# Entropy in compressing gas

1. Mar 28, 2008

### enricfemi

consider a adiabatic system, when we compressing the gas, i found its entropy is decreasing.

but it should increasing in adiabatic systems.

2. Mar 29, 2008

### Mapes

Yes, it should increase or stay constant.

3. Mar 29, 2008

### pterid

It's easy to tie yourself in knots about entropy and the Second Law. I do it all the time...

If you take an ideal gas cylinder, and expand or compress it "reversibly and adiabatically", then the change in the gas's entropy is exactly zero.

Why? Well, entropy is a function of state - if you know two other of the ideal gas's properties, e.g. its temperature $$T$$ and volume $$V$$, you can calculate its entropy directly:

$$S = S_0 + C_V \ln T + R \ln V$$

This is a standard result from classical thermodynamics, and comes out of a form of the First Law. At high-ish temperatures, we can assume that the heat capacity at constant volume, $$C_V$$, is a constant. So, if we move from an initial state $$(T_i, V_i)$$ to a final state $$(T_f, V_f)$$, then the entropy of the gas will change by an amount $$\Delta S$$:

$$\Delta S = C_V \ln \left( \frac{T_f}{T_i} \right) + R \ln \left( \frac{V_f}{V_i} \right)$$

For a reversible adiabatic compression/expansion, there is a link between the initial and final temperatures and volumes (you can derive it from the First Law):

$$\frac{T_f}{T_i} = \left( \frac{V_f}{V_i} \right)^{1 - \gamma}$$

where $$\gamma = C_p / C_V$$ is the ratio of heat capacities. Hence, the change in entropy as a result of the adiabatic compression is:

$$\Delta S = C_V \ln \left( \frac{T_f}{T_i} \right) + R \ln \left( \frac{V_f}{V_i} \right)$$

$$\quad = C_V \ln \left( \frac{V_f}{V_i} \right)^{1-\gamma} + R \ln \left( \frac{V_f}{V_i} \right)$$

$$\quad = C_V (1 - C_p/C_v) \ln \left( \frac{V_f}{V_i} \right) + R \ln \left( \frac{V_f}{V_i} \right)$$

$$\quad = (C_V - C_p + R) \ln \left( \frac{V_f}{V_i} \right)$$

and, as the heat capacities of an ideal gas are related by:

$$C_p = C_V + R$$

we see imediately that $$\Delta S = 0$$ regardless of whether the change was a compression or an expansion.

4. Mar 29, 2008

### olgranpappy

adiabatic and isentropic are not necessarily the same thing. "Adiabatic" mean that no heat flows. This does not mean that the entropy necessarily remains constant. One can only say that
$$\delta Q \leq T\delta S$$
the *equality* does not always hold--if the process is adiabatic *and* reversable than the entropy doesn't change.

The canonical example of an adiabatic but not-isentropic process is the free expansion of a thermally isolated gas. The entropy increases, but there is no change in energy and no heat flows.

5. Mar 31, 2008

### enricfemi

thanks for all your guys! especially petrid!

but i am still confused. i can agree with that:

and in the textbook, it says that the free expansion is always entropy-increased.

i just not sure why the free expansion is not reversibly, and how can i calculate its increased entropy in compressing gas.

Last edited: Mar 31, 2008
6. Mar 31, 2008

### Mapes

You can always calculate along the reversible path. Find out how much work would be extracted if the gas were expanded reversibly. Then re-inject that energy as heat, and calculate the increase in entropy.

Free expansion increases entropy because there is a gradient in pressure, and any process caused by a gradient (in material concentration, temperature, pressure, momentum, etc.) increases entropy. All the work that could have been extracted is instead dissipated in the system as heat in the turbulence during expansion.

7. Mar 31, 2008

### pterid

In my first post, I described a reversible adiabatic expansion. A "free expansion" of an ideal gas (e.g. a Joule expansion) is quite different.

The procedure for a Joule expansion goes something like this:
1. Start with a gas in a "room" of size $$V_i$$.
2. "Next door" to this "room" is another "room" of volume $$(V_f - V_i)$$ which is completely empty - it has no gas in.
3. The two rooms are separated by a partition, so the gas cannot get through.
4. Remove the partition.
5. The gas expands very quickly to fill the whole box.

In A Joule expansion, if the gas is ideal, then its temperature does not change:

$$T_f = T_i$$

(Why? Because its internal energy $$U$$ stays the same, and the internal energy of an ideal gas is a function of its temperature only $$U = U(T)$$. So $$T$$ also stays the same).

The entropy of the gas will change by $$\Delta S$$:

$$\Delta S = C_V \ln \left( \frac{T_f}{T_i} \right) + R \ln \left( \frac{V_f}{V_i} \right)$$

$$= R \ln \left( \frac{V_f}{V_i} \right)$$

because $$T_f / T_i = 1$$.

As you can see, this is very different from $$\Delta S = 0$$ in a reversible expansion!

8. Apr 3, 2008

### enricfemi

can i make it like this:

if we expand the ideal and adiabatic gas slowly enough, it is a reversible adiabatic expansion.

if we expand it more quickly, such as free expansion, its entropy increase.