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(15.5) A 1.0-L volume of air initially at 4.5 atm of pressure is allowed to expand isothermally until the pressure is 2.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

(15.5) A 1.0-L volume of air initially at 4.5 atm of pressure is allowed to expand isothermally until the pressure is 2.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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### Thermodynamics Exercises #### Problem 15.5 A 1.0-L volume of air initially at 4.5 atm of pressure is allowed to expand isothermally until the pressure is 2.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes. **Explanation of the Diagram:** 1. **Isothermal Expansion**: Initially at point A (1.0 L, 4.5 atm), the gas expands isothermally to point B (a larger volume, 2.0 atm). The path is a hyperbolic curve. 2. **Isobaric Compression**: From point B, the gas is compressed at constant pressure (2.0 atm) back to its initial volume (point C). 3. **Isochoric Heating**: Finally, the gas is brought back to the original pressure (4.5 atm) at constant volume, returning to point A. #### Problem 15.6 The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process, 465 kJ of heat left the gas. (a) How much work was done during this process? (b) What was the change in internal energy of the gas during this process? **Explanation:** Since the gas is kept in a container with rigid walls, the volume does not change: - **Work Done (W)**: With no volume change, the work done is zero (\( W = 0 \)). - **Change in Internal Energy (ΔU)**: According to the first law of thermodynamics (\( ΔU = Q - W \)), and since 465 kJ of heat left the system (\( Q = -465 \) kJ), the change in internal energy \( ΔU = -465 \) kJ. #### Problem 15.7 In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2630 J of work is done on the gas. (a) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall? **Explanation:** - **Heat Flow (Q)**: In an adiabatic
Transcribed Image Text:### Thermodynamics Exercises #### Problem 15.5 A 1.0-L volume of air initially at 4.5 atm of pressure is allowed to expand isothermally until the pressure is 2.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes. **Explanation of the Diagram:** 1. **Isothermal Expansion**: Initially at point A (1.0 L, 4.5 atm), the gas expands isothermally to point B (a larger volume, 2.0 atm). The path is a hyperbolic curve. 2. **Isobaric Compression**: From point B, the gas is compressed at constant pressure (2.0 atm) back to its initial volume (point C). 3. **Isochoric Heating**: Finally, the gas is brought back to the original pressure (4.5 atm) at constant volume, returning to point A. #### Problem 15.6 The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process, 465 kJ of heat left the gas. (a) How much work was done during this process? (b) What was the change in internal energy of the gas during this process? **Explanation:** Since the gas is kept in a container with rigid walls, the volume does not change: - **Work Done (W)**: With no volume change, the work done is zero (\( W = 0 \)). - **Change in Internal Energy (ΔU)**: According to the first law of thermodynamics (\( ΔU = Q - W \)), and since 465 kJ of heat left the system (\( Q = -465 \) kJ), the change in internal energy \( ΔU = -465 \) kJ. #### Problem 15.7 In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2630 J of work is done on the gas. (a) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall? **Explanation:** - **Heat Flow (Q)**: In an adiabatic
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