What is the role of heat and pressure in a slow thermodynamic process?

In summary, the conversation is about a cylinder and piston experiment where the volume and pressure are changed while keeping the temperature constant. The question is about the logical mistake in thinking that the pressure should be the same in both cases, and it is highlighted that the error is not defining what the piston is pushing against. It is suggested to reword the question to clarify the setup and the process. The original question is based on an example from a university physics book, and the conversation discusses different ways of carrying out the experiment. The concept of gauge pressure is brought up, and it is mentioned that work is done on the environment by the working gas.
  • #1
Nikhil Rajagopalan
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5
Dear experts,
Considering a cylinder and a piston where the volume is V and the pressure is P and the temperature being T, the gas is slowly allowed to expand accepting heat keeping the temperature constant at T to twice the volume. The pressure according to the ideal gas equation should go down to half of P.
Considering the initial and final states as two separate equilibrium, what is the logical mistake in thinking that the pressure in both cases should be the same. To keep the piston in place, it requires same amount of force, that is , the product of atmospheric pressure and area of piston.
 
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  • #2
Nikhil Rajagopalan said:
...what is the logical mistake in thinking that the pressure in both cases should be the same. To keep the piston in place, it requires same amount of force, that is , the product of atmospheric pressure and area of piston.
That's the error. You didn't define what the piston is pushing against at the start of the problem or what the pressure is inside, and this contradicts what you described to be happening.
 
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  • #3
Thank you for highlighting the error.

If the piston is pushing against the atmospheric pressure in both cases, will the question be valid?
 
  • #4
Nikhil Rajagopalan said:
If the piston is pushing against the atmospheric pressure in both cases, will the question be valid?
It depends on what, exactly, the re-worked question asks. Here's how I would word it:

A cylinder with a fixed piston has atmospheric pressure and temperature air inside and outside. When the piston is released, what happens?
 
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  • #5
Thank you,

My question is based on an example from a university physics book which explains how heat absorbed is path dependent. The example is of a cylinder and a piston, initially at 2 liters of volume subjected to 'slow, controlled isothermal expansion' to 5 liters at 300K temperature. It would be helpful to know how this process is actually carried out.
 
  • #6
Nikhil Rajagopalan said:
Thank you,

My question is based on an example from a university physics book which explains how heat absorbed is path dependent. The example is of a cylinder and a piston, initially at 2 liters of volume subjected to 'slow, controlled isothermal expansion' to 5 liters at 300K temperature. It would be helpful to know how this process is actually carried out.
Could you provide the actual example? Because with your setup and my wording of the question, the answer is "nothing happens". There must be another error in your description if the textbook describes something happening.
 
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  • #7
kuruman said:
With questions of this kind it often helps to specify a sequence of events to illustrate your process. In this case I would say forget atmospheric pressure. Start with a massless piston and mass 2m2m on top of it in contact with a heat reservoir. The initial pressure is pi=2mg/Api=2mg/A where AA is the area of the piston. Start removing mass from the piston by very small amounts, say a millionth of mm at a time and wait until enough heat flows into re-establish equilibrium. This makes the process quasi-static. Keep any mass that has been removed at the height where you removed it. Keep doing this until the mass on top of the piston is mm. If your system is the gass and the massless piston, the final state of your system will be the same temperature and a final pressure pf=mg/Apf=mg/A just as you described it. However, you will have a distribution of mass at dif
I think your post got cut off, but I like this construction of the problem...
 
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  • #8
Thank you,
The example was described just as i stated. In fact the explanation was comparing it with a free expansion from 2 liters to 5 liters by bursting a membrane inside the sealed chamber.
 
  • #9
Nikhil Rajagopalan said:
Thank you,
The example was described just as i stated. In fact the explanation was comparing it with a free expansion from 2 liters to 5 liters by bursting a membrane inside the sealed chamber.
Did it actually say the pressure inside was equal to atmospheric? Isn't it obvious that nothing would happen? Please type in the question word for word or post a photo.
 
  • #10
The idea of gauge pressure helps better. So, in fact what i perceive is that, for the above, slow controlled isothermal expansion to happen, a supply of heat alone will not work. From the post which was cut out earlier, i believe the process should be carried out by continuously and gradually varying the pressure along with heating.
 
