WEBVTT
00:00:00.528 --> 00:00:06.547
As an exercise, let's consider isothermal expansion of 2 moles of ideal gas
00:00:06.547 --> 00:00:12.768
from 0.5 m3 to 500 m3 at 300 K.
00:00:12.768 --> 00:00:18.347
We can do this process either reversibly or not reversibly,
00:00:18.347 --> 00:00:22.168
for example adiabatic expansion against vacuum.
00:00:22.168 --> 00:00:27.336
Let's first calculate entropy changes in the reversible process.
00:00:27.336 --> 00:00:35.024
The work done on the system is - resisting pressure times volume changes dV.
00:00:35.024 --> 00:00:41.289
For the reversible isothermal process, the resisting pressure is the same with the gas pressure,
00:00:41.289 --> 00:00:46.273
the system pressure, so it is nRT over V.
00:00:46.273 --> 00:00:51.445
The work is then the integration of nRT over V for the volume changes
00:00:51.445 --> 00:00:56.585
and it gives -1.15 X 10 to 4 Joule.
00:00:56.585 --> 00:01:02.057
For the isothermal process of ideal gas, the internal energy change is zero,
00:01:02.057 --> 00:01:07.401
since the internal energy of ideal gases is a function of temperature only.
00:01:07.401 --> 00:01:18.473
Thus from the first law, the heat delta Q is - delta W, and it is 1.15 x 10 to 4 Joule.
00:01:18.473 --> 00:01:24.369
Now, we can easily calculate the entropy change of the system .
00:01:24.369 --> 00:01:30.257
Since the process is reversible, the heat involved is the reversible heat.
00:01:30.257 --> 00:01:39.577
Inserting the heat just calculated into the definition of entropy gives 38.3 J/K
00:01:39.577 --> 00:01:48.129
For surrounding, the actual heat transferred is the reversible heat in this case, since the process is reversible.
00:01:48.129 --> 00:01:55.649
By the sign convention, the heat transferred “to the surrounding” is -delta Q reversible,
00:01:55.649 --> 00:02:01.185
so the entropy change of the surrounding is -38.3 J/K
00:02:01.185 --> 00:02:06.177
The total entropy change of the universe is the summation of delta S of the system
00:02:06.177 --> 00:02:09.105
and delta S of the surrounding, thus is zero.
00:02:09.105 --> 00:02:17.297
Therefore, we see here that for a reversible process, the total entropy change of the universe is zero,
00:02:17.297 --> 00:02:21.241
and it is the 2nd law of thermodynamics.
00:02:21.241 --> 00:02:29.378
Now let's consider irreversible case. The gas adiabatically expands against the vacuum.
00:02:29.378 --> 00:02:36.882
Since the resisting pressure is zero, the vacuum here, the work done by the system is zero.
00:02:36.882 --> 00:02:41.384
The heat is also zero since the process is adiabatic.
00:02:41.384 --> 00:02:46.400
Then by the 1st law of thermodynamics, the internal energy change is zero.
00:02:46.400 --> 00:02:51.640
Since internal energy of an ideal gas only depends on temperature,
00:02:51.640 --> 00:02:56.160
the temperature is invariant in this case, at 300K.
00:02:56.160 --> 00:03:00.588
So, we can see that this irreversible process and the case one,
00:03:00.588 --> 00:03:05.618
the reversible process have the same initial and final states.
00:03:05.618 --> 00:03:14.609
Here, we intuitively know that this expansion occur spontaneously since its against vacuum.
00:03:14.609 --> 00:03:21.033
The entropy change of the system does not depends on whether the process is reversible or not.
00:03:21.033 --> 00:03:27.825
It only depends on the initial and final states since entropy is a state function.
00:03:27.825 --> 00:03:31.793
If different processes have the same initial and final states,
00:03:31.793 --> 00:03:35.673
the entropy changes of those processes are all the same.
00:03:35.673 --> 00:03:39.858
It is the entropy change for the reversible process.
00:03:39.858 --> 00:03:46.845
So here the entropy change of the system is the same with that of the case I, the reversible process.
00:03:46.845 --> 00:03:51.614
The entropy change of the surrounding is now different from the case I.
00:03:51.614 --> 00:03:58.136
Since the entropy change of the surrounding is not a state function, and it depends on the actual heat.
00:03:58.136 --> 00:04:03.289
The actual heat here is zero, since the process is adiabatic.
00:04:03.289 --> 00:04:11.880
Now, the total entropy change of the universe is + 38.3 J/K, the positive value.
00:04:11.880 --> 00:04:15.008
So, for the spontaneous, irreversible process,
00:04:15.008 --> 00:04:21.160
the total entropy change of the universe is positive and it is the second law.
00:04:21.160 --> 00:04:31.961
Here we summarize the equations describing the 2nd law. The equations are entropy of the system.
00:04:31.961 --> 00:04:37.610
For the close system, the entropy of the system is Q reversible divided by temperature,
00:04:37.610 --> 00:04:43.906
and also the reversible heat is the same with Q actual - lost work.
00:04:43.906 --> 00:04:46.394
For the entropy change of the open system,
00:04:46.394 --> 00:04:53.213
we need to consider the entropy comes in and out accompanied with the materials in transit.
00:04:53.213 --> 00:04:58.309
So the entropy change of the open system is the entropy change for the closed system
00:04:58.309 --> 00:05:04.045
+ entropy goes in with the material - entropy goes out with the material.