Isothermal expansion of a gas: heat of surroundings

In summary, in an isothermal process for an expanding gas, ΔUsys=0 and Q=-W. To evaluate Qsurr, it is shown that Qsurr=-Qsys. This can be derived from the first principles by considering the combination of the system and surroundings as a new isolated system. Heat is not a property of a system, but rather a property of a process, and is the flow of thermal energy. It is important to understand the physical mechanisms by which heat energy can be transferred across the boundary between a system and its surroundings. It is not always assumed that work in the environment is zero, and temperature can vary. In general, Qenv=-Qsys and Wsys=-Wenv.
  • #1
santimirandarp
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In an isothermal process, for an expanding gas ##\Delta U_{sys}=0## and ##Q=-W## but then,
  • How can we evaluate ##Q_{surr} ##?
It should be ##Q_{surr}=-Q_{sys}##, but I don't know how to show it in equations.

If I try to get the result through the principles:

##\Delta U_{sys}=-\Delta U _{surr}=0## but then nothing appears.

Any help?
 
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  • #2
Heat is not a property of a system, it is a property of a process. Heat is flow of thermal energy, hence the energy leaving leaving the system as heat must end up somewhere, namely the surroundings.
 
  • #3
DrClaude said:
Heat is not a property of a system, it is a property of a process. Heat is flow of thermal energy, hence the energy leaving leaving the system as heat must end up somewhere, namely the surroundings.
I couldn't derive it from 1st principle.
 
  • #4
Well, if the heat that passes through the boundary between the system and the surroundings leaves the surroundings and enters the system, the system must be gaining that amount of heat and the surroundings must be losing it. Where else can it be coming from?
 
  • #5
santimirandarp said:
I couldn't derive it from 1st principle.
If you want to derive it from first principles, consider the combination of system and surroundings as a new isolated system, such that ##Q_{macro}=0##, ##W_{macro}=0##, and ##\Delta U_{macro}=0##. So, $$Q_{system}+Q_{surroundings}=Q_{macro}=0$$
$$W_{system}+W_{surroundings}=W_{macro}=0$$ and $$\Delta U_{system}+\Delta U_{surroundings}=\Delta U_{macro}=0$$
 
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Likes santimirandarp
  • #6
Chestermiller said:
Well, if the heat that passes through the boundary between the system and the surroundings leaves the surroundings and enters the system, the system must be gaining that amount of heat and the surroundings must be losing it. Where else can it be coming from?
Thanks, yes but there is something else. Suppose we say '30J are transferred from the system to the environment as heat' it is completely abstract to me. What do we mean by energy here? It's not something easy to picture as work is, it seems. What do we mean by 'energy transferred as heat' that makes it obvious that the system received 30J as heat too?

I think the confusion comes from not having a good picture about what heat is; and looking through the web isn't very helpful so far.
 
  • #7
santimirandarp said:
Thanks, yes but there is something else. Suppose we say '30J are transferred from the system to the environment as heat' it is completely abstract to me. What do we mean by energy here? It's not something easy to picture as work is, it seems. What do we mean by 'energy transferred as heat' that makes it obvious that the system received 30J as heat too?

I think the confusion comes from not having a good picture about what heat is; and looking through the web isn't very helpful so far.
Are you aware of the physical mechanisms by which heat energy can be transferred across the boundary between a system and its surroundings?
 
  • #8
Chestermiller said:
Are you aware of the physical mechanisms by which heat energy can be transferred across the boundary between a system and its surroundings?
Yes, I am, but still miss something -I'll try to better understand what it is- that makes difficult to see that energy released as heat is absorbed as heat in a system. For example, why isn't this heat diminished by causing some orderly motion in the system?

And also, why isn't it possibly that ##Q_{env}=-(Q+W)_{sys}##?
 
  • #9
santimirandarp said:
Yes, I am, but still miss something -I'll try to better understand what it is- that makes difficult to see that energy released as heat is absorbed as heat in a system. For example, why isn't this heat diminished by causing some orderly motion in the system?

And also, why isn't it possibly that ##Q_{env}=-(Q+W)_{sys}##?
Do you not remember from mechanics that if A does work on B, B does an equal and opposite amount of work on A?
 
  • #10
Chestermiller said:
Do you not remember from mechanics that if A does work on B, B does an equal and opposite amount of work on A?
I do. But isn't usually assumed that work in the environment is zero, and temperature is constant?

And also, do you mean always ##Q_{env}=-Q_{sys}## and ##W_{sys}=-W_{env}##?
 
  • #11
santimirandarp said:
I do. But isn't usually assumed that work in the environment is zero
No, the system can do work on the environment, and the environment can do work on the system.
, and temperature is constant?
No. Consider an ice bath that can receive or discharge heat without its temperature changing. Usually, the environment is assumed to have constant temperature in this sense. But, of course, not always.
And also, do you mean always ##Q_{env}=-Q_{sys}## and ##W_{sys}=-W_{env}##?
Yes, for a closed system.
 

1. What is isothermal expansion of a gas?

Isothermal expansion of a gas refers to the process in which a gas expands at a constant temperature, resulting in a change in volume.

2. How does isothermal expansion of a gas occur?

Isothermal expansion of a gas occurs when the gas is allowed to expand in a container that is in thermal equilibrium with its surroundings. This means that the temperature of the gas remains constant throughout the expansion process.

3. What is the heat of surroundings in isothermal expansion?

The heat of surroundings in isothermal expansion refers to the heat energy that is transferred between the gas and its surroundings during the expansion process. This heat is equal to the work done by the gas as it expands.

4. How is the heat of surroundings calculated in isothermal expansion?

The heat of surroundings in isothermal expansion can be calculated using the formula Q = W = nRTln(V2/V1), where Q is the heat, W is the work, n is the number of moles of gas, R is the gas constant, T is the temperature, and V1 and V2 are the initial and final volumes of the gas.

5. What factors affect the heat of surroundings in isothermal expansion?

The heat of surroundings in isothermal expansion is affected by the number of moles of gas, the gas constant, the temperature, and the change in volume of the gas. Additionally, the heat of surroundings can also be influenced by external factors such as pressure and the type of gas being expanded.

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