Question

A vessel containing 10 liters of an ideal gas at a pressure of 760 mm of Hg is connected to an evacuated 9 liter vessel. The resultant pressure is

• A. 400 mm of Hg
• B. 1440 mm of Hg
• C. 40 mm of Hg
• D. 760 mm of Hg

Hint:

In problems where a gas expands into an evacuated container, the final volume is the total volume of all connected containers. Assuming the temperature is constant (which is usually the case for ideal gas free expansion), Boyle's law is the direct way to find the final pressure.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An ideal gas initially in one container is allowed to expand into an empty (evacuated) container. We need to find the final pressure of the gas.

Step 2: Key Formula or Approach:
This process is a free expansion of an ideal gas. Since no heat is exchanged and no work is done, the temperature of the ideal gas remains constant. Therefore, we can apply Boyle's Law.
Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional:
\[ P_1 V_1 = P_2 V_2 \]

Step 3: Detailed Explanation:
Let's define the initial and final states of the gas.

Initial State:
- Initial pressure, \(P_1 = 760\) mm of Hg.
- Initial volume, \(V_1 = 10\) liters.

Final State:
- The gas expands to occupy both vessels. So, the final volume is the sum of the volumes of the two vessels.
- Final volume, \(V_2 = 10 \text{ liters} + 9 \text{ liters} = 19\) liters.
- Final pressure, \(P_2\), is what we need to find.
Apply Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
\[ (760 \text{ mm of Hg}) \times (10 \text{ L}) = P_2 \times (19 \text{ L}) \]
Solve for \(P_2\):
\[ P_2 = \frac{760 \times 10}{19} \]
\[ P_2 = \frac{7600}{19} \]
\[ P_2 = 400 \text{ mm of Hg} \]

Step 4: Final Answer:
The resultant pressure is 400 mm of Hg.