Textbook Question
0.10 mol of argon gas is admitted to an evacuated 50 cm3 container at 20°C. The gas then undergoes an isochoric heating to a temperature of 300°C. Show the process on a pV diagram. Include a proper scale on both axes.
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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 32a
0.10 mol of argon gas is admitted to an evacuated 50 cm3 container at 20°C. The gas then undergoes an isochoric heating to a temperature of 300°C. What is the final pressure of the gas?
Verified step by step guidance1
Step 1: Start by identifying the given values and converting them into appropriate SI units. The initial number of moles of argon gas is \( n = 0.10 \; \text{mol} \). The volume of the container is \( V = 50 \; \text{cm}^3 = 50 \times 10^{-6} \; \text{m}^3 \). The initial temperature is \( T_1 = 20^\circ \text{C} = 20 + 273.15 = 293.15 \; \text{K} \), and the final temperature is \( T_2 = 300^\circ \text{C} = 300 + 273.15 = 573.15 \; \text{K} \).
Step 2: Use the ideal gas law to find the initial pressure of the gas. The ideal gas law is \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant \( R = 8.314 \; \text{J/(mol·K)} \), and \( T \) is the temperature. Rearrange the equation to solve for \( P_1 \): \( P_1 = \frac{nRT_1}{V} \). Substitute the known values to calculate \( P_1 \).
Step 3: Recognize that the process is isochoric, meaning the volume remains constant. For an isochoric process, the pressure and temperature are related by the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Rearrange this equation to solve for the final pressure \( P_2 \): \( P_2 = P_1 \cdot \frac{T_2}{T_1} \).
Step 4: Substitute the value of \( P_1 \) from Step 2 and the temperatures \( T_1 \) and \( T_2 \) into the equation from Step 3 to calculate \( P_2 \). Ensure that all units are consistent (e.g., temperature in Kelvin).
Step 5: The final pressure \( P_2 \) is the result of the calculation in Step 4. Express the answer in Pascals (Pa) or another appropriate unit of pressure, depending on the context of the problem.
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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. In this scenario, it is essential for calculating the final pressure of the argon gas after heating, as it allows us to understand how changes in temperature affect pressure when volume is constant.
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Ideal Gases and the Ideal Gas Law
Isochoric Process
An isochoric process is one in which the volume of the gas remains constant. In this case, the argon gas is heated in a fixed volume container, meaning any increase in temperature will directly affect the pressure according to the Ideal Gas Law, without any change in volume.
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Temperature Conversion
Temperature must be expressed in Kelvin when using the Ideal Gas Law. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. This conversion is crucial for accurately calculating the final pressure after the gas is heated from 20°C to 300°C.
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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
A gas with initial state variables p1, V1, and T1 expands isothermally until V2 = 2V1. What are p2?
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Textbook Question
A gas with an initial temperature of 900°C undergoes the process shown in FIGURE EX18.35. What type of process is this?
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Textbook Question
A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 20 kg piston. The piston is 84 cm above the bottom when the gas temperature is 303°C. The air above the piston is at 1.00 atm pressure. What will the height of the piston be if the temperature is lowered to 15°C?
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Textbook Question
A gas with initial state variables p1, V1, and T1 expands isothermally until V2 = 2V1. What are T1?
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