Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into — States of Matter and Gaseous State Chemistry Question
Question
Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference, $\Delta P \ge 2.0\text{ atm}$. Initially there was a vacuum in one vessel while the other contained ideal gas at a temperature 300 K and pressure 4.0 atm. Then both vessels were heated to a temperature 600 K. Up to what value will be the pressure (in atm) in the first vessel (which had a vacuum initially) increase?
💡 Solution & Explanation
Let Vessel 2 have initial gas ($P_2=4.0$), Vessel 1 be vacuum ($P_1=0$). At 300 K, valve opens until $P_2 - P_1 = 2.0$. By mole conservation $P_1 + P_2 = 4.0$. Thus $P_2 = 3.0\text{ atm}$, $P_1 = 1.0\text{ atm}$. When heated to 600 K (doubling T), isolated pressures would double: $P'_2 = 6.0$, $P'_1 = 2.0$. $\Delta P = 4.0 > 2.0$, so valve opens again. It settles when $P''_2 - P''_1 = 2.0$. Mole conservation requires $P''_1 + P''_2 = 8.0$. Solving gives $P''_2 = 5.0\text{ atm}$ and $P''_1 = 3.0\text{ atm}$. The pressure in Vessel 1 increases to 3.0 atm. Therefore, correct answer is 3.