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

AI Thread Summary
The discussion revolves around the irreversibility of gas expansion and its implications for entropy changes. It establishes that while the initial and final states of the gas are the same, leading to a change in entropy of zero, the process is irreversible, suggesting an increase in total entropy. A key point of contention is whether heat exchange with the surroundings occurs during this irreversible expansion, with the assumption being that no heat is exchanged. The conversation highlights the importance of thermally insulating the system to maintain the same final state as in reversible processes. Ultimately, the first law of thermodynamics supports the conclusion that no work is done on the surroundings, resulting in a net heat exchange of zero.
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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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You can imagine the process to take place in a thermally insulating dewar.
 
@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.
 
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.
 
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