Thermodynamics: Work Under Isothermal Condition

In summary: Otherwise this thread might be left in the unanswered forum and never get looked at again. In summary, the conversation discusses the confinement of a 2-D ideal gas in an isotropic cone potential with a positive parameter lambda and the displacement vector r. The Helmholtz free energy A is determined at temperature T, and the work done (delta W) when varying lambda is computed following Einstein's formula. It is then shown that this work agrees with the variation of the Helmholtz free energy due to the same parameter change. The canonical partition function and Hamiltonian are mentioned, but there is uncertainty about how to compute the sum and a potential error in the calculation of delta W. Further clarification or assistance is requested.
  • #1
cox24
2
2

Homework Statement


[/B]
One mole of 2-D ideal gas is confined in an isotropic cone potential:

[itex] U = \lambda |r| [/itex]

where [itex] \lambda [/itex] is a positive parameter and [itex] r [/itex] is the displacement vector (2-dimensional) from the origin. The mass of each molecule is [itex] m [/itex].

(1) Determine the Helmholtz free energy [itex] A [/itex] of this gas confined in the potential at temperature [itex] T [/itex].

(2) If you want to vary [itex] \lambda → \lambda + \delta \lambda [/itex], you must do some work [itex] \delta W [/itex]. Compute [itex] \delta W [/itex], following Einstein.

(3) Show, as thermodynamics tells us, that [itex] \delta W [/itex] in (2) agrees with the variation of [itex] \delta A [/itex] of the Helmholtz free energy [itex] A [/itex] due to the same parameter change in (2).

Homework Equations



[itex] A = -k_{B}TlogZ [/itex]

[itex] Z = \sum e^{\frac{-H}{k_{B}T}} [/itex][/B]

H = Hamiltonian
[itex]k_{B}[/itex]=Boltzmann constant

Einstein: [itex] \delta W = < \delta H > [/itex]


The Attempt at a Solution


[/B]
(1) For an ideal gas, the system hamiltonian (with the potential term added) is:

[itex] H = \sum \frac{p_{i}^2}{2m} + \lambda |r| [/itex]

so,

[itex] Z = \sum e^{\frac{-(p_{x}^2 + p_{y}^{2})}{2mk_{B}T} - \frac{\lambda |r|}{k_{B}T}} [/itex]

How is this sum computed for the canonical partition function of a 2-D ideal gas with the hamiltonian included?

Also, for (2),

[itex] \delta W = < \delta \lambda H > = \delta \lambda < |r| > [/itex]

this answer is wrong, but I really don't see what I'm missing here.

Any help would be appreciated, thanks.
 
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  • #2
Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

Related to Thermodynamics: Work Under Isothermal Condition

1. What is the definition of work under isothermal condition in thermodynamics?

In thermodynamics, work under isothermal condition refers to the work done on or by a system while keeping its temperature constant. This means that the system is in thermal equilibrium with its surroundings throughout the process, and there is no change in its temperature.

2. How is work calculated under isothermal condition?

The work done under isothermal condition is calculated using the formula W = -nRT ln(Vf/Vi), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, Vf is the final volume of the gas, and Vi is the initial volume of the gas. This formula is derived from the first law of thermodynamics and the ideal gas law.

3. What is the significance of work under isothermal condition?

Work done under isothermal condition is important because it allows us to study the behavior of a system while keeping its temperature constant. This is particularly useful in understanding the behavior of gases, as it allows us to see how changes in volume affect the pressure and temperature of the gas.

4. Can work be negative under isothermal condition?

Yes, work done under isothermal condition can be negative. This occurs when the volume of the gas decreases, meaning that the system is doing work on the surroundings. In this case, the gas is losing energy, which results in a decrease in its temperature.

5. How does work under isothermal condition differ from work under adiabatic condition?

The main difference between work done under isothermal condition and work done under adiabatic condition is that in the former, the temperature of the system remains constant, while in the latter, there is no heat transfer between the system and its surroundings. This means that in an isothermal process, the work done is equal to the change in internal energy, while in an adiabatic process, the work done is equal to the change in internal energy plus the change in thermal energy.

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