This fixed mass system contains 1.07185388 kgs of water (m) and undergoes a reversible, polytropic process from state 1 to state 2. P₁ is 2000 kPa, P₂ is 1400 kPa, T₁ is 773.15 K and T₂ is 673.15 K. Find the work (₁W₂) in KJoules, the 1Q₂ in KJoules and then apply the 2nd Law to the process to determine if the process is possible or impossible.

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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### Thermodynamic Analysis of a Polytropic Process

#### Problem Statement

This fixed mass system contains 1.07185388 kg of water (m) and undergoes a reversible, polytropic process from state 1 to state 2. The given conditions are:

- Initial pressure \( P_1 = 2000 \) kPa
- Final pressure \( P_2 = 1400 \) kPa
- Initial temperature \( T_1 = 773.15 \) K
- Final temperature \( T_2 = 673.15 \) K

#### Tasks

1. **Work Calculation ( \(W_{1 \rightarrow 2}\) )**
   Determine the work done, \(W_{1 \rightarrow 2}\), in kilojoules.

2. **Heat Calculation ( \(Q_{1 \rightarrow 2}\) )**
   Calculate the heat transferred, \(Q_{1 \rightarrow 2}\), in kilojoules.

3. **Second Law Analysis**
   Apply the Second Law of Thermodynamics to verify if the process is possible or impossible.

---

#### Detailed Solution

1. **Work Calculation**

   The work done in a polytropic process can be calculated using the following formula:
   \[
   W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
   \]
   where \(n\) is the polytropic index. For this specific case, you will need to determine the exact values of \(V_1\) and \(V_2\) and the index \(n\) which fits the given conditions.

2. **Heat Calculation**

   The heat transferred in a polytropic process can be determined by the first law of thermodynamics:
   \[
   Q = \Delta U + W
   \]
   where \( \Delta U \) is the change in internal energy.

3. **Second Law Analysis**

   To verify the process:
   - Calculate the entropy change \( \Delta S \).
   - Check if the entropy change complies with the second law of thermodynamics.

#### Graphical Representation

A detailed graph typically includes:

- An isobaric process curve showing the pressure change from 2000 kPa to 1400 kPa.
- An isothermal curve representing temperature drop from 773.15 K to 673.15 K.
- Volume changes corresponding to the
Transcribed Image Text:### Thermodynamic Analysis of a Polytropic Process #### Problem Statement This fixed mass system contains 1.07185388 kg of water (m) and undergoes a reversible, polytropic process from state 1 to state 2. The given conditions are: - Initial pressure \( P_1 = 2000 \) kPa - Final pressure \( P_2 = 1400 \) kPa - Initial temperature \( T_1 = 773.15 \) K - Final temperature \( T_2 = 673.15 \) K #### Tasks 1. **Work Calculation ( \(W_{1 \rightarrow 2}\) )** Determine the work done, \(W_{1 \rightarrow 2}\), in kilojoules. 2. **Heat Calculation ( \(Q_{1 \rightarrow 2}\) )** Calculate the heat transferred, \(Q_{1 \rightarrow 2}\), in kilojoules. 3. **Second Law Analysis** Apply the Second Law of Thermodynamics to verify if the process is possible or impossible. --- #### Detailed Solution 1. **Work Calculation** The work done in a polytropic process can be calculated using the following formula: \[ W = \frac{P_2 V_2 - P_1 V_1}{1 - n} \] where \(n\) is the polytropic index. For this specific case, you will need to determine the exact values of \(V_1\) and \(V_2\) and the index \(n\) which fits the given conditions. 2. **Heat Calculation** The heat transferred in a polytropic process can be determined by the first law of thermodynamics: \[ Q = \Delta U + W \] where \( \Delta U \) is the change in internal energy. 3. **Second Law Analysis** To verify the process: - Calculate the entropy change \( \Delta S \). - Check if the entropy change complies with the second law of thermodynamics. #### Graphical Representation A detailed graph typically includes: - An isobaric process curve showing the pressure change from 2000 kPa to 1400 kPa. - An isothermal curve representing temperature drop from 773.15 K to 673.15 K. - Volume changes corresponding to the
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