Solution for Question 11
Problem Breakdown:
- Initial State: The bottle contains air at 2.0 atm. The balance is in equilibrium.
- Process: The bottle is opened. Air flows into the attached plastic bag until pressure equalizes to 1.0 atm.
- Physics Principle: We need to calculate the change in the effective weight on the left pan.
Calculations:
Let $V_0 = 2.0 \text{ L}$ be the volume of the bottle.
Initial Pressure $P_i = 2 \text{ atm}$.
Final Pressure $P_f = 1 \text{ atm}$.
When the valve opens, the total volume of the gas becomes $V_{total} = V_{bottle} + V_{bag}$. Using Boyle’s Law ($P_1V_1 = P_2V_2$):
$$2 \text{ atm} \times 2 \text{ L} = 1 \text{ atm} \times (2 \text{ L} + V_{bag})$$
$$4 = 2 + V_{bag} \Rightarrow V_{bag} = 2 \text{ L}$$
This means 2 Liters of air (at 1 atm density) has moved from the rigid bottle into the flexible bag.
Mass Analysis:
The mass of this transferred air is:
$$m = \text{Volume} \times \text{Density} = 2.0 \times 10^{-3} \text{ m}^3 \times 1.3 \text{ kg/m}^3$$
$$m = 2.6 \times 10^{-3} \text{ kg} = 2.6 \text{ g}$$
Buoyancy Effect:
- The bottle lost 2.6 g of air. This reduces the weight pressing down on the pan.
- The bag gained 2.6 g of air. However, the bag is immersed in the surrounding atmosphere. By Archimedes’ principle, the buoyant force on the inflated bag equals the weight of the displaced air. Since the air inside the bag is at the same density as the outside air (1 atm), the weight of the air in the bag is exactly cancelled by the buoyant force ($W_{air} – F_{buoyancy} = 0$).
Result: The bag contributes 0 effective weight to the scale. The bottle, however, has become lighter by 2.6 g. Therefore, the left pan moves up. To re-establish balance, we must add 2.6 g to the left pan (or remove from the right).