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Step-by-Step Solution
Step 1: Identify the Physical Process
The container is thermally insulated and no work is done on the gas. Hence, the total internal energy of the system remains constant. Since we are dealing with ideal gases, their internal energy depends only on temperature (and number of moles, with the same degrees of freedom assumed for both gases).
Step 2: Express the Conservation of Internal Energy
Let the final equilibrium temperature be $T$. For a monoatomic (or any ideal) gas, the internal energy $U$ is proportional to $n C_V T$, where $n$ is the number of moles, $C_V$ the molar heat capacity at constant volume, and $T$ the temperature. Conservation of internal energy implies:
$$(n_1 + n_2) C_V \, T \;=\; n_1 C_V \, T_1 \;+\; n_2 C_V \, T_2,$$
where:
$n_1$ is the number of moles in the first chamber; $T_1$ is its initial temperature.
$n_2$ is the number of moles in the second chamber; $T_2$ is its initial temperature.
$T$ is the final equilibrium temperature.
Step 3: Relate Number of Moles to Pressure and Volume
For an ideal gas, $P V = n R T$. Thus,
$n_1 = \dfrac{P_1 V_1}{R T_1}.$
$n_2 = \dfrac{P_2 V_2}{R T_2}.$
Step 4: Substitute into the Internal Energy Equation
Substituting $n_1$ and $n_2$ into the internal energy balance gives:
\[
\left(\dfrac{P_1 V_1}{R T_1} + \dfrac{P_2 V_2}{R T_2}\right) C_V \, T
\;=\;
\dfrac{P_1 V_1}{R T_1} \, C_V \, T_1
\;+\;
\dfrac{P_2 V_2}{R T_2} \, C_V \, T_2.
\]
The factor $C_V/R$ can be canceled out across all terms, yielding:
\[
\left(\dfrac{P_1 V_1}{T_1} + \dfrac{P_2 V_2}{T_2}\right) T
\;=\;
P_1 V_1 + P_2 V_2.
\]
Step 5: Solve for the Final Temperature
Rearrange to isolate $T$:
\[
T
\;=\;
\dfrac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}}
\;=\;
\dfrac{\bigl(P_1 V_1 + P_2 V_2\bigr)\,T_1\,T_2}{P_1 V_1\,T_2 + P_2 V_2\,T_1}.
\]
Final Answer
$\displaystyle T \;=\; \frac{T_1 \, T_2 \,\bigl(P_1 \, V_1 + P_2 \, V_2\bigr)}{P_1 V_1 \, T_2 + P_2 V_2 \, T_1}.
$
