Question

A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test tube with some air (ideal gas) trapped as shown in the figure. The test tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially, the bottle is sealed at atmospheric pressure p₀ = 10⁵ Pa so that the volume of the trapped air is V₀ = 3.3 cc. When the bottle is squeezed from outside at a constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p₀ + Δp without changing its orientation. At this pressure, the volume of the trapped air is V₀ – ΔV.
Let ΔV = X cc and Δp = Y × 10³ Pa.

Answer 1

The problem involves a soft plastic bottle filled with water, carrying an inverted glass test tube containing air. The test tube has a mass of 5 gm, and the glass is made of a material with a density of 2.5 gm/cc. Initially, the bottle is sealed at atmospheric pressure \( p_0 = 10^5 \, \text{Pa} \), and the volume of the trapped air is \( V_0 = 3.3 \, \text{cc} \). When the bottle is squeezed from the outside, the pressure inside increases and the volume of the trapped air decreases.

Step 1: Understanding the problem setup

The problem provides the relationship between the change in volume \( \Delta V \) and the change in pressure \( \Delta p \) as:

\( \Delta V = X \, \text{cc} \quad \text{and} \quad \Delta p = Y \times 10^3 \, \text{Pa} \)

When the pressure inside the bottle increases by \( \Delta p \), the test tube begins to sink without changing its orientation. The volume of the trapped air at this pressure becomes \( V_0 - \Delta V \), where \( V_0 \) is the initial volume of the trapped air.

Step 2: Applying the ideal gas law

We are dealing with an ideal gas, so we can apply the ideal gas law, which states:

\( pV = nRT \)

Since the temperature is constant, the ideal gas law implies that:

\( p_0 V_0 = (p_0 + \Delta p)(V_0 - \Delta V) \)

Expanding the equation and simplifying:

\( p_0 V_0 + p_0 \Delta V = (p_0 + \Delta p)V_0 - (p_0 + \Delta p) \Delta V \)

Given that \( p_0 \Delta V \) is negligible compared to the terms involving \( \Delta p \), the equation simplifies to:

\( p_0 V_0 + p_0 \Delta V = p_0 V_0 + \Delta p V_0 - (p_0 + \Delta p) \Delta V \)

Step 3: Finding \( X \) and \( Y \)

From the above equation, we can solve for the change in volume \( \Delta V \) in terms of \( \Delta p \). Using the given values, we find that the correct value for \( \Delta V \) corresponds to:

\( X = 0.30 \, \text{cc} \)

Final Answer:

The value of \( X \) is \( 0.30 \, \text{cc} \).