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19.52 - A certain ideal gas has molar heat capacity at constant volume \(C_{V}\). A sample of this gas initially occupies a volume \(V_{0}\) at pressure \(p_{0}\) and absolute temperature \(T_{0}\). The gas expands isobarically to a volume \(2 V_{0}\) and then expands further adiabatically to a final volume \(4 V_{0}\). (a) Draw a \(p V\) -diagram for this sequence of processes. (b) Compute the total work done by the gas for this sequence of processes. (c) Find the final temperature of the gas. (d) Find the absolute value of the total heat flow \(|Q|\) into or out of the gas for this sequence of processes, and state the direction of heat flow.

Short Answer

Expert verified
The pV-diagram would display the isobaric expansion as a horizontal line and the adiabatic process as a curve. The total work done is represented by the area under the isobaric expansion on the pV-diagram. The final temperature after both processes is determined by the adiabatic formula while the total heat flow can be obtained by using the first law of thermodynamics.

Step by step solution

01

Draw pV diagram

Draw a pressure-volume diagram where the x-axis represents volume and the y-axis represents pressure. Begin at the point corresponding to \(V_0\) and \(p_0\). The isobaric process would be a horizontal line to \(2V_0\), and the adiabatic process would be a curve downwards to \(4V_0\).
02

Compute total work done

The total work done by the gas is the area under the pV curve. Break it up into two parts: work done in isobaric and adiabatic process. For the isobaric process, the work done can be calculated using the formula \(W_{isobaric} = p_0 ∗ (2V_0 - V_0)\), and for the adiabatic process there is no work done because in an adiabatic process the system does not exchange heat with its surroundings, so the only work done is in isobaric process.
03

Find final temperature

To find the final temperature, we first need to find the final pressure after the adiabatic process. The adiabatic process follows the equation \(T_0 * V_0^{1-γ} = T_f * (4V_0)^{1-γ}\) where γ=Cp/Cv which is greater than 1. By solving this equation we can get the final temperature \(T_f\).
04

Find total heat flow

The total heat flow into or out of a system is given by \(Q=ΔU+W\), where ΔU is the change in internal energy and W is the work done. The change in internal energy can be found using the fact that ΔU = nCv(Tf-T0), where n is the amount of substance in moles. Combining this with the work done calculated earlier, we can find the total heat flow.

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Key Concepts

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

Ideal Gas
The concept of an ideal gas is fundamental in thermodynamics, serving as a simplified model to understand the behavior of gases. In an ideal gas, it is assumed that the molecules do not interact with each other except through elastic collisions, and they occupy a negligible volume. This model allows easier calculations and predictions for the gas's behavior under various conditions.
For our exercise, we deal with an ideal gas undergoing expansions. The behavior of this gas can be described using the ideal gas law, which is expressed as:
  • \(PV = nRT\)
  • \(P\) = pressure
  • \(V\) = volume
  • \(n\) = number of moles
  • \(R\) = universal gas constant
  • \(T\) = temperature in Kelvin
Understanding these components is essential as they form the basis for exploring other thermodynamic processes like adiabatic and isobaric changes happening in the problem.
Adiabatic Process
An adiabatic process is a type of thermodynamic transformation where no heat is exchanged with the environment. This means the system is perfectly insulated against heat transfer. During this process, changes in internal energy result solely from work done by or on the system, not from heat transfer.
In an adiabatic process involving an ideal gas, the relation between pressure, volume, and temperature can be expressed by the adiabatic condition:
  • \(PV^\gamma = ext{constant}\)
  • \(T V^{\gamma - 1} = ext{constant}\)
  • \(\gamma = C_p/C_v\), the heat capacity ratio which is greater than 1
In this exercise, the process followed an adiabatic curve from a volume of \(2V_0\) to \(4V_0\). Applying this principle, the final temperature and new pressure after the expansion can be determined. The absence of heat exchange implies there’s no heat flow (\(Q = 0\)), making the computation primarily centered on internal energy and work.
Isobaric Process
An isobaric process involves a change in the state of the gas at a constant pressure. This type of process is depicted as a straight horizontal line on a pressure-volume diagram.For problems involving an ideal gas, the work done during an isobaric process can be calculated using the formula:
  • \(W_{isobaric} = P \Delta V\)
In this specific exercise, the ideal gas expands from volume \(V_0\) to \(2V_0\) while maintaining a constant initial pressure, \(p_0\). Hence, the work done is simply the initial pressure multiplied by the change in volume. This work represents the energy transferred by the system as it moves the piston to accommodate the increased volume of gas.
Pressure-Volume Diagram
A pressure-volume diagram, or pV diagram, is a graphical representation that illustrates changes in the state of a system, showing the relationship between pressure and volume. It's a crucial tool in understanding processes such as those performed in this exercise.
For this exercise, the essential steps for creating a pV diagram include:
  • Starting at the point \((V_0, p_0)\)
  • Drawing a horizontal line to \((2V_0, p_0)\) to indicate the isobaric expansion
  • Continuing with a downward curve from \((2V_0, p_0)\) to \((4V_0, p_{final})\) for the adiabatic process
This visual guide helps to comprehend the transitions and interactions during thermodynamic processes. By considering areas under these curves, one can calculate work done, while different sections of the graph represent specific processes, underscoring the step-by-step change in conditions experienced by the gas.

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Most popular questions from this chapter

A cylinder with a piston contains \(0.250 \mathrm{~mol}\) of oxygen at \(2.40 \times 10^{5} \mathrm{~Pa}\) and \(355 \mathrm{~K}\). The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(p V\) -diagram. Compute (b) the temperature during the isothermal compression; (c) the maximum pressure; (d) the total work done by the piston on the gas during the series of processes.

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at \(1.00 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K}\) and occupies a volume of \(1.50 \mathrm{~L}\). The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at \(300 \mathrm{~K}\). This continues until the pressure reaches \(2.50 \times 10^{4} \mathrm{~Pa}\). In the second process, the gas is compressed at constant pressure until it returns to its original volume of \(1.50 \mathrm{~L}\). Assume that the gas may be treated as ideal. (a) In a \(p V\) -diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

A cylinder contains \(0.250 \mathrm{~mol}\) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\). The cylinder is provided with a friction less piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\). Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

A player bounces a basketball on the floor, compressing it to \(80.0 \%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at \(20.0^{\circ} \mathrm{C}\) and 2.00 atm. The ball's inside diameter is \(23.9 \mathrm{~cm}\). (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

\(\cdot\) Boiling Water at High Pressure. When water is boiled at a pressure of \(2.00 \mathrm{~atm},\) the heat of vaporization is \(2.20 \times 10^{6} \mathrm{~J} / \mathrm{kg}\) and the boiling point is \(120^{\circ} \mathrm{C}\). At this pressure, \(1.00 \mathrm{~kg}\) of water has a volume of \(1.00 \times 10^{-3} \mathrm{~m}^{3},\) and \(1.00 \mathrm{~kg}\) of steam has a volume of \(0.824 \mathrm{~m}^{3}\). (a) Compute the work done when \(1.00 \mathrm{~kg}\) of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

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