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During an adiabatic expansion the temperature of 0.450 mol of argon (Ar) drops from 66.0C to 10.0C. The argon may be treated as an ideal gas. (a) Draw a pV-diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

Short Answer

Expert verified
The gas does 485.4 J of work, and the change in internal energy is -485.4 J.

Step by step solution

01

Convert Temperature to Kelvin

First, convert the initial and final temperatures from Celsius to Kelvin. Use the formula T(K)=T(\degreeC)+273.15. For the initial temperature, Ti=66+273.15=339.15K. For the final temperature, Tf=10+273.15=283.15K.
02

Draw a pV-Diagram

In an adiabatic process, the pressure p and volume V of an ideal gas change without heat exchange with the surroundings. On a pV-diagram, this process is represented by a curve that slopes downward from left to right, as volume increases and pressure decreases.
03

Apply the Ideal Gas Law

Use the relation PVγ=constant for adiabatic processes, where γ=Cp/Cv. For a monoatomic gas like argon, γ=5/3. This relationship helps in understanding the shape and behavior of the curve in the pV-diagram.
04

Calculate Work Done by the Gas

For an adiabatic process, work done W can be calculated using the formula for reversible adiabatic processes: W=nRγ1(TiTf). Substituting n=0.450 mol, R=8.314 J/mol K, γ=5/3, Ti=339.15 K, and Tf=283.15 K into the equation: W=0.450×8.314(5/3)1(339.15283.15). Solve this to find W=485.4 J.
05

Calculate the Change in Internal Energy

The change in internal energy ΔU for a monatomic ideal gas can be found using the formula ΔU=32nR(TfTi). Substituting the known values: ΔU=32×0.450×8.314×(283.15339.15). Solve this to find ΔU=485.4 J, which matches the work done but opposite in sign, as expected due to energy conservation in adiabatic processes.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics. It is used to relate the pressure, volume, and temperature of a gas, assuming the gas behaves ideally. This means we assume the gas particles have no significant interactions and occupy no volume themselves.

The Ideal Gas Law is formulated as: PV=nRT where:
  • P is the pressure of the gas
  • V is the volume of the gas
  • n is the amount of substance in moles
  • R is the ideal gas constant, approximately equal to 8.314 J/mol K
  • T is the temperature of the gas in Kelvin
In adiabatic processes, like the one described in the exercise, there is no heat exchange with the surroundings. Therefore, the product PVγ remains constant, where γ=CpCv is specific to the type of gas. For monoatomic gases like argon, γ is typically 53. This constant relationship influences the pressure-volume behavior during the adiabatic expansion or compression.
Understanding the Ideal Gas Law and its application in adiabatic processes allows one to predict how gases respond to changes in temperature, volume, and pressure.
Internal Energy
Internal energy is the total energy contained within a system due to microscopic motions and interactions of its molecules. For ideal gases, internal energy is primarily dependent on the kinetic energy of the molecules since they are assumed to have no potential interactions.

In an adiabatic process, the change in internal energy ΔU is directly related to the change in temperature. This is calculated using the formula: ΔU=32nR(TfTi) for a monoatomic ideal gas. Here, the number of moles n, the ideal gas constant R, and the initial and final temperatures Ti and Tf are used as inputs.

As seen in the exercise, the internal energy change ΔU for the argon gas was calculated to be 485.4 J. This implies that the system lost energy, reflecting the decrease in temperature through the adiabatic expansion. No heat was absorbed or emitted, affirming that work done by the system led to its lowering internal energy.
Work Done by Gas
Work done by gas in an adiabatic process depends on the initial and final states of the system. More specifically, it is the energy transferred due to the expansion or compression of the gas without heat exchange with the surroundings.

For a reversible adiabatic process, we compute the work done W using W=nRγ1(TiTf). This incorporates the number of moles n, the ideal gas constant R, and other specific parameters including the type-specific heat capacity ratio γ. For the argon gas in the exercise, this resulted in W=485.4 J.

Notice how the work done matches in magnitude with the change in internal energy but is opposite in sign. This highlights the principle of conservation of energy, confirming that, in an adiabatic process, the work done by the gas is equal in magnitude but opposite to the energy change, since no heat is transferred.

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Most popular questions from this chapter

A cylinder contains 0.250 mol of carbon dioxide (CO2) gas at a temperature of 27.0C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0C. Assume that the CO2 may be treated as an ideal gas. (a) Draw a pV-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?

Three moles of argon gas (assumed to be an ideal gas) originally at 1.50 × 104 Pa and a volume of 0.0280 m3 are first heated and expanded at constant pressure to a volume of 0.0435 m3, then heated at constant volume until the pressure reaches 3.50 × 104 Pa, then cooled and compressed at constant pressure until the volume is again 0.0280 m3, and finally cooled at constant volume until the pressure drops to its original value of 1.50 × 104 Pa. (a) Draw the pV-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?

A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 × 105 Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a pV-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

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 × 105 Pa and 300 K and occupies a volume of 1.50 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 K. This continues until the pressure reaches 2.50 × 104 Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a pV-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?

Propane gas (C3H8) behaves like an ideal gas with Υ = 1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant pressure.

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