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:

When gas expands into an evacuated vessel, total final volume is the sum of both vessel volumes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves the expansion of an ideal gas into a vacuum. Since the gas is ideal and the process happens in a way that suggests no temperature change (free expansion), we can use Boyle's Law. Boyle's Law states that for a fixed amount of gas at constant temperature, the product of pressure and volume is constant.
Step 2: Key Formula or Approach:
Boyle's Law: $P_1V_1 = P_2V_2$. - $P_1$ and $V_1$ are the initial pressure and volume. - $P_2$ and $V_2$ are the final pressure and volume. We need to correctly identify the initial and final states.
Step 3: Detailed Explanation:
Initial State: - The gas is contained in a 10-liter vessel. So, the initial volume is $V_1 = 10$ liters. - The initial pressure is $P_1 = 760$ mm of Hg. Final State: - The gas expands to fill both the original 10-liter vessel and the connected 9-liter vessel. - So, the final total volume is $V_2 = 10 \text{ liters} + 9 \text{ liters} = 19$ liters. - We need to find the final pressure, $P_2$. Apply Boyle's Law: \[ P_1V_1 = P_2V_2 \] \[ (760 \text{ mm of Hg})(10 \text{ L}) = P_2 (19 \text{ L}) \] 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 of the gas after expansion is 400 mm of Hg. Therefore, option (A) is correct.