  • #11
...maybe you interpreted the question wrong and the intent was that the starting pressure is above atmospheric?
 
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  • #12
With questions of this kind I find that it often helps to specify a sequence of events to illustrate your process. In this case I would say forget atmospheric pressure. Start with a massless piston and mass 2m on top of it in contact with a heat reservoir. The initial pressure is pi=2mg/Api=2mg/A where A is the area of the piston. Start removing mass from the piston by very small amounts, say a millionth of m at a time and wait until enough heat flows into re-establish equilibrium. This makes the process quasi-static. Keep any mass that has been removed at the height where you removed it. Keep doing this until the mass on top of the piston is ##m##. If your system is the gas and the massless piston, the final state of your system will be the same temperature and a final pressure pf=mg/Apf=mg/A just as you described it. However, you will have a distribution of mass at different heights such that the work done by the gas in this isothermal expansion is equal to the gain in potential energy of the mass on the piston.

To @russ_watters: I did not get cut off, I pushed "Post reply" too soon and had to delete the post. Yes, the example is about gauge pressure. I wanted to show how work is done on the environment by the working gas. Atmospheric pressure can be added to both ##p_i## and ##p_f## in the example shown in which case the mass removed will not be half of the initial mass. But these are details.
 
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  • #13
Nikhil Rajagopalan said:
To keep the piston in place, it requires same amount of force, that is , the product of atmospheric pressure and area of piston.
No. To keep the piston in place requires twice as much force initially as in the end. I don’t see how that pressure could be provided by the atmosphere.
 
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  • #14
Thank you, @russ_watters and @kuruman , I could not have asked for a better explanation. The explanation I feel fitting is that, for a combination of initial and final states (P,V,T ) and (P/2, 2V, T) , a combination of heat supply and controlled reduction of gauge pressure is the way to execute the process.
 
  • #15
Nikhil Rajagopalan said:
Thank you,

My question is based on an example from a university physics book which explains how heat absorbed is path dependent. The example is of a cylinder and a piston, initially at 2 liters of volume subjected to 'slow, controlled isothermal expansion' to 5 liters at 300K temperature. It would be helpful to know how this process is actually carried out.
As you slowly add heat, the pressure inside the cylinder does not ever become more than only differentially higher than the external (atmospheric) pressure during the entire expansion. To bring this slow heat addition about, one needs to raise the temperature of the reservoir with which the gas is in contact to slightly above 300 K (say 300 + dT) so the heat can flow to the gas at essentially constant temperature. So the work done by the gas on the piston to push back the surrounding atmosphere is ##p_{atm}\Delta V=3\ liter-atm##.
 
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FAQ: What is the role of heat and pressure in a slow thermodynamic process?

1. What is a slow thermodynamic process?

A slow thermodynamic process is a type of thermodynamic process where the system changes gradually over a long period of time. This means that the system is in a state of equilibrium at each step of the process, and the changes are small and continuous.

2. How is a slow thermodynamic process different from a fast thermodynamic process?

A fast thermodynamic process, also known as an irreversible process, is a type of thermodynamic process where the system changes rapidly and is not in equilibrium at each step. In contrast, a slow thermodynamic process is a reversible process where the system is always in equilibrium.

3. What are some examples of slow thermodynamic processes?

Some examples of slow thermodynamic processes include adiabatic expansion of a gas, isothermal expansion of a gas, and reversible changes in temperature or pressure.

4. What are the implications of a slow thermodynamic process?

A slow thermodynamic process has important implications in terms of energy transfer and efficiency. Since the system is always in equilibrium, there is no energy loss during the process, making it more efficient. Additionally, slow processes allow for better control and measurement of the system.

5. How is a slow thermodynamic process used in practical applications?

Slow thermodynamic processes are commonly used in practical applications such as refrigeration and heat engines. In refrigeration, a slow process allows for the efficient transfer of heat from a colder region to a hotter region. In heat engines, a slow process allows for the conversion of heat energy into mechanical work with high efficiency.

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