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 3 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 C 3P1 2P1 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;V/(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. S.

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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:**

The graph is a \( P \) (pressure) vs. \( V \) (volume) graph with the following key points:
- Point \( A \) at \((V_1, P_1)\)
- Point \( B \) at \((6V_1, P_1)\)
- Point \( C \) at \((6V_1, 3P_1)\)
- Point \( D \) at \((V_1, 3P_1)\)

The path follows:
- From \( A \) to \( B \): Horizontal line at \( P = P_1 \)
- From \( B \) to \( C \): Vertical line at \( V = 6V_1 \)
- From \( C \) to \( D \): Horizontal line at \( P = 3P_1 \)
- From \( D \) to \( A \): Vertical line at \( V = V_1 \)

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 \( P = 3P_1 V_1 / (V - 3V_1) \), 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 \( \delta W = -PdV \) is not an exact differential.

\[ \text{Problem 3a-3c} \]
Transcribed Image Text: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:** The graph is a \( P \) (pressure) vs. \( V \) (volume) graph with the following key points: - Point \( A \) at \((V_1, P_1)\) - Point \( B \) at \((6V_1, P_1)\) - Point \( C \) at \((6V_1, 3P_1)\) - Point \( D \) at \((V_1, 3P_1)\) The path follows: - From \( A \) to \( B \): Horizontal line at \( P = P_1 \) - From \( B \) to \( C \): Vertical line at \( V = 6V_1 \) - From \( C \) to \( D \): Horizontal line at \( P = 3P_1 \) - From \( D \) to \( A \): Vertical line at \( V = V_1 \) 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 \( P = 3P_1 V_1 / (V - 3V_1) \), 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 \( \delta W = -PdV \) is not an exact differential. \[ \text{Problem 3a-3c} \]
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