Solution 18
We need to construct the $p-T$ graph for the gas. The key behavior of the flexible, non-stretchable bottle is that it maintains its maximum volume $V_1$ only if the internal pressure $P \ge P_{atm}$. If $P < P_{atm}$, it collapses to zero volume (vanishingly small), effectively transferring all gas to the rigid vessel.
Initial State (Before Mixing):
Bottle: $V_1 = 1.0$ L, $P_1 = 2.0 p_0$.
Vessel: $V_2 = 4.0$ L, $P_2 = 0.1 p_0$.
Temperature $T_1 = 240$ K.
Process A: Valve Opens (Mixing at $T=240$K)
The total moles are conserved. Let’s calculate the equilibrium pressure $P_{mix}$ assuming the total volume is $V_1 + V_2 = 5.0$ L.
$$P_{mix}(V_1 + V_2) = P_1 V_1 + P_2 V_2$$
$$P_{mix}(5) = (2 p_0)(1) + (0.1 p_0)(4) = 2.4 p_0$$
$$P_{mix} = \frac{2.4 p_0}{5} = 0.48 p_0$$
Since $0.48 p_0 < p_0$ (atmospheric pressure), the flexible bottle collapses. The gas is compressed by the atmosphere into the rigid vessel.
Corrected State A: Volume = $V_2 = 4.0$ L.
$$P_A (4) = 2.4 p_0 \implies P_A = 0.6 p_0$$
Start point: $(T=240, P=0.6 p_0)$.
Process A $\to$ B: Heating from 240 K (Isochoric)
The gas heats up in the 4.0 L vessel. The bottle remains collapsed until pressure reaches $p_0$.
$$P \propto T \implies \frac{0.6 p_0}{240} = \frac{1.0 p_0}{T_B}$$
$$T_B = \frac{240}{0.6} = 400\,\text{K}$$
Path: Straight line from $(240, 0.6p_0)$ to $(400, p_0)$.
Process B $\to$ C: Bottle Inflation (Isobaric)
Once $P = p_0$, the bottle starts inflating against the atmosphere. Pressure remains constant at $p_0$ until the bottle is fully expanded ($V_{total} = 5.0$ L).
$$\frac{V_B}{T_B} = \frac{V_C}{T_C} \implies \frac{4.0}{400} = \frac{5.0}{T_C}$$
$$T_C = 500\,\text{K}$$
Path: Horizontal line from $(400, p_0)$ to $(500, p_0)$.
Process C $\to$ D: Heating to 600 K (Isochoric)
The system is now rigid again at max volume 5.0 L.
$$\frac{p_0}{500} = \frac{P_D}{600} \implies P_D = 1.2 p_0$$
Path: Straight line from $(500, p_0)$ to $(600, 1.2p_0)$.