A gas expands according to the pressure-volume graph shown in the drawing. How much work W is done by the gas in going from A to B? Be sure to include the correct algebraic sign. Pressure ( 10³ Pa) 0.30 B 0.60 0.75 Volume, m²

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### Work Done by Gas During Expansion

#### Problem Statement:
A gas expands according to the pressure-volume graph shown in the drawing. How much work \( W \) is done by the gas in going from point A to point B? Be sure to include the correct algebraic sign.

#### Graph Analysis:
The provided graph is a Pressure-Volume (P-V) diagram, where:
- The y-axis represents the Pressure (in \( \times 10^5 \, \text{Pa} \)).
- The x-axis represents the Volume (in \( \text{m}^3 \)).

The process described involves a gas expanding:
- From point A (4.0 \( \times 10^5 \, \text{Pa} \), 0.30 \( \text{m}^3 \))
- To point B (2.0 \( \times 10^5 \, \text{Pa} \), 0.75 \( \text{m}^3 \))

The graph consists of two linear segments:
1. From (4.0 \( \times 10^5 \, \text{Pa} \), 0.30 \( \text{m}^3 \)) to (2.0 \( \times 10^5 \, \text{Pa} \), 0.60 \( \text{m}^3 \))
2. From (2.0 \( \times 10^5 \, \text{Pa} \), 0.60 \( \text{m}^3 \)) to (2.0 \( \times 10^5 \, \text{Pa} \), 0.75 \( \text{m}^3 \))

#### Calculation:
To find the work done \( W \) by the gas, we calculate the area under the P-V curve.

1. **First Segment (A to midpoint):**
   - Pressure \( P \) changes linearly from 4.0 \( \times 10^5 \, \text{Pa} \) to 2.0 \( \times 10^5 \, \text{Pa} \).
   - Volume \( V \) changes from 0.30 \( \text{m}^3 \) to 0.60 \( \text{m}^3 \).

   Area of this trapezoid:
   \[
   W_1
Transcribed Image Text:### Work Done by Gas During Expansion #### Problem Statement: A gas expands according to the pressure-volume graph shown in the drawing. How much work \( W \) is done by the gas in going from point A to point B? Be sure to include the correct algebraic sign. #### Graph Analysis: The provided graph is a Pressure-Volume (P-V) diagram, where: - The y-axis represents the Pressure (in \( \times 10^5 \, \text{Pa} \)). - The x-axis represents the Volume (in \( \text{m}^3 \)). The process described involves a gas expanding: - From point A (4.0 \( \times 10^5 \, \text{Pa} \), 0.30 \( \text{m}^3 \)) - To point B (2.0 \( \times 10^5 \, \text{Pa} \), 0.75 \( \text{m}^3 \)) The graph consists of two linear segments: 1. From (4.0 \( \times 10^5 \, \text{Pa} \), 0.30 \( \text{m}^3 \)) to (2.0 \( \times 10^5 \, \text{Pa} \), 0.60 \( \text{m}^3 \)) 2. From (2.0 \( \times 10^5 \, \text{Pa} \), 0.60 \( \text{m}^3 \)) to (2.0 \( \times 10^5 \, \text{Pa} \), 0.75 \( \text{m}^3 \)) #### Calculation: To find the work done \( W \) by the gas, we calculate the area under the P-V curve. 1. **First Segment (A to midpoint):** - Pressure \( P \) changes linearly from 4.0 \( \times 10^5 \, \text{Pa} \) to 2.0 \( \times 10^5 \, \text{Pa} \). - Volume \( V \) changes from 0.30 \( \text{m}^3 \) to 0.60 \( \text{m}^3 \). Area of this trapezoid: \[ W_1
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