3. For a mole of a perfect monoatomic gas, the internal energy, U, can be expressed as a function of the pressure and volume as U = U(P,V) = PV a) Calculate explicitly the line integral of dU along the closed path ABCD shown as a black trace in the P – V graph below. 3P1 2P1 S. h P1 A V V1 2V1 3V1 4V1 5V1 6V1 Problem 3a-3c b) Compute the following line integrals between the points B and C in the figure above: 1. S, PdV, along the path, h, described by P = 3P,Vi/(V – 3V1), shown in red in the figure above. 2. S, PdV, along the path, s, shown in black in the figure above. Use these results to demonstrate that SW = -PdV is not an exact differential.
3. For a mole of a perfect monoatomic gas, the internal energy, U, can be expressed as a function of the pressure and volume as U = U(P,V) = PV a) Calculate explicitly the line integral of dU along the closed path ABCD shown as a black trace in the P – V graph below. 3P1 2P1 S. h P1 A V V1 2V1 3V1 4V1 5V1 6V1 Problem 3a-3c b) Compute the following line integrals between the points B and C in the figure above: 1. S, PdV, along the path, h, described by P = 3P,Vi/(V – 3V1), shown in red in the figure above. 2. S, PdV, along the path, s, shown in black in the figure above. Use these results to demonstrate that SW = -PdV is not an exact differential.
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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![The problem presented is related to thermodynamics, specifically focusing on a monoatomic gas. It involves calculating the change in internal energy and work done during a cyclic process represented in a pressure-volume (P-V) diagram.
### Problem Statement:
**3.** For a mole of a perfect monoatomic gas, the internal energy, \( U \), can be expressed as a function of the pressure and volume as
\[ U = U(P,V) = \frac{3}{2} PV \]
**a)** Calculate explicitly the line integral of \( dU \) along the closed path \( ABCD \) shown as a black trace in the \( P-V \) graph below.
**Graph Explanation:**
- **Axes:**
- **Horizontal axis (V):** Volume. Marked as \( V_1, 2V_1, 3V_1, 4V_1, 5V_1, 6V_1 \).
- **Vertical axis (P):** Pressure. Levels \( P_1, 2P_1, 3P_1 \) are marked.
- **Path \( ABCD \):**
- **A to B:** Horizontal line at pressure level \( P_1 \), from \( V_1 \) to \( 6V_1 \).
- **B to C:** Curved path (h), from \( P_1, 6V_1 \) to \( 3P_1, V_1 \).
- **C to D:** Horizontal line at \( 3P_1 \), from \( V_1 \) to some unspecified volume between \( V_1 \) and \( 6V_1 \).
- **D to A:** Vertical line returning to \( P_1, V_1 \).
**b)** Compute the following line integrals between the points \( B \) and \( C \) in the figure above:
1. \( \int_h PdV \), along the path, \( h \), described by \( P = \frac{3P_1 V_1}{(V - 3V_1)} \), shown in red in the figure above.
2. \( \int_s PdV \), along the path, \( s \), shown in black in the figure above.
Use these results to demonstrate that \( \delta W = -PdV \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f6d2f58-ebc5-487b-abb4-91cd16bfc1c1%2Ff3490d05-cf05-4ff4-9cb6-233e72b724ce%2Fgho2w9n_processed.png&w=3840&q=75)
Transcribed Image Text:The problem presented is related to thermodynamics, specifically focusing on a monoatomic gas. It involves calculating the change in internal energy and work done during a cyclic process represented in a pressure-volume (P-V) diagram.
### Problem Statement:
**3.** For a mole of a perfect monoatomic gas, the internal energy, \( U \), can be expressed as a function of the pressure and volume as
\[ U = U(P,V) = \frac{3}{2} PV \]
**a)** Calculate explicitly the line integral of \( dU \) along the closed path \( ABCD \) shown as a black trace in the \( P-V \) graph below.
**Graph Explanation:**
- **Axes:**
- **Horizontal axis (V):** Volume. Marked as \( V_1, 2V_1, 3V_1, 4V_1, 5V_1, 6V_1 \).
- **Vertical axis (P):** Pressure. Levels \( P_1, 2P_1, 3P_1 \) are marked.
- **Path \( ABCD \):**
- **A to B:** Horizontal line at pressure level \( P_1 \), from \( V_1 \) to \( 6V_1 \).
- **B to C:** Curved path (h), from \( P_1, 6V_1 \) to \( 3P_1, V_1 \).
- **C to D:** Horizontal line at \( 3P_1 \), from \( V_1 \) to some unspecified volume between \( V_1 \) and \( 6V_1 \).
- **D to A:** Vertical line returning to \( P_1, V_1 \).
**b)** Compute the following line integrals between the points \( B \) and \( C \) in the figure above:
1. \( \int_h PdV \), along the path, \( h \), described by \( P = \frac{3P_1 V_1}{(V - 3V_1)} \), shown in red in the figure above.
2. \( \int_s PdV \), along the path, \( s \), shown in black in the figure above.
Use these results to demonstrate that \( \delta W = -PdV \
Transcribed Image Text:### Problem Statement
**c)** Use \( dU = \delta Q - PdV \) to compute the heat, \( Q_{D \rightarrow A} \), associated with the process \( D \rightarrow A \) along the black vertical line shown in the figure above.
**d)** Calculate the line integral of \( dU \) along the path shown below.
### Diagram Analysis
The diagram is a pressure-volume (P-V) graph with marked axes labeled \( P \) (pressure) and \( V \) (volume). On the x-axis, the volume is marked at intervals of \( 0 \), \( V1 \), and \( 2V1 \). On the y-axis, the pressure is marked as \( P1 \), \( 2P1 \), and \( 3P1 \).
- **Path Description:**
- A spiral path is shown in red, indicating a process involving varying pressure and volume, ultimately moving inward and ending at a specific point on the diagram near the center.
This visualization is essential for calculating changes in internal energy using thermodynamic equations.
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