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Chapter 11: Problem 62

A fluid expands by \(0.42 \mathrm{~m}^{3}\) at a pressure of \(121 \mathrm{kPa}\). How much work is done by the fluid?

Short Answer

Expert verified
The work done by the fluid is 50,820 J.

Step by step solution

01

Understand the Formula

Work done by or on a fluid during expansion or compression at constant pressure is calculated as the change in volume multiplied by the pressure. This is given by the formula: \[ W = P \Delta V \]where \(W\) is the work done, \(P\) is the pressure, and \(\Delta V\) is the change in volume.
02

Convert Units

The pressure \(P\) given is in kilopascals (\(\mathrm{kPa}\)), and to maintain consistency with units in scientific calculations, we need to convert it to pascals (\(\mathrm{Pa}\)). Since \(1\, \mathrm{kPa} = 1000\, \mathrm{Pa}\), the pressure in pascals is:\[ P = 121 \times 1000 = 121,000\, \mathrm{Pa} \]
03

Calculate the Work Done

Now, substitute the pressure in pascals and the change in volume into the formula for work:\[ W = 121,000\, \mathrm{Pa} \times 0.42\, \mathrm{m}^{3} \]Calculate to find the work done:\[ W = 50,820\, \mathrm{J} \]
04

Express the Result

The work done by the fluid during the expansion is \(50,820\, \mathrm{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure-Volume Work
When a fluid expands, it performs work, which can be understood using the concept of pressure-volume work. Imagine a container with a piston fixed on top. As the fluid in the container heats up, it expands, pushing the piston upward. This movement is the essence of work done by the fluid. The formula used to calculate this work is:
  • W = P \Delta V
  • \(W\) represents the work done,
  • \(P\) is the constant pressure the fluid is under,
  • \(\Delta V\) is the change in volume.
The process of expansion under constant pressure is central to many applications like engines and respiratory systems. As the fluid expands and occupies more space, energy is transferred, allowing mechanical work to occur.
Unit Conversion
In the scientific world, ensuring that all units are consistent is crucial, which is where unit conversion comes in. For instance, pressure is often given in kilopascals (\(\mathrm{kPa}\)), while scientific calculations typically use pascals (\(\mathrm{Pa}\)). Converting units helps maintain consistency and avoid errors. Here鈥檚 how you convert \(\mathrm{kPa}\) to \(\mathrm{Pa}\):
  • Know the conversion factor: \(1\, \mathrm{kPa} = 1000\, \mathrm{Pa}\).
  • Multiply the given value in \(\mathrm{kPa}\) by 1000 to get \(\mathrm{Pa}\).
For example, if pressure is \(121\, \mathrm{kPa}\), converting to pascals gives us \(121,000\, \mathrm{Pa}\). This conversion ensures our calculations for work done are accurate and precise. Unit conversion is essential in engineering, physics, and many other fields where precision is key.
Fluid Dynamics
Fluid dynamics is the study of how fluids (liquids and gases) move. It's a branch of physics that deals with the motion of fluids and the forces involved. Understanding fluid dynamics is essential for calculating work done by fluids during expansion or contraction. Here are key principles in fluid dynamics:
  • **Continuity Equation**: Ensures mass conservation, stating that the mass flow rate is constant in a pipe with variable diameters.
  • **Bernoulli's Equation**: Describes energy conservation in a fluid flow, vital for understanding pressure changes in a flowing fluid.
  • **Laminar vs. Turbulent Flow**: Determines how smoothly or chaotically a fluid moves through a space.
Fluid dynamics is applied in various fields including aviation, weather prediction, and hydraulic systems. By understanding these principles, one can better grasp how fluids perform work when engaging in processes like expansion or compression.

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Most popular questions from this chapter

Predict \& Explain (a) If you rub your hands together, does the entropy of the universe increase, decrease, or stay the same? (b) Choose the best explanation from among the following: A. Rubbing the hands together draws heat from the surroundings and therefore lowers the entropy. B. No mechanical work is done by the rubbing and hence the entropy does not change. C. The heat produced by the rubbing raises the temperature of the hands and the air, which increases the entropy.

Rank Determine the thermal energy changes of systems A through D, described below, and then rank them in order of increasing change in thermal energy. Indicate ties where appropriate. $$ \begin{array}{|c|c|c|c|c|} \hline \text { System } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \boldsymbol{W}(\boldsymbol{J}) & 10 & -10 & 30 & -20 \\ \hline \boldsymbol{Q}(\boldsymbol{J}) & 20 & -20 & -50 & -10 \\ \hline \end{array} $$

Think \& Calculate A gas is at a constant pressure of \(115 \mathrm{kPa}\). (a) If work is done on the gas, does its volume increase or decrease? Explain. (b) Find the change in volume of the gas if \(62 \mathrm{~J}\) of work is done on it.

An 88-kg parachutist descends through a vertical height of \(360 \mathrm{~m}\) with constant speed. Find the increase in entropy produced by the parachutist, assuming that the air temperature is \(21^{\circ} \mathrm{C}\).

Calculate A system's thermal energy decreases by \(20 \mathrm{~J}\) while the system performs \(10 \mathrm{~J}\) of work. How much heat was added to the system?
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