Solution 33
1. Final Equilibrium Temperature
Let the number of moles in the left and right parts be $n_1$ and $n_2$ respectively. Since the system is thermally insulated from the outside, the total internal energy of the system is conserved.
Using the Ideal Gas Law $PV = nRT$, we can express the initial moles as:
$$ n_1 = \frac{p_0 V}{R T_1}, \quad n_2 = \frac{p_0 V}{R T_2} $$Let the final common temperature be $T$. By conservation of energy ($\Delta U_{system} = 0$):
$$ \frac{3}{2} n_1 R (T – T_1) + \frac{3}{2} n_2 R (T – T_2) = 0 $$ $$ n_1 (T – T_1) + n_2 (T – T_2) = 0 $$ $$ T (n_1 + n_2) = n_1 T_1 + n_2 T_2 $$ $$ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} $$Substituting the expressions for $n_1$ and $n_2$:
$$ T = \frac{\frac{p_0 V}{R T_1} T_1 + \frac{p_0 V}{R T_2} T_2}{\frac{p_0 V}{R T_1} + \frac{p_0 V}{R T_2}} = \frac{2}{\frac{1}{T_1} + \frac{1}{T_2}} = \frac{2 T_1 T_2}{T_1 + T_2} $$2. Heat Transferred
Let’s calculate the heat transferred to the gas on the left (initially at $T_1$, assuming $T_1 < T_2$). For a system where pressure balances continuously while heat is exchanged slowly (quasi-static), the heat transfer for one side in this specific coupled system corresponds to the enthalpy change, using $C_p$.
$$ \Delta Q = n_1 C_p (T – T_1) $$For a monoatomic gas, $C_p = \frac{5}{2}R$.
$$ \Delta Q = \left( \frac{p_0 V}{R T_1} \right) \left( \frac{5}{2} R \right) (T – T_1) $$ $$ \Delta Q = \frac{5 p_0 V}{2 T_1} \left( \frac{2 T_1 T_2}{T_1 + T_2} – T_1 \right) $$ $$ \Delta Q = \frac{5 p_0 V}{2 T_1} \left( \frac{2 T_1 T_2 – T_1^2 – T_1 T_2}{T_1 + T_2} \right) = \frac{5 p_0 V}{2 T_1} \left( \frac{T_1 (T_2 – T_1)}{T_1 + T_2} \right) $$