Recognizing Thermodynamic Scenarious (Work Done)

In summary, the problem discusses a scenario where a gas confined in a box does work on a barrier that is released. There are three parts to the problem, each with different conditions. The first part deals with an adiabatic expansion, the second part with an isothermal expansion, and the third part with an isobaric expansion. The problem is unclear on whether or not the barrier has thermal inertia, and if it does, how the kinetic energy of the barrier gets converted to internal energy. The third part also involves a change in the number of moles of gas.
  • #1
PhDeezNutz
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Homework Statement
See Below
Relevant Equations
See Below
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Let me start out by saying it's been a LONG time since I've touched any thermodynamics but I'm starting to think that the answer for all 3 parts are the exact same (at least for work)

Namely

##W = \int_{\frac{L^3}{2}}^{L^3} P(V) \, dV = NkT_0 \int_{\frac{L^3}{2}}^{L^3} \frac{1}{V} \, dV = NkT_0 \ln (2)##

My probably wrong reasoning is as follows

for part (a) The only exchanged quantity is volume. There is no source of temperature difference that can cause the temperature of the gas to change so ##T## is constant. ##PV = NkT_0 \Rightarrow P =\frac{NkT_0}{V}## and we just integrate from initial to final volume.

for part (b) I contend that the temperature is also constant. If the right wall is held constant at ##T_0## and the gas is initially at ##T_0## then I don't see why the temperature of the gas would change.

for part (c) If the right side is at equilibrium with the surroundings on the right then I also don't see why temperature would change.

I'd imagine that I'm way off base so I'd just like to thank posters for their patience in advance.
 
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  • #2
PhDeezNutz said:
for part (a) The only exchanged quantity is volume. There is no source of temperature difference that can cause the temperature of the gas to change so ##T## is constant. ##PV = NkT_0 \Rightarrow P =\frac{NkT_0}{V}## and we just integrate from initial to final volume.
I guess the internal barrier can, when released, slide (like a piston in a syringe for example). So the work done on the barrier equals its gain in kinetic energy.

In terms of the motion of the constituent particles, what determines the temperature of a gas?
If a thermally insulated gas does work (e.g. pushing a piston outwards) should the gas's temperature change?
 
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  • #3
Steve4Physics said:
I guess the internal barrier can, when released, slide (like a piston in a syringe for example). So the work done on the barrier equals its gain in kinetic energy.

In terms of the motion of the constituent particles, what determines the temperature of a gas?
If a thermally insulated gas does work (e.g. pushing a piston outwards) should the gas's temperature change?
Now that you mention it. Due to the gas doing work (imparting kinetic energy to the barrier) it would lose kinetic energy and therefore temperature would decrease?

In that case T is not constant. V is certainly not constant. By process of elimination that would make P constant?

I can’t think of why physically that would be the case (P being constant).
 
  • #4
PhDeezNutz said:
Now that you mention it. Due to the gas doing work (imparting kinetic energy to the barrier) it would lose kinetic energy and therefore temperature would decrease?
Yes. To be a bit more specific, the average kinetic energy per particle drops; the average particle speed is reduced.

PhDeezNutz said:
In that case T is not constant. V is certainly not constant. By process of elimination that would make P constant?

I can’t think of why physically that would be the case (P being constant).
Will T increase or decrease when the barrier is pushed? What does that direction of temperature-change tend to do to the pressure?

Will V increase or decrease when the barrier is pushed? What does that direction of volume-change tend to do to the pressure?

So if T and V simultaneously change, as above, will these two effects tend to cancel or to re-enforce?

I don’t want to give too much away, but the process being described in part a) of the question is called an adiabatic expansion. I’m not sure if you are familiar with the term.

There are standard equations used when dealing with adiabatic changes. You are probably expected to recognise part a) is an adiabatic expansion and then use standard equations to answer the question.

You might want to do some background reading on adiabatic changes. (Also isothermal and isobaric ones.) There’s plenty information online (including some good YouTube videos) but try this to one to get the flavour: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html
 
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  • #5
This is a very poorly posed problem. What book is this from?

