Let's solve the problem step by step! ### **Step 1: Understanding the Work Done by the Gas** The **work done by a gas** during expansion or compression is the **area under the PV curve**. #### **Given:** - The x-axis (Volume) is in increments of **0.5 m³**. - The y-axis (Pressure) values are in **Pa**. - The points: - \( i \): \( (1.0, 6.0 \times 10^6) \) - \( f \): \( (2.0, 2.0 \times 10^6) \) #### **Path from i to f:** - **i to the right:** Horizontal line at \( 6 \times 10^6 \) Pa, from \( 1.0\,m^3 \) to \( 1.5\,m^3 \) - **Slope down:** \( 6 \times 10^6 \) Pa to \( 2 \times 10^6 \) Pa, from \( 1.5\,m^3 \) to \( 2.0\,m^3 \) --- ### **Step 2: Calculate the Area Under Each Segment** #### **Segment 1 (i to the right):** - Horizontal segment at \( 6 \times 10^6 \) Pa from \( 1.0 \) to \( 1.5 \) \( m^3 \) - **Area = Pressure × ΔV** - \( A_1 = 6 \times 10^6 \times (1.5 - 1.0) = 6 \times 10^6 \times 0.5 = 3.0 \times 10^6 \) J #### **Segment 2 (Downward Slope):** - From \( (1.5\,m^3, 6\times10^6\,Pa) \) to \( (2.0\,m^3, 2\times10^6\,Pa) \) - Area under this segment is a **trapezoid**: - Height = \( 2.0 - 1.5 = 0.5\,m^3 \) - Bases = \( 6 \times 10^6 \) Pa and \( 2 \times 10^6 \) Pa - \( A_2 = \frac{1}{2} \times (6 \times 10^6 + 2 \times 10^6) \times 0.5 \) - \( = \frac{1}{2} \times 8 \times 10^6 \times 0.5 \) - \( = 4 \times 10^6 \times 0.5 = 2.0 \times 10^6 \) J --- ### **Step 3: Total Work Done (i to f)** \[ W_{i \to f} = A_1 + A_2 = (3.0 \times 10^6) + (2.0 \times 10^6) = 5.0 \times 10^6 \text{ J} \] Convert to MJ: \[ W_{i \to f} = \frac{5.0 \times 10^6}{10^6} = 5.0 \text{ MJ} \] --- ### **Step 4: Work Done on the Gas (f to i)** - The work done **on** the gas is the **negative** of the work done **by** the gas (since the area is the same, but the volume is decreasing): \[ W_{f \to i} = -5.0 \text{ MJ} \] --- ## **Final Answers:** ### (a) Work done on a gas that expands from \(i\) to \(f\): \[ \boxed{5.0 \text{ MJ}} \] ### (b) Work performed on the gas if it is compressed from \(f\) to \(i\) along the same path: \[ \boxed{-5.0 \text{ MJ}} \] Let me know if you need a drawing or further breakdown!

Certainly! Let's go through each part in detail with step-by-step calculations. --- ## **Part (a): Work done on the gas during expansion from i to f** ### **Step 1: Understand the problem** - The process occurs along a PV diagram. - The pressure is given in Pa, and the volume in m³. - The points: - \( i: V_i = 1.0\,m^3,\quad P_i = 6 \times 10^6\,Pa \) - \( f: V_f = 2.0\,m^3,\quad P_f = 2 \times 10^6\,Pa \) - The path from \( i \) to \( f \): 1. From \( V = 1.0\,m^3 \) to \( 1.5\,m^3 \) at constant pressure \( 6 \times 10^6\,Pa \). 2. From \( 1.5\,m^3 \) to \( 2.0\,m^3 \) with pressure decreasing from \( 6 \times 10^6\,Pa \) to \( 2 \times 10^6\,Pa \). --- ### **Step 2: Break the process into segments** The work done \( W \) is the **area under the PV curve** between the initial and final points. **Work done during a process:** \[ W = \int_{V_i}^{V_f} P\, dV \] **In practice**, for piecewise linear paths, this is calculated as the sum of areas of geometric shapes (rectangles, trapezoids). --- ### **Step 3: Calculate the area for each segment** #### **Segment 1: Horizontal line from \( V=1.0\,m^3 \) to \( V=1.5\,m^3 \) at \( P=6 \times 10^6\,Pa \)** - **Pressure \( P = 6 \times 10^6\,Pa \) (constant)** - **Volume change \( \Delta V = 1.5 - 1.0 = 0.5\,m^3 \)** **Work done (area):** \[ A_1 = P \times \Delta V = 6 \times 10^6 \times 0.5 = 3.0 \times 10^6\,J \] --- #### **Segment 2: Sloped line from \( V=1.5\,m^3 \) to \( V=2.0\,m^3 \)** - **Starting point:** \( (1.5\,m^3, 6 \times 10^6\,Pa) \) - **Ending point:** \( (2.0\,m^3, 2 \times 10^6\,Pa) \) This is a straight line, forming a trapezoid under the curve. - **Base volume change:** \( 0.5\,m^3 \) - **Pressure at start:** \( P_1 = 6 \times 10^6\,Pa \) - **Pressure at end:** \( P_2 = 2 \times 10^6\,Pa \) **Area of a trapezoid:** \[ A_2 = \frac{1}{2} \times (P_1 + P_2) \times \Delta V \] Plugging in the values: \[ A_2 = \frac{1}{2} \times (6 \times 10^6 + 2 \times 10^6) \times 0.5 \] \[ A_2 = \frac{1}{2} \times 8 \times 10^6 \times 0.5 \] \[ A_2 = 4 \times 10^6 \times 0.5 = 2.0 \times 10^6\,J \] --- ### **Step 4: Total work done** \[ W_{i \to f} = A_1 + A_2 = 3.0 \times 10^6 + 2.0 \times 10^6 = 5.0 \times 10^6\,J \] Expressed in megajoules: \[ \boxed{ W_{i \to f} = \mathbf{5.0\,MJ} } \] --- ## **Part (b): Work performed on the gas during compression from \( f \) to \( i \)** - **Work done on the gas is negative of the work done by the gas during expansion**, assuming the same path but in reverse. \[ W_{f \to i} = - W_{i \to f} = -5.0\,MJ \] --- ## **Final summarized results:** | Part | Work done (MJ) | Explanation | |--------|------------------|--------------------------------------------------------| | (a) | **+5.0 MJ** | Work **done by** the gas during expansion | | (b) | **-5.0 MJ** | Work **done on** the gas during compression | --- Let me know if you'd like a detailed diagram or further clarifications!

