7. Air obeying ideal gas law undergoes change of state from P₁= 1.0135 x105 N/m² to 10.135 x 10³ N/m² and V₁ = 10.16m3 to V2 = 1.0165m3 by following the reversible processes. i. Compression at constant temperature ii. Adiabatic compression followed by isobaric cooling. iii. Isochoric heating followed by isobaric cooling. Calculate AU, W. Q, AH for each of the process.

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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**Problem 7: Thermodynamic Process Analysis**

Air, which obeys the ideal gas law, undergoes a change of state from the initial pressure \( P_1 = 1.0135 \times 10^5 \, \text{N/m}^2 \) to the final pressure \( P_2 = 10.135 \times 10^5 \, \text{N/m}^2 \). The initial volume is \( V_1 = 10.16 \, \text{m}^3 \) and the final volume is \( V_2 = 1.0165 \, \text{m}^3 \). The process follows these reversible paths:

i. **Compression at Constant Temperature:**
   - The gas is compressed while maintaining the same temperature throughout the process.

ii. **Adiabatic Compression Followed by Isobaric Cooling:**
   - The gas is compressed without any heat exchange with the surroundings (adiabatic process) and then cooled at constant pressure.

iii. **Isochoric Heating Followed by Isobaric Cooling:**
   - The gas is heated at a constant volume initially and then cooled at constant pressure.

**Tasks:**

Calculate the following thermodynamic quantities for each process:
- \( \Delta U \) (Change in Internal Energy)
- \( W \) (Work Done)
- \( Q \) (Heat Transfer)
- \( \Delta H \) (Change in Enthalpy)

**Note:** There is no graphical or diagrammatic content in the text to describe.
Transcribed Image Text:**Problem 7: Thermodynamic Process Analysis** Air, which obeys the ideal gas law, undergoes a change of state from the initial pressure \( P_1 = 1.0135 \times 10^5 \, \text{N/m}^2 \) to the final pressure \( P_2 = 10.135 \times 10^5 \, \text{N/m}^2 \). The initial volume is \( V_1 = 10.16 \, \text{m}^3 \) and the final volume is \( V_2 = 1.0165 \, \text{m}^3 \). The process follows these reversible paths: i. **Compression at Constant Temperature:** - The gas is compressed while maintaining the same temperature throughout the process. ii. **Adiabatic Compression Followed by Isobaric Cooling:** - The gas is compressed without any heat exchange with the surroundings (adiabatic process) and then cooled at constant pressure. iii. **Isochoric Heating Followed by Isobaric Cooling:** - The gas is heated at a constant volume initially and then cooled at constant pressure. **Tasks:** Calculate the following thermodynamic quantities for each process: - \( \Delta U \) (Change in Internal Energy) - \( W \) (Work Done) - \( Q \) (Heat Transfer) - \( \Delta H \) (Change in Enthalpy) **Note:** There is no graphical or diagrammatic content in the text to describe.
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