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Heat, Work, and Internal Energy As a gas cools, it is compressed from 2.50 L to 1.25 L under a constant pressure of \(1.01 \times 10^{5}\) Pa. Calculate the work (in J) required to compress the gas.

Short Answer

Expert verified
The work required to compress the gas is 1.2625 Joules.

Step by step solution

01

Understanding Work in Thermodynamics

In thermodynamics, work done on or by a system during volume changes can be calculated using the formula for work done at constant pressure:\[ W = -P \Delta V \]where \( W \) is the work done, \( P \) is the pressure, and \( \Delta V \) is the change in volume. The negative sign indicates that work done on the system (compression) is considered positive.
02

Calculate Change in Volume

First, determine the change in volume. The initial volume \( V_i = 2.50 \, \text{L} \) and the final volume \( V_f = 1.25 \, \text{L} \). The change in volume \( \Delta V \) is calculated as:\[ \Delta V = V_f - V_i = 1.25 \, \text{L} - 2.50 \, \text{L} = -1.25 \, \text{L} \]Since 1 L = 0.001 m³, the change in volume in cubic meters is \( -1.25 \, \text{L} \times 0.001 = -0.00125 \, \text{m}^3 \).
03

Calculate the Work Done

Now, use the work formula to calculate the work done:\[ W = -P \Delta V \]Substitute the given pressure \( P = 1.01 \times 10^{5} \, \text{Pa} \) and the calculated change in volume \( \Delta V = -0.00125 \, \text{m}^3 \).\[ W = -(1.01 \times 10^{5} \, \text{Pa}) (-0.00125 \, \text{m}^3) = 1.2625 \, \text{J} \]
04

Conclusion

The calculated work required to compress the gas is 1.2625 Joules, taking into account the conventions for work done in thermodynamics.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy
In thermodynamics, internal energy is a crucial concept that refers to the total energy contained within a system. This energy includes various forms such as kinetic energy from particle movements and potential energy from molecular positions. When a system undergoes a change, like cooling or being compressed, its internal energy can also change.

Internal energy is not measured directly, but rather inferred by observing changes in state properties such as temperature and volume.
  • It is a state function, meaning it depends only on the current state and not on the path taken to reach that state.
  • Changes in internal energy can be due to work done on or by the system, or heat exchanged with the surroundings.
Understanding internal energy allows us to predict how a system will respond to external changes, providing insights into energy conservation and conversion processes.
Work in Thermodynamics
Work in thermodynamics often relates to the energy transfer resulting from a force applied over a distance, or, in the context of gases, a change in volume at constant pressure.
The formula for calculating work when a gas is compressed or expanded is:
  • \( W = -P \Delta V \)
where \( W \) is the work done, \( P \) is pressure, and \( \Delta V \) is the change in volume. The negative sign indicates that if the volume decreases (compression), the work is positive as it is done on the system.

In cases of gas compression, work represents the energy supplied to condense the gas, against the natural tendency for gases to expand.
This concept is foundational as it connects mechanical actions with thermodynamic processes, explaining how energy can be stored or released in forms that can be converted to or from heat.
Volume Change
Volume change is an essential aspect of thermodynamics, especially when dealing with gases. It influences how energies like work and internal energy evolve in a system. In thermodynamic processes, volume change can drastically affect a system's behavior and energy interactions.

Consider a simple example of compressing or expanding a gas:
  • Initial and final volumes are noted, say, \( V_i \) (initial) and \( V_f \) (final).
  • Change in volume, \( \Delta V \), is calculated as \( V_f - V_i \).
This change reflects how much a system has contracted or expanded.
For calculations like work done during compression or expansion, it's important to convert this volume change into compatible units, such as from liters to cubic meters, to match other parameters like pressure.Recognizing the role of volume change is pivotal in energy balance equations and helps in precisely calculating work and energy transfer.
Pressure
Pressure is a measure of force exerted per unit area. It's a central parameter in thermodynamics, especially for processes involving gases.

When dealing with pressure, consider:
  • It is often given in Pascals (Pa), defined as 1 Newton per square meter.
  • In thermodynamic calculations, pressure often remains constant (isobaric processes) or varies depending on the conditions.
Pressure dictates how systems respond to changes, such as in volume or temperature. In the context of work, a constant pressure precisely measures the resistance needed for volume change in a gas.
Pressure **\( P \)** often links with volume **\( V \)** and temperature **\( T \)** through the ideal gas law relation, enhancing our understanding of how gases behave under different conditions. Understanding pressure and its role in thermodynamics equips us to handle calculations involving work, energy changes, and even predict material behavior under various environmental conditions.

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Most popular questions from this chapter

A balloon does 324 J of work on the surroundings as it expands under a constant pressure of \(7.33 \times 10^{4} \mathrm{Pa}\). What is the change in volume (in L) of the balloon?

Calculate the quantity of energy required to convert \(60.1 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) at \(0.0^{\circ} \mathrm{C}\) to \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100.0^{\circ} \mathrm{C} .\) The enthalpy of fusion of ice at \(0^{\circ} \mathrm{C}\) is \(333 \mathrm{J} / \mathrm{g}\) i the enthalpy of vaporization of liquid water at \(100^{\circ} \mathrm{C}\) is \(2256 \mathrm{J} / \mathrm{g}.\)

A A piece of gold \(\left(10.0 \mathrm{g}, C_{\mathrm{Au}}=0.129 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is heated to \(100.0^{\circ} \mathrm{C} .\) A piece of copper (also \(10.0 \mathrm{g}\) \(\left.C_{\alpha_{i}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is chilled in an ice bath to \(0^{\circ} \mathrm{C} .\) Both pieces of metal are placed in a beaker containing \(150 . \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C} .\) Will the temperature of the water be greater than or less than \(20^{\circ} \mathrm{C}\) when thermal equilibrium is reached? Calculate the final temperature.

Nitrogen gas is confined in a cylinder with a movable piston under a constant pressure of \(9.95 \times 10^{4}\) Pa. When \(695 \mathrm{J}\) of energy in the form of heat is transferred from the gas to the surroundings, its volume decreases by 1.88 L. What is the change in internal energy of the gas?

A You have the six pieces of metal listed below, plus a beaker of water containing \(3.00 \times 10^{2} \mathrm{g}\) of water. The water temperature is \(21.00^{\circ} \mathrm{C}.\) $$\begin{array}{|l|c|c|} \hline \text { Metals } & \text { Specific Heat }(J / g K) & \text { Mass }(g) \\\ \hline 1 . A 1 & 0.9002 & 100.0 \\ 2 . A 1 & 0.9002 & 50.0 \\ 3 . A u & 0.1289 & 100.0 \\ 4 . A u & 0.1289 & 50.0 \\ 5 . Z n & 0.3860 & 100.0 \\ 6 . Z n & 0.3860 & 50.0 \\ \hline \end{array}$$ (a) In your first experiment you select one piece of metal and heat it to \(100^{\circ} \mathrm{C},\) and then select a second piece of metal and cool it to \(-10^{\circ} \mathrm{C}\) Both pieces of metal are then placed in the beaker of water and the temperatures equilibrated. You want to select two pieces of metal to use, such that the final temperature of the water is as high as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water? (b) The second experiment is done in the same way as the first. However, your goal now is to cause the temperature to change the least, that is, the final temperature should be as near to \(21.00^{\circ} \mathrm{C}\) as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water?

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