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An amount of n moles of a monatomic ideal gas in a conducting container with a movable piston is placed in a large thermal heat bath at temperature \(T_{1}\) and the gas is allowed to come to equilibrium. After the equilibrium is reached, the pressure on the piston is lowered so that the gas expands at constant temperature. The process is continued quasi-statically until the final pressure is \(4 / 3\) of the initial pressure \(p_{1}\). (a) Find the change in the internal energy of the gas. (b) Find the work done by the gas. (c) Find the heat exchanged by the gas, and indicate, whether the gas takes in or gives up heat.

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: (a) The change in internal energy of the gas is 0. (b) The work done by the gas is \(nRT\ln{\frac{4}{3}}\). (c) The heat exchanged by the gas is \(nRT\ln{\frac{4}{3}}\), and the gas takes in heat.

Step by step solution

01

Determine the Initial and Final Volumes of the Gas

Given that the final pressure is \(4/3\) times the initial pressure, we have: \[p_2 = \frac{4}{3}p_1\] Since this is an isothermal process (i.e., the temperature remains constant throughout the process), we can apply the Ideal Gas Law, which states: \[pV = nRT\] where - \(p\) is the pressure, - \(V\) is the volume, - \(n\) is the number of moles of the gas, - \(R\) is the ideal gas constant, and - \(T\) is the temperature. By substituting the initial and final pressures into the Ideal Gas Law, we get the following equations: \(p_1V_1 = nRT_1\) (for the initial state) \(p_2V_2 = nRT_1\) (for the final state) Dividing the second equation by the first equation, we get: \[\frac{p_2V_2}{p_1V_1} = 1\] Now, substituting \(p_2 = \frac{4}{3}p_1\) into the above equation, we can solve for the ratio of the final volume to the initial volume: \[\frac{\frac{4}{3}p_1V_2}{p_1V_1} = 1\] \[V_2 = \frac{4}{3} V_1\]
02

Calculate the Change in Internal Energy

Since the gas is monatomic and ideal, its internal energy depends only on its temperature. As this is an isothermal process where the temperature remains constant, the change in internal energy (\(Δ±«\)) of the gas is zero: \[Δ±« = 0\]
03

Calculate the Work Done by the Gas

The work done by the gas in an isothermal process can be calculated using the following formula: \[W = nRT\ln{\frac{V_2}{V_1}}\] We have already determined the relationship \(V_2 = \frac{4}{3} V_1\), so we can substitute it into the formula: \[W = nRT\ln{\frac{4V_1}{3V_1}}\] \[W = nRT\ln{\frac{4}{3}}\]
04

Calculate the Heat Exchanged by the Gas

To determine the heat exchanged by the gas, we can apply the First Law of Thermodynamics, which states: \[Δ±« = Q - W\] Where \(Δ±«\) is the change in internal energy, \(Q\) is the heat exchanged, and \(W\) is the work done by the gas. We determined that \(Δ±« = 0\) and calculated the work done by the gas as \(W = nRT\ln{\frac{4}{3}}\). Therefore, we can solve for the heat exchanged (\(Q\)): \[0 = Q - nRT\ln{\frac{4}{3}}\] \[Q = nRT\ln{\frac{4}{3}}\] Since the value of \(Q\) is positive, this means that the gas takes in heat during the isothermal process. In summary: (a) The change in internal energy of the gas is 0. (b) The work done by the gas is \(nRT\ln{\frac{4}{3}}\). (c) The heat exchanged by the gas is \(nRT\ln{\frac{4}{3}}\), and the gas takes in heat.

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Key Concepts

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

Ideal Gas Law
Understanding the Ideal Gas Law is fundamental for studying the behavior of gases under various conditions. In its simplest form, this law is represented by the equation pV = nRT, where p stands for pressure, V indicates volume, n is the number of moles, R is the universal gas constant, and T is the temperature in kelvins.

This relationship is crucial when analyzing an isothermal process, like the one in our exercise, where a gas expands or contracts while maintaining a constant temperature. By keeping T constant, any changes in pressure are directly related to changes in volume — a concept described by Boyle's Law, which is a special case of the Ideal Gas Law for isothermal processes.

Applying the Ideal Gas Law to an Isothermal Expansion

Using the Ideal Gas Law, we can compare the initial and final states of the gas during an isothermal expansion. By setting up equations for each state and dividing them, we can eliminate common terms, simplifying the equation to help us understand the relationship between pressure and volume throughout the process.
Internal Energy of Gas
The concept of internal energy, denoted as Δ±«, is integral when examining thermodynamic systems. For an ideal monatomic gas, internal energy is solely dependent on temperature and is related to the kinetic energy of the randomly moving gas particles. If the temperature of the gas remains unchanged, as in an isothermal process, there's no alteration in the average random kinetic energy of particles, implying that the internal energy remains constant.

Zero Change in Internal Energy for Isothermal Process

It's essential to recognize that even though a gas might perform work or have work done on it during an isothermal expansion or compression, if there is no temperature change as described in our exercise, the internal energy does not vary. This principle allows us to calculate other thermodynamic quantities knowing that Δ±« = 0, simplifying the evaluation of heat exchange and work done in such processes.
First Law of Thermodynamics
The First Law of Thermodynamics is a cornerstone for understanding energy conservation within a thermodynamic system. It states that the change in internal energy of a system (Δ±«) is equal to the heat added to the system (Q) minus the work done by the system (W): Δ±« = Q - W. This law highlights the concept of energy conservation, meaning that energy cannot be created or destroyed, only transferred or transformed.

When we consider an isothermal process where the internal energy does not change (Δ±« = 0), the amount of heat added to the system is exactly equal to the work performed by it, as illustrated in our exercise. As a result, we can calculate the heat exchanged by equating it to the work done.

Determining Heat Exchange in an Isothermal Process

In the context of our example, this law helps us deduce if the gas absorbs or releases heat. Since we've derived that Δ±« = 0 and have the work done by the gas, we can ascertain the heat exchange. If the calculated value for Q is positive, as it is in this case, the gas has absorbed heat from its surroundings, thus following the natural tendency of heat flow from a hotter region to a cooler one.

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Most popular questions from this chapter

Two moles of helium gas are placed in a cylindrical container with a piston. The gas is at room temperature \(25^{\circ} \mathrm{C}\) and under a pressure of \(3.0 \times 10^{5} \mathrm{Pa} .\) When the pressure from the outside is decreased while keeping the temperature the same as the room temperature, the volume of the gas doubles. (a) Find the work the external agent does on the gas in the process. (b) Find the heat exchanged by the gas and indicate whether the gas takes in or gives up heat. Assume ideal gas behavior.

A bullet of mass \(10 \mathrm{g}\) is traveling horizontally at 200 \(\mathrm{m} / \mathrm{s}\) when it strikes and embeds in a pendulum bob of mass 2.0 kg. (a) How much mechanical energy is dissipated in the collision? (b) Assuming that \(C_{v}\) for the bob plus bullet is 3R, calculate the temperature increase of the system due to the collision. Take the molecular mass of the system to be \(200 \mathrm{g} / \mathrm{mol}\).

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