Textbook Question
0.0050 mol of gas undergoes the process 1→2→3 shown in FIGURE EX18.37. What are temperature T1?
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Ch 18: A Macroscopic Description of Matter
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 18, Problem 37c
0.0050 mol of gas undergoes the process 1→2→3 shown in FIGURE EX18.37. What is the volume V3?
Verified step by step guidance1
Step 1: Identify the given data for each state in the process. From the graph, we know: At point A: Pₐ = 6.00 atm, Vₐ = 3.00 L, Tₐ = 320 K. At point B: Pᵦ = 6.00 atm, Tᵦ = 940 K. At point C: P꜀ = 3.40 atm, T꜀ = 884 K.
Step 2: Use the ideal gas law, PV = nRT, to calculate the volume at point B (Vᵦ). Rearrange the formula to solve for Vᵦ: Vᵦ = (nRTᵦ) / Pᵦ. Substitute the values: n = 0.0050 mol, R = 0.0821 L·atm/(mol·K), Tᵦ = 940 K, and Pᵦ = 6.00 atm.
Step 3: Use the ideal gas law again to calculate the volume at point C (V꜀). Rearrange the formula to solve for V꜀: V꜀ = (nRT꜀) / P꜀. Substitute the values: n = 0.0050 mol, R = 0.0821 L·atm/(mol·K), T꜀ = 884 K, and P꜀ = 3.40 atm.
Step 4: Perform unit consistency checks for each calculation. Ensure that the gas constant R is in the correct units (L·atm/(mol·K)) and that pressure, temperature, and volume are consistent with these units.
Step 5: Interpret the results. The calculated volumes Vᵦ and V꜀ represent the spatial dimensions of the gas at points B and C, respectively, under the given conditions of pressure and temperature.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Ideal Gas Law
The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is fundamental for understanding gas behavior under varying conditions. In this context, it can be used to calculate unknown variables such as volume or pressure at different states of the gas.
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Ideal Gases and the Ideal Gas Law
Thermodynamic Processes
Thermodynamic processes describe how a gas changes state, including isothermal, isobaric, and adiabatic processes. The transitions between points A, B, and C in the provided graph represent different thermodynamic processes, which can affect the gas's pressure and volume. Understanding these processes is crucial for analyzing the gas's behavior during the transitions.
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Pressure-Volume Relationship
The pressure-volume relationship, often illustrated by Boyle's Law, states that for a given amount of gas at constant temperature, the pressure and volume are inversely related. This concept is essential for interpreting the graph provided, as it shows how pressure changes with volume during the gas's transitions, allowing for the calculation of unknown volumes or pressures.
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Related Practice
Textbook Question
A gas with an initial temperature of 900°C undergoes the process shown in FIGURE EX18.35. How many moles of gas are there?
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Textbook Question
0.0050 mol of gas undergoes the process 1→2→3 shown in FIGURE EX18.37. What are pressure p2?
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Textbook Question
A sealed container holds 3.2 g of oxygen at 1 atm pressure and 20°C. The gas first undergoes an isobaric process that doubles the absolute temperature, then an isothermal process that halves the pressure. What is the final volume of the gas in L?
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