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This fixed mass system contains air with mass m = 13.947 kgs. The air is ideal gas and you must assume constant specific heats. The properties of air given are Cvo = 0.717 KJ/K/Kg, R = 0.287 KJ/K/Kg, and K-Cpo/Cvo = 1.4. Then a reversible, constant volume process occurs. We are given To = 200K (Surroundings temp) and Entropy change of the surroundings (S2-S1) surrounding = +4.0 KJ/K. Find the 1Q2 in KJoules, (Heat transfer from/to the system, sign + or -), find work done on/by the system (1W₂) and then find the initial and final temperatures of the system in K.

This fixed mass system contains air with mass m = 13.947 kgs. The air is ideal gas and you must assume constant specific heats. The properties of air given are Cvo = 0.717 KJ/K/Kg, R = 0.287 KJ/K/Kg, and K-Cpo/Cvo = 1.4. Then a reversible, constant volume process occurs. We are given To = 200K (Surroundings temp) and Entropy change of the surroundings (S2-S1) surrounding = +4.0 KJ/K. Find the 1Q2 in KJoules, (Heat transfer from/to the system, sign + or -), find work done on/by the system (1W₂) and then find the initial and final temperatures of the system in K.

Introduction to Chemical Engineering Thermodynamics
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
Section: Chapter Questions
Problem 1.1P
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**Thermodynamics Problem: Fixed Mass System Containing Air** This fixed mass system contains air with mass \( m = 13.947 \) kgs. The air is assumed to be an ideal gas, and you must assume constant specific heats. The properties of air given are: - \( C_{v0} = 0.717 \, \text{KJ/K/Kg} \) - \( R = 0.287 \, \text{KJ/K/Kg} \) - \( K = \frac{C_{p0}}{C_{v0}} = 1.4 \) Then a reversible, constant volume process occurs. We are given: - \( T_0 = 200 \, \text{K} \) (Surroundings temperature) - Entropy change of the surroundings (\( S_2 - S_1 \))\(_\text{surrounding} = +4.0 \, \text{KJ/K} \) **Tasks:** 1. **Find the heat transfer** (\( Q_{12} \)) in KJoules (Heat transfer from/to the system, sign + or -). 2. **Determine the work** done on/by the system (\( W_{12} \)). 3. **Find the initial and final temperatures** of the system in K. **Explanation:** - \( m \): Mass of air (13.947 kgs). - \( C_{v0} \): Specific heat at constant volume (0.717 KJ/K/Kg). - \( R \): Specific gas constant (0.287 KJ/K/Kg). - \( K \): Ratio of specific heats (\( \frac{C_p}{C_v} \) = 1.4). - \( T_0 \): Temperature of surroundings (200 K). - \( \Delta S_{surr} \): Entropy change of surroundings (+4.0 KJ/K). ### Analysis Steps: 1. **Heat Transfer (\( Q_{12} \))**: \[ Q_{12} = \Delta S_{surr} \cdot T_0 \] 2. **Work Done (\( W_{12} \))**: Since the process is at constant volume, \( W = 0 \). 3. **Temperatures**: - For initial and final temperatures,
Transcribed Image Text:**Thermodynamics Problem: Fixed Mass System Containing Air** This fixed mass system contains air with mass \( m = 13.947 \) kgs. The air is assumed to be an ideal gas, and you must assume constant specific heats. The properties of air given are: - \( C_{v0} = 0.717 \, \text{KJ/K/Kg} \) - \( R = 0.287 \, \text{KJ/K/Kg} \) - \( K = \frac{C_{p0}}{C_{v0}} = 1.4 \) Then a reversible, constant volume process occurs. We are given: - \( T_0 = 200 \, \text{K} \) (Surroundings temperature) - Entropy change of the surroundings (\( S_2 - S_1 \))\(_\text{surrounding} = +4.0 \, \text{KJ/K} \) **Tasks:** 1. **Find the heat transfer** (\( Q_{12} \)) in KJoules (Heat transfer from/to the system, sign + or -). 2. **Determine the work** done on/by the system (\( W_{12} \)). 3. **Find the initial and final temperatures** of the system in K. **Explanation:** - \( m \): Mass of air (13.947 kgs). - \( C_{v0} \): Specific heat at constant volume (0.717 KJ/K/Kg). - \( R \): Specific gas constant (0.287 KJ/K/Kg). - \( K \): Ratio of specific heats (\( \frac{C_p}{C_v} \) = 1.4). - \( T_0 \): Temperature of surroundings (200 K). - \( \Delta S_{surr} \): Entropy change of surroundings (+4.0 KJ/K). ### Analysis Steps: 1. **Heat Transfer (\( Q_{12} \))**: \[ Q_{12} = \Delta S_{surr} \cdot T_0 \] 2. **Work Done (\( W_{12} \))**: Since the process is at constant volume, \( W = 0 \). 3. **Temperatures**: - For initial and final temperatures,
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