Certainly! Let's go through the problem in a very detailed manner, step by step, with clear calculations and explanations. --- ## **Given Data:** - The PV diagram shows a process from point **i** to point **f**. - The **x-axis (volume V)** is in **meters cubed (m³)**, with increments of 0.5 m³. - The **pressure P** is in **Pa**. - **Points:** - \( i: V_i = 1.0\,m^3,\quad P_i = 6 \times 10^6\,Pa \) - \( f: V_f = 2.0\,m^3,\quad P_f = 2 \times 10^6\,Pa \) --- ## **Part (a): Calculate the work done during expansion from i to f** ### **Step 1: Understand the path** The process from i to f appears to be two segments: 1. **i to an intermediate point** at \( V=1.5\,m^3 \), where pressure remains constant at \( 6 \times 10^6\,Pa \). 2. **From volume 1.5 m³ to 2.0 m³**, where pressure decreases linearly from \( 6 \times 10^6\,Pa \) to \( 2 \times 10^6\,Pa \). This is typical of a piecewise linear process on a PV diagram. --- ### **Step 2: Calculate the work for each segment** **Work done** is the **area under the PV curve** between the initial and final points: \[ W = \int_{V_i}^{V_f} P\, dV \] Since the process is piecewise linear, we break it into: - **Segment 1:** Horizontal line at \( P=6 \times 10^6\,Pa \), from \( V=1.0\,m^3 \) to \( 1.5\,m^3 \). - **Segment 2:** Linear decrease from \( P=6 \times 10^6\,Pa \) to \( 2 \times 10^6\,Pa \), from \( 1.5\,m^3 \) to \( 2.0\,m^3 \). --- ### **Step 3: Calculate the area for each segment** #### **Segment 1: Horizontal line at constant pressure** - **Pressure:** \( P = 6 \times 10^6\,Pa \) - **Volume change:** \( \Delta V = 1.5 - 1.0 = 0.5\,m^3 \) **Work done:** \[ A_1 = P \times \Delta V = 6 \times 10^6 \times 0.5 = 3.0 \times 10^6\,J \] --- #### **Segment 2: Linear decrease from \( P=6 \times 10^6 \) to \( 2 \times 10^6 \) Pa** - **Starting point:** \( (V=1.5\,m^3, P=6 \times 10^6\,Pa) \) - **Ending point:** \( (2.0\,m^3, 2 \times 10^6\,Pa) \) This is a straight line, forming a trapezoid when we consider the pressure-volume relationship. - **Pressure at start:** \( P_1 = 6 \times 10^6\,Pa \) - **Pressure at end:** \( P_2 = 2 \times 10^6\,Pa \) - **Volume change:** \( \Delta V = 0.5\,m^3 \) **Average pressure:** \[ P_{avg} = \frac{P_1 + P_2}{2} = \frac{6 \times 10^6 + 2 \times 10^6}{2} = 4 \times 10^6\,Pa \] **Work done:** \[ A_2 = P_{avg} \times \Delta V = 4 \times 10^6 \times 0.5 = 2.0 \times 10^6\,J \] *(Alternatively, for a linear pressure change, area can be computed as trapezoid:)* \[ A_2 = \frac{1}{2} \times (P_1 + P_2) \times \Delta V \] which matches the calculation above. --- ### **Step 4: Total work done** \[ W_{i \to f} = A_1 + A_2 = 3.0 \times 10^6 + 2.0 \times 10^6 = 5.0 \times 10^6\,J \] **Expressed in Megajoules:** \[ \boxed{ \textbf{Work done by the gas} = \mathbf{5.0\,MJ} } \] --- ## **Part (b): Work **done on** the gas during compression from f to i** - Since the process is the same path in reverse, the work **on** the gas is the **negative** of the work **done by** the gas: \[ W_{f \to i} = - W_{i \to f} = -5.0\,MJ \] --- ## **Final Results:** | Part | Description | Value | Units | |---------|----------------|--------|--------| | (a) | Work **done by** the gas during expansion | **+5.0** | MJ | | (b) | Work **done on** the gas during compression | **–5.0** | MJ | --- ## **Summary:** - **Work done during expansion (from i to f):** **5.0 MJ** - **Work done on the gas (from f to i):** **–5.0 MJ** This detailed approach ensures clarity in how the areas under the PV curve translate to work in thermodynamic processes. --- Let me know if you need further clarifications or a graphical representation!