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Chapter 2: Problem 18

For a process taking place in a closed system containing gas, the volume and pressure relationship is \(p V^{1.4}=\) constant. The process starts with initial conditions, \(p_{1}=1.5\) bar, \(V_{1}=0.03 \mathrm{~m}^{3}\) and ends with final volume, \(V_{2}=0.05 \mathrm{~m}^{3}\). Determine the work done by the gas.

Short Answer

Expert verified
The work done by the gas is approximately 2370 Joules.

Step by step solution

01

- Understand the given relationship

The problem states that the relationship between pressure (p) and volume (V) for the process is given by: \[ p V^{1.4} = \text{constant} \]
02

- Find the constant

We start by using the initial conditions to find the constant. Given \(p_{1} = 1.5\) bar and \(V_{1} = 0.03 \, m^3\), we plug these into the equation: \[ 1.5 \times (0.03)^{1.4} = \text{constant} \] Solving this, we have: \[ \text{constant} = 1.5 \times (0.03)^{1.4} \approx 0.0249526 \]
03

- Express final pressure in terms of volume

Using the constant we found, express the final pressure (\(p_{2}\)) in terms of the final volume (\(V_{2}\)). The relationship gives us: \[ p_{2} \times (V_{2})^{1.4} = \text{constant} \] \[ p_{2} \times (0.05)^{1.4} = 0.0249526 \] Solving for \(p_{2}\): \[ p_{2} = \frac{0.0249526}{(0.05)^{1.4}} \approx 0.7106 \, \text{bar} \]
04

- Calculate the work done (W)

The work done by the gas during an expansion or compression process described by \(pV^n = \text{constant}\) can be calculated using the formula: \[ W = \frac{p_{1} V_{1} - p_{2} V_{2}}{1.4 - 1} \] Substituting the values, we have: \[ W = \frac{1.5 \times 0.03 - 0.7106 \times 0.05}{1.4 - 1} \] Simplifying: \[ W = \frac{0.045 - 0.03553}{0.4} \approx 0.0237 \, \text{bar} \, m^3 \]
05

- Convert units (optional)

If necessary, convert the work from bar \(\cdot m^3\) to more common units such as Joules. Recall that 1 bar \(\cdot m^3 = 10^5\) Joules: \[ W = 0.0237 \times 10^5 = 2370 \, \text{Joules} \]

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Key Concepts

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

Closed System Thermodynamics
In thermodynamics, a closed system refers to a system where no matter enters or leaves the system's boundary. However, energy can still be exchanged with the surroundings, typically in the form of heat or work. Think of it as a sealed container where the gas inside can be heated or compressed, but no gas gets in or out.

When studying closed systems, it's essential to understand that only energy changes occur, not mass changes. This is crucial for analyzing processes like the one in our exercise, where the volume and pressure of the gas change, but the quantity of gas remains constant. Understanding this helps us apply the correct formulas and concepts, like the one used to establish the pressure-volume relationship in the next section.
Pressure-Volume Relationship
The relationship between pressure and volume in a closed system often follows a polytropic process, expressed as: in an equation like this: This is a common relationship in closed system thermodynamics, often encountered in processes where the specific heat ratio (gamma) is applied. By knowing the initial conditions of a system, we can determine the constant value using the formula: where p1 is the initial pressure, and V1 is the starting volume. In the given exercise, these relationships help us understand how the pressure changes as the volume changes while keeping the product of pressure and volume raised to a power constant. This is foundational for calculating work done by or on the gas during the process.
Adiabatic Process
An adiabatic process is one where no heat is exchanged with the surroundings. This means all the work done by the system results solely from changes within the system.

In our exercise, the process follows an adiabatic pathway where the relationship between pressure and volume is given by: where represents the adiabatic index, often 1.4 for air. In the adiabatic process, we can use the initial and final states to calculate changes in pressure and work done. Knowing the volume and pressure at both states, we apply the formula: Using this equation, we can find out how much work the gas performs during expansion or compression. This step is vital because it allows us to calculate the energy transferred as work without needing to account for heat exchange, making the adiabatic process simpler to analyze for many gas dynamics problems.

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Most popular questions from this chapter

\(.\) In a rigid insulated container of volume \(0.8 \mathrm{~m}^{3}, 2.5 \mathrm{~kg}\) of air is filled. A paddle wheel is fitted in the container and it transfers energy to the contained air at a constant rate of \(12 \mathrm{~W}\) for a period of \(1 \mathrm{~h}\). There is no change in the potential or kinetic energy of the system. Determine the energy transfer by the wheel to the air per \(\mathrm{kg}\) of air.

\(.\) An electric generator supplies electricity to a storage battery at a rate of \(15 \mathrm{~kW}\) for a period of 12 hours. During this \(12-\mathrm{h}\) period there also is heat transfer from the battery to the surroundings at a rate of \(1.5 \mathrm{~kW}\). Then, during the next 12 -h period the battery discharges electricity to an external load at a rate of \(5 \mathrm{~kW}\) while heat transfer from the battery to the surroundings occurs at a rate of \(0.5 \mathrm{~kW}\). (a) For the first 12 -h period, determine, in \(\mathrm{kW}\), the time rate of change of energy stored within the battery. (b) For the second 12 -h period, determine, in \(\mathrm{kW}\), the time rate of change of energy stored in the battery. (c) For the full 24-h period, determine, in \(\mathrm{kJ}\), the overall change in energy stored in the battery.

\(\mathrm{~A}\) gas is compressed from \(V_{1}=0.3 \mathrm{~m}^{3}, p_{1}=1\) bar to \(V_{2}=0.1 \mathrm{~m}^{3}, p_{2}=3\) bar. Pressure and volume are related linearly during the process. For the gas, find the work, in \(\mathrm{kJ}\).

\(.\mathrm{~A}\) gas expands in a piston-cylinder assembly from \(p_{1}=8\) bar, \(V_{1}=0.02 \mathrm{~m}^{3}\) to \(p_{2}=2\) bar in a process during which the relation between pressure and volume is \(p V^{1.2}=\) constant. The mass of the gas is \(0.25 \mathrm{~kg}\). If the specific internal energy of the gas decreases by \(55 \mathrm{~kJ} / \mathrm{kg}\) during the process, determine the heat transfer, in \(\mathrm{kJ}\). Kinetic and potential energy effects are negligible.

A chip with surface area of upper face of \(25 \mathrm{~mm}^{2}\) is exposed to a convection environment that has a temperature equal to \(25^{\circ} \mathrm{C}\). An electrical input of \(0.325 \mathrm{~W}\) is provided to the chip and under steady state heat is transferred from the upper face of the chip into the surrounding. The convective heat transfer coefficient between the chip and the surrounding is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the surface temperature of the chip.
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