The pressure inside a balloon is proportional to the square of its diameter. It contains 2 kg of water at 150 kPa with 85% quality. The balloon and water are now heated so that a final pressure of 600 kPa is reached. The process undergone by the water is given by p-v equation as:
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 \(P V^{-\frac{2}{3}} = \rm{constant}\).
Important Questions from Thermodynamics System and Processes
At constant pressure, heat energy of 1800 J supplied to 2 kg of fluid, which cause arise in temperature of 15 degree C. The average specific heat of the fluid is
Work done during a process can be determined by \(\smallint {\rm{pdv}}\) when the process is
Pick the correct extensive property of a thermodynamic system from the following options.
- Cold water flowing at 0.1 kg/s is heated from $20^\circ C$ to $70^\circ C$ in a counter-flow type heat exchanger by a hot water stream flowing at 0.1 kg/s and entering at $90^\circ C$. The specific heat of water is $4200 \text{ J/(kg K)} $ and density is $1000 \text{ kg/m}^3 $. If the overall heat transfer coefficient U for the heat exchanger is $2000 \text{ W/(m}^2 \text{ K)} $, the required heat exchange area (in $m^2$) is
A reversible polytrophic process can be described by the equation
