1. x
Thermodynamics

2. Thermodynamics
Laws of Thermodynamics
Zero Law
First Law
Second Law
Third Law
State
𝑃[𝑃𝑎], 𝑉[𝑚3
], 𝑇[𝐾]
1st Law
𝑈 = 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
To change 𝑈1 to 𝑈2, there are 2 ways;
(i) Change Q
(ii) Change W
***When energy (𝑄) is supplied to a
system, the system will do work (𝑊)
In reality, the work done is lesser than
the energy supplied.
𝑄 𝑖𝑠 𝑠𝑢𝑝𝑝𝑙𝑦 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑄 > 0
𝑊 𝑖𝑠 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑊 > 0
If the system releases the energy
𝑄 < 0
If work is done on the system
𝑊 < 0
𝑠𝑦𝑠𝑡𝑒𝑚
𝑄 𝑊
∆𝑈 = 𝑄 − 𝑊
𝑈 =
𝐷𝑂𝐹
2
𝑁𝑘𝑇
Internal energy
Work done
Work done by gas; 𝑊 > 0
Work done on gas; W < 0
Since, ∆𝑊 = 𝐹∆𝑠
𝐹 =
𝑑𝑊
𝑑𝑠
Since, 𝑃 =
𝐹
𝐴
, 𝐹 = 𝑃𝐴
𝑑𝑊 = 𝑃 𝐴𝑑𝑠
𝑑𝑊 = 𝑃𝑑𝑉
𝑊 = 𝑑𝑊 = 𝑃𝑑𝑉
𝑉 𝑓
𝑉 𝑖
𝑊 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑠𝑦𝑠𝑡𝑒𝑚
𝑷-𝑽 diagram
***the area under the graph is work
down
𝑊 = 𝑃𝑑𝑉
𝑉 𝑓
𝑉 𝑖
𝑃
𝑉
𝑉𝑓𝑉𝑖
𝑊 = 𝑃(𝑉𝑓 − 𝑉𝑖)

3. ∆𝑈 = 𝑄 − 𝑊
Process Condition
Isochoric
Isovolumetric
Constant volume
𝑊 = 0
Path dependent for 𝑸 & 𝑾
; although the final and initial states
(points) are the same, work done may
not be equal.
Path independent for ∆𝑼
; it is state function
; a state function is a property of a
system that depends only on the current
state of the system, not on the way in
which the system acquired that state
(independent of path)
∆𝑈 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑝𝑎𝑡ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
𝑊𝑖→𝑓→𝑖 = 𝑊𝑖→𝑓 + 𝑊𝑓→𝑖
𝑃
𝑉
𝑖
𝑓
𝑊𝑓→𝑖 < 0
𝑃
𝑉
𝑖
𝑓
𝑊𝑖→𝑓 > 0
𝑃
𝑉
𝑖
𝑓
𝑊𝑖→𝑓→𝑖 > 0
𝑊 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑔𝑎𝑠
= 𝑎𝑟𝑒𝑎 𝑖𝑛𝑠𝑖𝑑𝑒
𝑊 > 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑊 < 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑎𝑛𝑡𝑖 − 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑃
𝑉
𝑖
𝑓
𝑃
𝑉
∆𝑈 = 𝑄

4. ∆𝑈 = 𝑄 − 𝑊
Process Condition
Isobaric
Constant pressure
∆𝑃 = 0
Isothermal
Constant
temperature
∆𝑈 = 0
𝑃𝑉 = 𝑁𝑘𝑇 → 𝑃 =
𝑁𝑘𝑇
𝑉
𝑊 = 𝑃𝑑𝑉
𝑉 𝑓
𝑉𝑖
=
𝑁𝑘𝑇
𝑉
𝑑𝑉
𝑉 𝑓
𝑉𝑖
= NkT
1
𝑉
𝑑𝑉
𝑉 𝑓
𝑉𝑖
= 𝑁𝑘𝑇𝑙𝑛
𝑉𝑓
𝑉𝑖
= 𝑁𝑘𝑇𝑙𝑛
𝑃𝑖
𝑃𝑓
𝑃
𝑉
𝑊 = 𝑃∆𝑉
𝑃
𝑉
∆𝑈 = 0, Q = W
𝑇1
𝑇2
𝑇3
𝑊 = 𝑁𝑘𝑇 𝑙𝑛
𝑉𝑓
𝑉𝑖
= 𝑁𝑘𝑇 𝑙𝑛
𝑃𝑖
𝑃𝑓
∆𝑈 = 𝑄 − 𝑊
Process Condition
Adiabatic
Constant heat
𝑄 = 0
Heat capacity
𝐽
𝐾
Specific heat capacity
𝐽
𝑘𝑔 𝐾
Molar heat capacity
𝐽
𝑚𝑜𝑙 𝐾
Specific heat capacity at constant V, (𝒄 𝑽)
∆𝑄 = 𝑀𝑐∆𝑇
𝑐 =
1
𝑀
∆𝑄
∆𝑇
=
1
𝑀
𝜕𝑄
𝜕𝑇
At constant V, ∆𝑈 = 𝑄
𝑐 𝑉 =
1
𝑀
𝜕𝑄
𝜕𝑇 𝑉
=
1
𝑀
𝑑𝑈
𝑑𝑇
***For an ideal gas,
𝑈 =
3
2
𝑁𝑘𝑇
∴ 𝑐 𝑉 =
1
𝑀
𝑑𝑈
𝑑𝑇
=
1
𝑀
𝑑
𝑑𝑇
3
2
𝑁𝑘𝑇
=
3𝑁𝑘
2𝑀
=
3𝑁𝑘
2 𝑁𝑚
=
3𝑘
2𝑚
𝑀 = 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙𝑒
Specific heat capacity at constant P, (𝒄 𝑷)
∆𝑄 = 𝑀𝑐 𝑉∆𝑇
𝑐 𝑃 =
1
𝑀
𝜕𝑄
𝜕𝑇 𝑃
=
1
𝑀
𝜕
𝜕𝑇
𝑈 + 𝑊
𝑃
=
1
𝑀
𝑑𝑈
𝑑𝑇
+
𝜕
𝜕𝑇
𝑃𝑉
𝑇3 > 𝑇2 > 𝑇1
𝑐 𝑉 =
3𝑘
2𝑚

