- #1
zenterix
- 619
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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
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.
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
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.