The barrier is supposed to be very heavy, so it is assumed to accelerate and move very slowly. This implies that the work that the gas does on the barrier is assumed to take place reversibly during the barrier movement. So, in case (a) the work is adiabatic reversible, in case (b) the work is isothermally reversible, and in case (c) the work is isobaricly reversible.

But that is not the end of what happens. At the end of the barrier stroke, the barrier has kinetic energy, but at the final state, the barrier is no longer moving. We are not told whether the barrier has heat capacity to store internal energy. If it does, then part of the kinetic energy of the barrier will be converted to internal energy of both the barrier and the gas. If the barrier does not have thermal inertia, then all the kinetic energy imparted to the barrier during the initial stroke will end up as internal energy of the gas.

How does the kinetic energy of the barrier get converted to internal energy? Well, if the collision of the barrier with the far wall is inelastic, it will be converted to internal energy during the collision. If the collision is elastic, then the barrier will bounce off, and viscous forces in the gas will dissipate the kinetic energy. In this later case, the force of the gas on the barrier will be varying during the rebounds and re-collisions, and, in the end the work done by the gas on the barrier will be zero. So, in cases (a) and (b), the final state of the gas will be the same.

In case (c) the number of moles of gas in the box changes. How would you dope this out?
 
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  • #6
Chestermiller said:
This is a very poorly posed problem. What book is this from?

The barrier is supposed to be very heavy, so it is assumed to accelerate and move very slowly. This implies that the work that the gas does on the barrier is assumed to take place reversibly during the barrier movement. So, in case (a) the work is adiabatic reversible, in case (b) the work is isothermally reversible, and in case (c) the work is isobaricly reversible.

But that is not the end of what happens. At the end of the barrier stroke, the barrier has kinetic energy, but at the final state, the barrier is no longer moving. We are not told whether the barrier has heat capacity to store internal energy. If it does, then part of the kinetic energy of the barrier will be converted to internal energy of both the barrier and the gas. If the barrier does not have thermal inertia, then all the kinetic energy imparted to the barrier during the initial stroke will end up as internal energy of the gas.

How does the kinetic energy of the barrier get converted to internal energy? Well, if the collision of the barrier with the far wall is inelastic, it will be converted to internal energy during the collision. If the collision is elastic, then the barrier will bounce off, and viscous forces in the gas will dissipate the kinetic energy. In this later case, the force of the gas on the barrier will be varying during the rebounds and re-collisions, and, in the end the work done by the gas on the barrier will be zero. So, in cases (a) and (b), the final state of the gas will be the same.

In case (c) the number of moles of gas in the box changes. How would you dope this out?

I’m not sure it’s from any book. We use Schroeder Thermal Physics in undergrad an as far as I can tell this problem is not in it.
 

Related to Recognizing Thermodynamic Scenarious (Work Done)

1. What is work in the context of thermodynamics?

In thermodynamics, work is defined as the energy transferred from one system to another due to a force acting over a distance. It is a measure of the energy required to move an object against a force.

2. How is work related to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. Work is one of the ways in which energy can be transferred from one system to another, and is therefore closely related to the first law.

3. What are some common scenarios in which work is done in thermodynamics?

There are several common scenarios in which work is done in thermodynamics, such as when a gas expands or compresses, when a piston moves in a cylinder, or when a liquid is pumped from one location to another. Work can also be done in electrical, magnetic, and gravitational systems.

4. How is work calculated in thermodynamics?

The formula for calculating work in thermodynamics is W = F x d, where W is work, F is the force applied, and d is the distance over which the force is applied. In some cases, the force or distance may need to be calculated using other equations, such as the ideal gas law or the Bernoulli equation.

5. What is the difference between work and heat in thermodynamics?

Work and heat are both forms of energy transfer, but they differ in how they are transferred. Work involves the transfer of energy due to a force acting over a distance, while heat involves the transfer of energy due to a temperature difference. Additionally, work is a reversible process while heat is not, meaning that work can be completely converted back into its original form of energy while heat cannot.

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