Understanding the Balloon Process Relationship

This question asks us to determine the specific thermodynamic process equation relating pressure ($P$) and volume ($V$) for water inside a balloon. We are given that the pressure inside the balloon is proportional to the square of its diameter ($P \propto D^2$). We also know the initial and final pressures and the initial state of the water.

Relating Pressure, Diameter, and Volume

The key is to link the given pressure-diameter relationship to a pressure-volume relationship. We know the volume of a sphere (the balloon) is related to its diameter ($D$) by the formula:

\(V = \frac{\pi}{6} D^3\)

From this, we can express the diameter in terms of volume:

• Rearrange the volume formula to solve for $D^3$: \(D^3 = \frac{6V}{\pi}\)
• Take the cube root of both sides to find $D$: \(D = \left( \frac{6V}{\pi} \right)^{\frac{1}{3}}\)

Deriving the P-V Equation

The problem states that pressure is proportional to the square of the diameter:

\(P \propto D^2\)

Now, substitute the expression for $D$ in terms of $V$ into this proportionality:

\(P \propto \left[ \left( \frac{6V}{\pi} \right)^{\frac{1}{3}} \right]^2\)

Simplify the exponent:

\(P \propto \left( \frac{6V}{\pi} \right)^{\frac{2}{3}}\)

Separate the terms involving $V$ and the constants:

\(P \propto V^{\frac{2}{3}} \left( \frac{6}{\pi} \right)^{\frac{2}{3}}\)

Since \((\frac{6}{\pi})^{\frac{2}{3}}\) is a constant value, we can absorb it into the proportionality constant. This gives us:

\(P \propto V^{\frac{2}{3}}\)

This proportionality can be written in the form of an equation where $PV^k$ equals a constant. To match the standard form \(PV^k = \rm{constant}\), we rewrite the relationship:

\(P V^{-\frac{2}{3}} = \rm{constant}\)

Matching with the Options

Comparing our derived equation, \(P V^{-\frac{2}{3}} = \rm{constant}\), with the given options:

• Option 1: \(P{V^{ - \frac{2}{3}}} = \rm{constant}\) - This matches our result.
• Option 2: \(P{V^{ \frac{2}{3}}} = \rm{constant}\)
• Option 3: \(P{V^{ \frac{2}{5}}} = \rm{constant}\)
• Option 4: \(P{V^{ - \frac{2}{5}}} = \rm{constant}\)

The derived relationship directly corresponds to Option 1.

Final Answer Confirmation

The process followed by the water, based on the relationship between the balloon's pressure and its diameter (which dictates volume), is represented by the equation \(P V^{-\frac{2}{3}} = \rm{constant}\).