and reach thermal equilibrium,the total internal energy is conserved.
$U_{total} = U_1 + U_2$
$\frac{f}{2} P(V_1 + V_2) = \frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2$
Canceling $\frac{f}{2}$ from both sides,we get:
$P(V_1 + V_2) = P_1 V_1 + P_2 V_2$
Solving for the final pressure $P$:
$P = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$
Substituting the values:
$P = \frac{(1.00 \times 2.0) + (2.00 \times 3.0)}{2.0 + 3.0} \, atm$
$P = \frac{2.0 + 6.0}{5.0} \, atm$
$P = \frac{8.0}{5.0} \, atm = 1.60 \, atm$
The final pressure of the mixture is $1.60 \, atm$.