Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure $$p_i$$ and temperature $$T_1$$ are connected through a narrow tube of negligible volume, as shown in the figure below. The temperature of one of the bulbs is then raised to $$T_2$$. The final pressure $$P_f$$ is:

Initially, both bulbs have a volume $$V$$, pressure $$p_i$$, and temperature $$T_1$$.

Using the Ideal Gas Law ($$PV = nRT$$), the number of moles in each bulb is:

$$n_1 = \frac{p_i V}{R T_1}$$, $$n_2 = \frac{p_i V}{R T_1}$$

$$n_{total} = n_1 + n_2 = \frac{2 p_i V}{R T_1}$$

In the final state, the bulbs reach a common pressure $$P_f$$. Bulb 1 remains at temperature $$T_1$$, while bulb 2 is raised to temperature $$T_2$$.

$$n'_1 = \frac{P_f V}{R T_1}$$, $$n'_2 = \frac{P_f V}{R T_2}$$

The total number of moles remains the same:

$$n_{total} = n'_1 + n'_2$$

$$n_{total} = \frac{P_f V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right)$$

$$n_{total} = \frac{P_f V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$$

$$\frac{2 p_i V}{R T_1} = \frac{P_f V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$$

$$P_f = 2 p_i \left( \frac{T_2}{T_1 + T_2} \right)$$