/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Oxygen \(\left(\mathrm{O}_{2}\ri... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 30

Oxygen \(\left(\mathrm{O}_{2}\right)\) gas within a piston-cylinder assembly undergoes an expansion from a volume \(V_{1}=0.01 \mathrm{~m}^{3}\) to a volume \(V_{2}=0.03 \mathrm{~m}^{3}\). The relationship between pressure and volume during the process is \(p=A V^{-1}+B\), where \(\mathrm{A}=\) \(0.06\) bar \(\cdot \mathrm{m}^{3}\) and \(B=3.0\) bar. For the \(\mathrm{O}_{2}\), determine (a) the initial and final pressures, each in bar, and (b) the work, in \(\mathrm{kJ}\).

Short Answer

Expert verified
Initial pressure is 9.0 bar, final pressure is 5.0 bar, and work done is 0.0126 kJ.

Step by step solution

01

- Initial Pressure Calculation

Use the given formula for pressure where initial volume is \( V_1 = 0.01 \, \text{m}^3 \). We need to find the initial pressure \( p_1 \): \[ p = A V^{-1} + B \]Substitute \( V_1 = 0.01 \, \text{m}^3 \, A = 0.06 \, \text{bar} \cdot \text{m}^3 \, \text{and} \, B = 3.0 \, \text{bar} \): \[ p_1 = 0.06 \cdot (0.01)^{-1} + 3.0 \]Solve for \( p_1 \): \[ p_1 = 0.06 \cdot 100 + 3.0 = 6.0 + 3.0 = 9.0 \, \text{bar} \]
02

- Final Pressure Calculation

Use the same formula for pressure where the final volume is \( V_2 = 0.03 \, \text{m}^3 \).We need to find the final pressure \( p_2 \): \[ p = A V^{-1} + B \]Substitute \( V_2 = 0.03 \, \text{m}^3 \, A = 0.06 \, \text{bar} \cdot \text{m}^3 \, \text{and} \, B = 3.0 \, \text{bar} \): \[ p_2 = 0.06 \cdot (0.03)^{-1} + 3.0 \]Solve for \( p_2 \): \[ p_2 = 0.06 \cdot \frac{1}{0.03} + 3.0 = 0.06 \cdot 33.33 + 3.0 \approx 2.0 + 3.0 = 5.0 \, \text{bar} \]
03

- Work Done Calculation

Use the integral definition of work for variable pressure \( W = \int_{V_1}^{V_2} p \ dV \). Using \( p = A V^{-1} + B \), \[ W = \int_{V_1}^{V_2} (A V^{-1} + B) \, dV = A \int_{V_1}^{V_2} V^{-1} \, dV + B \int_{V_1}^{V_2} \, dV \]Evaluate the integrals: \[ W = A [\ln(V)]_{V_1}^{V_2} + B [V]_{V_1}^{V_2} \]Substitute the given values: \[ W = 0.06 \cdot [\ln(0.03) - \ln(0.01)] + 3.0 \cdot (0.03 - 0.01) \]Solve for \( W \): \[ W = 0.06 \cdot \ln(3) + 0.06 = 0.06 \cdot 1.0986 + 0.06 \approx 0.066 + 0.06 = 0.126 \, \text{bar} \cdot \text{m}^3 = 0.0126 \, \text{kJ} \]Note: Convert from bar \( \cdot \, \text{m}^3 \) to kJ by using the fact that \( 1 \, \text{bar} \cdot \text{m}^3 = 0.1 \, \text{kJ} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

