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. D 3P1 2P1 h P1 -> A B 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 &W = -PdV is not an exact differential.

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### Problem 3a-3c: Internal Energy of a Perfect Monoatomic Gas

#### Problem Statement:
For a mole of a perfect monoatomic gas, the internal energy \( U \) can be expressed as a function of pressure and volume:
\[ U = U(P, V) = \frac{3}{2}PV \]

#### Tasks:
a) **Calculate the Line Integral of \( dU \):**  
Calculate explicitly the line integral of \( dU \) along the closed path \( ABCD \) shown as a black trace in the \( P-V \) graph.

b) **Compute the Line Integrals Between Points B and C:**  
1. \(\int_h PdV\), along the path \( h \), described by \( P = \frac{3P_1V_1}{(V - 3V_1)} \), shown in red in the figure.
2. \(\int_s PdV\), along the path \( s \), shown in black.

#### Graph Description:
- **Axes:** The graph displays a pressure-volume (\( P-V \)) plot.
- **Coordinates:**
  - \( P \): Pressure axis (vertical)
  - \( V \): Volume axis (horizontal)
- **Points:**
  - \( A \): Starts at \( P_1, V_1 \)
  - \( B \): Ends at \( P_1, 6V_1 \)
  - \( C \): At \( 3P_1, 6V_1 \)
  - \( D \): At \( 3P_1, V_1 \)
- **Paths:**
  - **AB**: Horizontal, from \( A \) to \( B \)
  - **BC**: Vertical, from \( B \) to \( C \)
  - **CD**: Horizontal, from \( C \) to \( D \)
  - **DA**: Ascending diagonally, from \( D \) back to \( A \)

The closed cycle \( ABCD \) encompasses these paths.

#### Analysis:
Use the given functions to calculate the integrals and demonstrate that \( \delta W = -PdV \) is not an exact differential, highlighting the significance in thermodynamics.
Transcribed Image Text:### Problem 3a-3c: Internal Energy of a Perfect Monoatomic Gas #### Problem Statement: For a mole of a perfect monoatomic gas, the internal energy \( U \) can be expressed as a function of pressure and volume: \[ U = U(P, V) = \frac{3}{2}PV \] #### Tasks: a) **Calculate the Line Integral of \( dU \):** Calculate explicitly the line integral of \( dU \) along the closed path \( ABCD \) shown as a black trace in the \( P-V \) graph. b) **Compute the Line Integrals Between Points B and C:** 1. \(\int_h PdV\), along the path \( h \), described by \( P = \frac{3P_1V_1}{(V - 3V_1)} \), shown in red in the figure. 2. \(\int_s PdV\), along the path \( s \), shown in black. #### Graph Description: - **Axes:** The graph displays a pressure-volume (\( P-V \)) plot. - **Coordinates:** - \( P \): Pressure axis (vertical) - \( V \): Volume axis (horizontal) - **Points:** - \( A \): Starts at \( P_1, V_1 \) - \( B \): Ends at \( P_1, 6V_1 \) - \( C \): At \( 3P_1, 6V_1 \) - \( D \): At \( 3P_1, V_1 \) - **Paths:** - **AB**: Horizontal, from \( A \) to \( B \) - **BC**: Vertical, from \( B \) to \( C \) - **CD**: Horizontal, from \( C \) to \( D \) - **DA**: Ascending diagonally, from \( D \) back to \( A \) The closed cycle \( ABCD \) encompasses these paths. #### Analysis: Use the given functions to calculate the integrals and demonstrate that \( \delta W = -PdV \) is not an exact differential, highlighting the significance in thermodynamics.
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