Problem 44 - University Physics with Modern Physics

Question

Three moles of argon gas (assumed to be an ideal gas) originally at a pressure of \(1.50 \times 10^{4}\) Pa and a volume of 0.0280 \(\mathrm{m}^{3}\) are first heated and expanded at constant pressure to a volume of 0.0435 \(\mathrm{m}^{3}\) , then heated at constant volume until the pressure reaches \(3.50 \times 10^{4} \mathrm{Pa}\) , then cooled and compressed at constant pressure until the volume is again \(0.0280 \mathrm{m}^{3},\) and finally cooled at constant volume until the pressure drops to its original value of \(1.50 \times 10^{4}\) Pa. (a) Draw the \(p V\) -diagram for this cycle. (b) Calculate the total work done by (or on) the gas during the cycle. (c) Calculate the net heat exchanged with the surroundings. Does the gas gain or lose heat overall?

Step-by-Step Solution

Answer

(a) Draw a rectangular cycle on the PV diagram. (b) The net work is \(-230.25\, J\). (c) The net heat exchange is \(-230.25\, J\); the gas loses heat.

Step 1: Sketch the PV Diagram

Start by plotting the initial point on a PV diagram where pressure \( P = 1.50 \times 10^4 \) Pa and volume \( V = 0.0280 \) m\(^3\). As the gas is heated at constant pressure to \( V = 0.0435 \) m\(^3\), draw a horizontal line to the right to this new volume. Then, at constant volume, increase the pressure vertically to \( P = 3.50 \times 10^4 \) Pa. Next, compress the gas at constant pressure back to \( V = 0.0280 \) m\(^3\) with a horizontal leftward line. Finally, decrease the pressure at constant volume back to the original pressure, closing the cycle on the PV diagram with a vertical line downward.

Step 2: Calculate Work Done During Each Process

Work done by or on the gas can be calculated for each process using the formula \( W = P \Delta V \) when pressure is constant. For the first expansion at constant pressure, \( W_1 = 1.50 \times 10^4 \times (0.0435 - 0.0280) \). For the compression at constant pressure, \( W_3 = -1.50 \times 10^4 \times (0.0435 - 0.0280) \). There is no work done during the constant volume processes \( W_2 = 0 \) and \( W_4 = 0 \). Calculate \( W_1 \) and \( W_3 \) and sum them to find the total work over the cycle.

Step 3: Calculate Net Heat Exchange

Use the first law of thermodynamics \( \Delta U = Q - W \) for each stage. We need the net heat change over the whole cycle, which relates to \( Q = \Delta U + W \). Since this is a cyclic process, the internal energy change \( \Delta U = 0 \) over one complete cycle. Therefore, the net heat exchange \( Q \) is equal to the total work \( W \). Compute \( Q = W_1 + W_3 \). If \( Q > 0 \), heat is gained; if \( Q < 0 \), heat is lost.

Step 4: Interpret Heat Exchange Result

Based on the calculated net work and the sign of \( Q \), determine whether the gas gained or lost heat during the cycle. If the result of \( Q \) is negative, the gas loses heat. If it is positive, the gas gains heat.

Key Concepts

PV Diagram

A PV diagram is a powerful tool for visualizing processes in thermodynamics, especially cycles involving gases. In terms of the Ideal Gas Laws, it plots the pressure (P) on the vertical axis and volume (V) on the horizontal axis. Each point on this diagram represents a different state of the gas, defined by a specific pressure and volume.

When we sketch the PV diagram for a thermodynamic cycle, such as the one described in the exercise, different processes are illustrated using distinct paths. Horizontal lines represent changes at constant pressure, and vertical lines represent changes at constant volume. For the given cycle, the journey on the PV diagram starts with an expansion at constant pressure, seen as a horizontal line to the right.

Followed by this, a vertical line indicates increasing pressure with no change in volume. The next step retraces back to the lower volume at constant pressure—a horizontal line but to the left. Finally, the pressure returns to its original value, depicted by a downward vertical line. This completes the cycle, showing the gas returning to its initial state.

Thermodynamic Cycles

Thermodynamic cycles are sequences of processes that return a system to its initial state. These cycles are fundamental in understanding engines and refrigerators, where they perform work or transfer heat. In a cycle involving an ideal gas, like the one in our exercise, different stages of heating, cooling, expansion, and compression are illustrated.

The significance of thermodynamic cycles lies in the net effect they produce. Starting from the initial state, the gas undergoes an expansion at a constant pressure, absorbing energy. Then, it moves to constant volume processes, followed by compression, releasing energy. Eventually, it returns to the initial state, completing the cycle.

• In each phase, the type of process defines how thermodynamic properties like pressure and volume change.
• The complete cycle can carry out work or absorb heat according to these changes.

The completion of a cycle means all internal energy changes cancel out, emphasizing net work or heat transfer as the cycle’s real output.

Work Done by Gas

In thermodynamics, work done by or on a gas is a central focus, especially in processes at constant pressure or volume. Work is calculated using the formula, \( W = P \Delta V \), for processes at constant pressure. This formula signifies that work involves changing the volume of the gas.

In the problem described, when the gas expands from 0.0280 m³ to 0.0435 m³ at a constant pressure of \( 1.50 \times 10^4 \) Pa, it does positive work on the surroundings. Conversely, compressing the gas, the surroundings do work on it, which is negative work.

• When volume change, \( \Delta V \), increases, the gas does work.
• When \( \Delta V \) decreases, the gas receives work.

Using these calculations, the net work done over the entire cycle results from adding up contributions from all phases. Constant volume stages mean no volume change occurs, so no work is done during these parts.

Net Heat Exchange

Net heat exchange during a thermodynamic cycle is determined using the principles of the first law of thermodynamics, which is stated as \( \Delta U = Q - W \). Here, \( \Delta U \) is the change in internal energy, \( Q \) is the net heat added to the system, and \( W \) is the work done by the system.

In a complete cycle, since the system returns to its initial state, the internal energy change, \( \Delta U \), is zero. Thus, the net heat exchange \( Q \) is directly equal to the total work done by the gas, \( W \). Therefore, the formula simplifies to \( Q = W \).

• If \( Q > 0 \), it indicates that the gas has gained heat from its surroundings.
• If \( Q < 0 \), the gas has lost heat.

The net heat exchange before closing the cycle assists in comprehending the energy transaction within the system. This understanding is pivotal for engineering applications like heat engines, where efficient work extraction from heat is desired.