Explaination for Ngoc Duc’s questions:
1. Let’s observe the diagram:
The 𝑃1 and 𝑃2 here is the pressure and 𝑃2 > 𝑃1 . So the left diagram is the expansion process and the
right one is the compression. In an adiabatic compression process, as the presure increases, the
temperature thus also increases. However, in a real adiabactic compression process, due to internal
friction, the temperature of gas at the end of process will be higher than that of a reversible process.
And the same thing happens when a gas expands adiabatically in real situation.
Due to adiabatic condition, no heat transfer with the surroundings occurs. However, there is heat
generation due to internal friction. Therefore, the entropy is increased.
This can be explained using concept of ideal gas as below:
In reversible adiabatical process:
𝑇1,𝑟𝑒𝑣
𝑃1,𝑟𝑒𝑣
(
)=(
)
𝑇2,𝑟𝑒𝑣
𝑃2,𝑟𝑒𝑣
With 𝛾 =
𝑐𝑃
𝑐𝑣
𝛾−1
𝛾
is the specific heat ratio.
𝑅 = 𝑐𝑃 − 𝑐𝑣 = 𝑐𝑃 −
𝑐𝑃
1
𝛾−1
= 𝑐𝑃 (1 − ) = 𝑐𝑃 (
)
𝛾
𝛾
𝛾
Entropy variation of ideal gas:
𝑇2
𝑃2
Δ𝑠 = 𝑐𝑝 ln ( ) − 𝑅𝑙𝑛 ( )
𝑇1
𝑃1
with: P2 = 𝑃2,𝑟𝑒𝑣 𝑎𝑛𝑑 𝑃1 = 𝑃1,𝑟𝑒𝑣 . (We fix these values)
Now
𝑃
𝑅𝑙𝑛 (𝑃2 )
1
=
𝑃
𝛾−1
𝑐𝑃 ( 𝛾 ) ln (𝑃2,𝑟𝑒𝑣 )
1,𝑟𝑒𝑣
𝑃2,𝑟𝑒𝑣
= 𝑐𝑝 ln (𝑃
1,𝑟𝑒𝑣
𝛾−1
𝛾
)
𝑇
= 𝑐𝑃 ln(𝑇2,𝑟𝑒𝑣 )
1,𝑟𝑒𝑣
In reversible process:
𝑇2 = 𝑇2,𝑟𝑒𝑣
𝑇1 = 𝑇1,𝑟𝑒𝑣
so Δ𝑠 = 0
In present of internal friction: 𝑇2 > 𝑇2,𝑟𝑒𝑣 so ln (𝑇
𝑇2
1,𝑟𝑒𝑣
2.
𝑇
) > ln (𝑇2,𝑟𝑒𝑣 ) → Δ𝑠 > 0
1,𝑟𝑒𝑣
For any kind of irreversibilities that would lead to the increment of entropy of reversible? (it is
always or it only occurs in some cases?)
I do not totally understand your question. However I try to answer it:
During an adiabatic reversible process, entropy variation is zero. Entropy variation of a reversible
process may be not zero. Let’s consider the reversible adiabatic compression as in your first question.
If any kind of irreversibilities happen, the entropy of your isolated system increases. It is a bit
abstract here because you have to consider an isolated system. Let’s try to observe question 3.
Assuming that you have your first system undergoes a reversible compression. However, due to heat
transfer (an irreversibility), it may lose energy under heat form, and thus, its entropy is decreased. But
this leads to the increase in entropy of second system. The the total change in entropy is greater than
zero.
3. What does it mean about the term '' energetically isolated sys.'' + term '' at price of an increase at
least as large as that of another sys.'' ( i mean what is another sys. here, sir? )
Here, to reduce the entropy from a system to bring it back to its initial value, for example, you take
heat from this system in a heat transfer process. Assuming that this heat is added to second system
and these two system combine and form an isolated system. To done this without additional work, the
temperature of your first system must be higher than it of second system. You can refind this idea
from slide 35. Now, the magnitude of entropy decreased in your first system is less than the
magnitude of entropy increased in second system. The price here is the entropy increase in the second
system.
4. At the previous lecture, you explained about the integral of real entropy > integral of reversible
entropy so the system can only do at the reverse direction. Could you kindly explain more
about this?
𝛿𝑞
Actually, I means the comparison integral ∫( 𝑇 ) along the real path and the entropy variation
𝛿𝑞
calcualated by: Δ𝑠 = ∫ ( 𝑇 )
𝑟𝑒𝑣
. Note that the entropy variation is only calculated along a reversible
path not a real path.
Let’s consider two state 1 and 2 of a closed system.
Assuming the direction is from 1 to 2.
𝛿𝑞
𝑇 𝑟𝑒𝑣
The entropy variation: d𝑠 = ( )
First law : 𝑑𝑢 = 𝛿𝑞𝑟𝑒𝑣 + 𝛿𝑤𝑟𝑒𝑣 = 𝛿𝑞𝑟𝑒𝑎𝑙 + 𝛿𝑤𝑟𝑒𝑎𝑙
So
𝛿𝑞𝑟𝑒𝑣
𝑇
−
𝛿𝑞𝑟𝑒𝑎𝑙
𝑇
=
𝛿𝑤𝑟𝑒𝑎𝑙
𝑇
𝑑𝑠 −
𝛿𝑞𝑟𝑒𝑎𝑙 𝛿𝑤𝑟𝑒𝑎𝑙 − 𝛿𝑤𝑟𝑒𝑣 1
1
=
= (−𝑃𝑒𝑥𝑡 𝑑𝑉 − (−𝑃𝑑𝑉)) = (𝑃 − 𝑃𝑒𝑥𝑡 )𝑑𝑉
𝑇
𝑇
𝑇
𝑇
−
𝛿𝑤𝑟𝑒𝑣
𝑇
If 𝑃 > 𝑃𝑒𝑥𝑡 , system expands, 𝑑𝑉 > 0 then 𝑑𝑠 −
𝛿𝑞𝑟𝑒𝑎𝑙
𝑇
If 𝑃 < 𝑃𝑒𝑥𝑡 system compresses, 𝑑𝑉 < 0 then 𝑑𝑠 −
If 𝑃 = 𝑃𝑒𝑥𝑡 , reversible process then 𝑑𝑠 −
Therefore for any process: 𝑑𝑠 ≥
So if you find 𝑑𝑠 <
𝛿𝑞𝑟𝑒𝑎𝑙
𝑜𝑟
𝑇
𝛿𝑞𝑟𝑒𝑎𝑙
𝑇
>0
𝛿𝑞𝑟𝑒𝑎𝑙
𝑇
>0
=0
𝛿𝑞𝑟𝑒𝑎𝑙
𝑇
𝛿𝑞
Δ𝑠 < ∫( 𝑇 ), this process can not occur, or the system can not evolve in
this direction. In reverse direction: Δ𝑠𝑟𝑒𝑣𝑒𝑟𝑠𝑒 = −Δ𝑠 > − ∫
direction.
𝛿𝑞
𝑇
=∫
𝛿𝑞𝑟𝑒𝑣𝑒𝑟𝑠𝑒
.
𝑇
So system evolves in this