pressure-volume relationship
In thermodynamics, the relationship between pressure and volume during gas expansion or compression is critical. This relationship helps us understand how gases behave under different conditions. In our exercise, the formula provided is: \[ p = A V^{-1} + B \] Here, 'p' is the pressure, 'V' is the volume, 'A' is a coefficient related to the specific gas, and 'B' is another constant. This equation shows us that as the volume decreases, the pressure increases if 'A' and 'B' are constants. The term 'A' divided by volume (V^{-1}s) means that pressure is inversely proportional to volume. Meanwhile, 'B' adds a constant pressure throughout the process. Understanding this relationship is crucial for solving problems involving gas behaviors within confined spaces, such as a piston-cylinder assembly.
work done by gas
Work is done by a gas when it expands or compresses within a system like a piston-cylinder assembly. In thermodynamics, work done by a gas during expansion or compression can be calculated using the formula: \[ W = \frac{F}{A} \times \text{displacement} \] But when dealing with variable pressures as volumes change, simply multiplying force by displacement won't work. Here, we need integration using the pressure-volume relationship described earlier. Work done (W) by the gas in this case is found by integrating the pressure over the change in volume: \[ W = \text{Integral from V_1 to V_2} \, p \, dV \] This integral accounts for the variation in pressure as the gas expands or compresses. Calculating work using this integral offers a precise measurement of energy transferred during the process.
integral definition of work
To find the work done by a gas during a process where pressure varies with volume, we use the integral definition of work. The mathematical expression is: \[ W = \text{Integral from V_1 to V_2} \, p(V) \, dV \] Substituting our pressure equation from the exercise, \[ p = A V^{-1} + B \] gives: \[ W = \text{Integral from V_1 to V_2} \, (A V^{-1} + B) \, dV \] Splitting the integral: \[ W = A \text{Integral from V_1 to V_2} \, V^{-1} \, dV + B \text{Integral from V_1 to V_2} \, dV \] Solving these integrals separately, we evaluate: \[ A [\text{ln}(V)]_{V_1}^{V_2} + B [V]_{V_1}^{V_2} \] This approach integrates the pressure over a changing volume, providing a precise calculation of work done by the gas, detailed in the solution step by step.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Vehicle crumple zones are designed to absorb energy during an impact by deforming to reduce transfer of energy to occupants. How much kinetic energy, in Btu, must a crumple zone absorb to fully protect occupants in a 3000 -lb vehicle that suddenly decelerates from \(10 \mathrm{mph}\) to \(0 \mathrm{mph}\) ?

Helium gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of \(1 \mathrm{~kW}\). Heat transfer from the gas to its surroundings occurs at a rate of \(5 t\) watts, where \(t\) is time, in minutes. Plot the change in energy of the helium, in kJ, for \(t \geq 0\) and comment.

A closed system of mass \(10 \mathrm{~kg}\) undergoes a process during which there is energy transfer by work from the system of \(0.147 \mathrm{~kJ}\) per \(\mathrm{kg}\), an elevation decrease of \(50 \mathrm{~m}\), and an increase in velocity from \(15 \mathrm{~m} / \mathrm{s}\) to \(30 \mathrm{~m} / \mathrm{s}\). The specific internal energy decreases by \(5 \mathrm{~kJ} / \mathrm{kg}\) and the acceleration of gravity is constant at \(9.7 \mathrm{~m} / \mathrm{s}^{2}\). Determine the heat transfer for the process, in \(\mathrm{kJ}\).

Gaseous \(\mathrm{CO}_{2}\) is contained in a vertical piston-cylinder assembly by a piston of mass \(50 \mathrm{~kg}\) and having a face area of \(0.01 \mathrm{~m}^{2}\). The mass of the \(\mathrm{CO}_{2}\) is \(4 \mathrm{~g}\). The \(\mathrm{CO}_{2}\) initially occupies a volume of \(0.005 \mathrm{~m}^{3}\) and has a specific internal energy of \(657 \mathrm{~kJ} / \mathrm{kg}\). The atmosphere exerts a pressure of 100 \(\mathrm{kPa}\) on the top of the piston. Heat transfer in the amount of \(1.95 \mathrm{~kJ}\) occurs slowly from the \(\mathrm{CO}_{2}\) to the surroundings, and the volume of the \(\mathrm{CO}_{2}\) decreases to \(0.0025 \mathrm{~m}^{3}\). Friction between the piston and the cylinder wall can be neglected. The local acceleration of gravity is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\). For the \(\mathrm{CO}_{2}\) determine (a) the pressure, in \(\mathrm{kPa}\), and (b) the final specific internal energy, in \(\mathrm{kJ} / \mathrm{kg}\).

An object of mass \(1000 \mathrm{~kg}\), initially having a velocity of \(100 \mathrm{~m} / \mathrm{s}\), decelerates to a final velocity of \(20 \mathrm{~m} / \mathrm{s}\). What is the change in kinetic energy of the object, in \(\mathrm{kJ}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Engineering Physics

Read Explanation

Fluids

Read Explanation

Rotational Dynamics

Read Explanation

Famous Physicists

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.