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For the process shown in the pV diagram below, if p₁ = 1 × 105 Pa ,P2 = 2 × 105 Pa ,p3 = 3 × 105 Pa, V₁ = 1 m³ and V/₂ = 4 m³, the total work in going from a to d along the path shown is p (Pax 105) b ○ 9x 105 J O 15x 105 J 8x 105 J O 1× 105 J P3 P₂ P₁ a V₁₂ C d V₂ V(m³)

For the process shown in the pV diagram below, if p₁ = 1 × 105 Pa ,P2 = 2 × 105 Pa ,p3 = 3 × 105 Pa, V₁ = 1 m³ and V/₂ = 4 m³, the total work in going from a to d along the path shown is p (Pax 105) b ○ 9x 105 J O 15x 105 J 8x 105 J O 1× 105 J P3 P₂ P₁ a V₁₂ C d V₂ V(m³)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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## Understanding the pV Diagram In this problem, we are analyzing a pV (pressure-volume) diagram to determine the total work done along a specified path from point \( a \) to point \( d \). ### Problem Statement - Given: - \( p_1 = 1 \times 10^5 \, \text{Pa} \) - \( p_2 = 2 \times 10^5 \, \text{Pa} \) - \( p_3 = 3 \times 10^5 \, \text{Pa} \) - \( V_1 = 1 \, \text{m}^3 \) - \( V_2 = 4 \, \text{m}^3 \) - Objective: Calculate the total work done in moving from point \( a \) to point \( d \) along the path \( a \rightarrow b \rightarrow c \rightarrow d \). ### Analysis of the Diagram #### Diagram Explanation The diagram is a pV plot with volume \( V \) on the x-axis measured in cubic meters (m³) and pressure \( p \) on the y-axis measured in Pascals (Pa) multiplied by \( 10^5 \). - The process path is: 1. \( a \rightarrow b \): Vertical path (constant volume \( V_1 \)) 2. \( b \rightarrow c \): Horizontal path (constant pressure \( p_3 \)) 3. \( c \rightarrow d \): Vertical path (constant volume \( V_2 \)) #### Calculating Work Done 1. **\( a \rightarrow b \):** Vertical path, no volume change, thus work done is 0. 2. **\( b \rightarrow c \):** Horizontal path at pressure \( p_3 = 3 \times 10^5 \, \text{Pa} \). - Work done, \( W = p \Delta V = p_3 (V_2 - V_1) \) - \( W = 3 \times 10^5 \, \text{Pa} \times (4 \, \text{m}^3 - 1 \, \text{m}^3) = 3 \times 10^5 \, \text{Pa} \times 3 \, \text{
Transcribed Image Text:## Understanding the pV Diagram In this problem, we are analyzing a pV (pressure-volume) diagram to determine the total work done along a specified path from point \( a \) to point \( d \). ### Problem Statement - Given: - \( p_1 = 1 \times 10^5 \, \text{Pa} \) - \( p_2 = 2 \times 10^5 \, \text{Pa} \) - \( p_3 = 3 \times 10^5 \, \text{Pa} \) - \( V_1 = 1 \, \text{m}^3 \) - \( V_2 = 4 \, \text{m}^3 \) - Objective: Calculate the total work done in moving from point \( a \) to point \( d \) along the path \( a \rightarrow b \rightarrow c \rightarrow d \). ### Analysis of the Diagram #### Diagram Explanation The diagram is a pV plot with volume \( V \) on the x-axis measured in cubic meters (m³) and pressure \( p \) on the y-axis measured in Pascals (Pa) multiplied by \( 10^5 \). - The process path is: 1. \( a \rightarrow b \): Vertical path (constant volume \( V_1 \)) 2. \( b \rightarrow c \): Horizontal path (constant pressure \( p_3 \)) 3. \( c \rightarrow d \): Vertical path (constant volume \( V_2 \)) #### Calculating Work Done 1. **\( a \rightarrow b \):** Vertical path, no volume change, thus work done is 0. 2. **\( b \rightarrow c \):** Horizontal path at pressure \( p_3 = 3 \times 10^5 \, \text{Pa} \). - Work done, \( W = p \Delta V = p_3 (V_2 - V_1) \) - \( W = 3 \times 10^5 \, \text{Pa} \times (4 \, \text{m}^3 - 1 \, \text{m}^3) = 3 \times 10^5 \, \text{Pa} \times 3 \, \text{
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