No heat exchange with the surroundings in an irreversible expansion of an ideal gas?

In summary, an irreversible expansion of an ideal gas occurs without heat exchange with the surroundings, meaning the process is adiabatic. During this expansion, the gas does work against external pressure, leading to a decrease in internal energy and temperature. The lack of heat transfer highlights the non-equilibrium nature of the process, distinguishing it from reversible expansions, where both heat and work can be exchanged with the surroundings.
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Homework Statement
1) Half a mole of an ideal as expands isothermally and reversibly at 298.15K from a volume of 10L to a volume of 20L.

a) What is the change in entropy of the gas?
b) How much work is done on the gas?
c) What is ##q_{surr}##, the heat change of the surroundings?
d) What is the change in the entropy of the surroundings?
e) What is the change in entropy of the system plus surroundings?
Relevant Equations
2) Now consider that the expansion in the preceding question occurs irreversibly by simply opening a stopcock and allowing the gas to rush into an evacuated bulb of 10L volume.

a) What is the change in entropy of the gas?
b) How much work is done on the gas?
c) What is ##q_{surr}##, the heat change of the surroundings?
d) What is the change in the entropy of the surroundings?
e) What is the change in entropy of the system plus surroundings?
My doubts are about the second question above, ie the irreversibly expansion.

For the first question, we have

a)

$$dS=\frac{dq_{rev}}{T}=\frac{nR}{V}dV$$

$$\implies \Delta S=nR\ln{\frac{V_2}{V_1}}=2.88\mathrm{\frac{J}{K}}$$

b)

$$q_{rev}=T\Delta S=298.15\text{K}\cdot 2.88\mathrm{\frac{J}{K}}=859\text{J}$$

c)

$$q_{surr}=-q_{rev}=-859\text{J}$$

d)

$$dS_{surr}=-\frac{dq_{rev}}{T}$$

$$\Delta S_{surr}=-\int\frac{dq_{rev}}{T}=-\Delta S=-2.88\mathrm{\frac{J}{K}}$$

e)

$$\Delta S=0$$

For the second question, which is where I have issues, we have

a)

Though the process now is irreversible, the start and end states seem to be the same and so the change in entropy is the same, namely $\Delta S=0$.

b)

When the gas expands without any external pressure working against it, the work done is zero.

My doubts start at this point, namely, item (c) of question 2.

The book I am reading says

c) No heat is exchanged with the surroundings.

d) The entropy of the surroundings does not change.

e) The entropy of the system plus surroundings increases by 2.88 ##\mathrm{J\cdot K^{-1}}##. Since this is an irreversible process we expect the entropy to increase.

How do we know no heat is exchanged with the surroundings?

If we accept that no heat is exchanged, then of course the entropy of the surroundings does not change.
 
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  • #2
You can imagine the process to take place in a thermally insulating dewar.
 
  • #3
@DrDu I can certainly imagine an apparatus used to insulate the system. But I don't see why there is the assumption that there is such an apparatus in the setup of question 2. After all, in question 1 there was no such apparatus.
 
  • #5
In question 2, it is assumed that the stopcock can exchange heat between the two sections so that the final state of the system is the same as in question 1. The combined system is rigid, so no work is done on the surroundings, and , since the temperature does not change, the change in internal energy of the combined system is zero. So, from the first law of thermodynamics, Q=0.
 

FAQ: No heat exchange with the surroundings in an irreversible expansion of an ideal gas?

What does it mean for there to be no heat exchange with the surroundings during an irreversible expansion?

No heat exchange with the surroundings means that the system is adiabatic, meaning that there is no transfer of heat energy into or out of the system during the process. In the case of an irreversible expansion of an ideal gas, the gas expands without absorbing or releasing heat, which can happen in a perfectly insulated system.

How does an irreversible expansion differ from a reversible expansion?

An irreversible expansion occurs spontaneously and does not follow a thermodynamic equilibrium process, meaning it cannot be reversed without changing the surroundings. In contrast, a reversible expansion is carried out infinitely slowly, allowing the system to remain in equilibrium at all times, and can be reversed without any net change to the surroundings.

What happens to the temperature of the ideal gas during an irreversible expansion with no heat exchange?

During an irreversible expansion of an ideal gas with no heat exchange, the temperature of the gas typically decreases if it expands against a vacuum or into a larger volume without doing work on the surroundings. This is because the internal energy of the gas decreases, leading to a drop in temperature, as the gas does work to expand.

Can the first law of thermodynamics be applied to this process?

Yes, the first law of thermodynamics can still be applied to this process. The first law states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. In the case of an adiabatic irreversible expansion, since there is no heat exchange (Q = 0), the change in internal energy is equal to the negative of the work done by the gas.

What are the implications of this process in real-world applications?

The implications of no heat exchange during an irreversible expansion of an ideal gas are significant in various fields such as engineering and thermodynamics. It helps in understanding processes like free expansion, where gases expand into a vacuum, and plays a role in designing engines and refrigeration systems where efficiency and heat management are critical factors.

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