5. ∆𝑈 = 𝑄 − 𝑊
Process Condition
Adiabatic
Constant heat
𝑄 = 0
Specific heat capacity at constant P, (𝒄 𝑷)
∆𝑄 = 𝑀𝑐 𝑉∆𝑇
𝑐 𝑃 =
1
𝑀
𝑑𝑈
𝑑𝑇
+
𝜕
𝜕𝑇
𝑃𝑉
𝑐 𝑃 =
1
𝑀
𝑑𝑈
𝑑𝑇
+
𝜕
𝜕𝑇
𝑁𝑘𝑇
𝑐 𝑃 =
1
𝑀
𝑑𝑈
𝑑𝑇
+ 𝑁𝑘 =
1
𝑀
𝑑𝑈
𝑑𝑇
+
𝑁𝑘
𝑁𝑚
Since 𝑐 𝑉 =
1
𝑀
𝑑𝑈
𝑑𝑇
,
𝒄 𝑷 = 𝒄 𝑽 +
𝒌
𝒎
For an ideal gas, 𝑐 𝑉 =
3𝑘
2𝑚
,
𝑐 𝑃 =
3𝑘
2𝑚
+
𝑘
𝑚
***We define 𝜸 =
𝒄 𝑷
𝒄 𝑽
, (adiabatic index)
γ =
𝑐 𝑉 +
𝑘
𝑚
𝑐 𝑉
= 1 +
𝑘
𝑚𝑐 𝑉
For an ideal gas,
γ =
3𝑘
2𝑚
+
𝑘
𝑚
3𝑘
2𝑚
=
5
3
𝑐 𝑃 =
5𝑘
2𝑚
𝛾 =
𝑐 𝑃
𝑐 𝑉
∆𝑈 = 𝑄 − 𝑊
Process Condition
Adiabatic
Constant heat
𝑄 = 0
Since 𝑄 = 0,
∆𝑈 = −𝑊
And since ∆𝑈 is a state function (path
independent),
∆𝑈 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐= ∆𝑈𝑖𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐 when the
initial states and final states of the 2
processes are the same.
∴ 𝑑𝑈 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 = 𝑑𝑈𝑖𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐 = 𝑄
= 𝑀𝑐 𝑉 𝑑𝑇
𝑑𝑈 = −𝑑𝑊
𝑀𝑐 𝑉 𝑑𝑇 = −𝑃𝑑𝑉
𝑑𝑇 = −
𝑃
𝑀𝐶 𝑣
𝑑𝑉
From an ideal gas equation, 𝑃𝑉 = 𝑁𝑘𝑇
𝑑 𝑃𝑉 = 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑁𝑘𝑑𝑇
∴ 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑁𝑘 −
𝑃
𝑀𝐶 𝑣
𝑑𝑉
Since 𝑀 = 𝑁𝑚,
𝑃𝑑𝑉 + 𝑉𝑑𝑃 = −
𝑘
𝑚𝐶 𝑣
𝑃𝑑𝑉
Since, γ =
𝑐 𝑉+
𝑘
𝑚
𝑐 𝑉
= 1 +
𝑘
𝑚𝑐 𝑉
𝑘
𝑚𝑐 𝑉
= 𝛾 − 1
∴ 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = − 𝛾 − 1 𝑃𝑑𝑉
𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 1 − 𝛾 𝑃𝑑𝑉
𝑉𝑑𝑃 = −𝛾𝑃𝑑𝑉
𝑑𝑃
𝑃
= −𝛾
𝑑𝑉
𝑉
∴ ln 𝑃 = −𝛾 ln 𝑉 + 𝑐
ln 𝑃 + 𝛾 ln 𝑉 = 𝑐

6. ∆𝑈 = 𝑄 − 𝑊
Process Condition
Adiabatic
Constant heat
𝑄 = 0
ln 𝑃 + 𝛾 ln 𝑉 = 𝑐
ln 𝑃𝑉 𝛾
= 𝑐
Since 𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
∴ 𝑃𝑉 𝛾
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Since 𝑃𝑉 = 𝑁𝑘𝑇, 𝑃 =
𝑁𝑘𝑇
𝑉
𝑁𝑘𝑇
𝑉 𝑖
𝑉𝑖
𝛾
=
𝑁𝑘𝑇
𝑉 𝑓
𝑉𝑓
𝛾
***the graph of adiabatic process is
more steeper than the graph of
isothermal process.
𝑃𝑖 𝑉𝑖
𝛾
= 𝑃𝑓 𝑉𝑓
𝛾
𝑇𝑖 𝑉𝑖
𝛾−1
= 𝑇𝑓 𝑉𝑓
𝛾−1
𝑃
𝑉
𝑖
𝑓
𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙
𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐
∆𝑈 = 𝑄 − 𝑊
Process Condition
Free expansion ∆𝑈 = 𝑄 = 𝑊 = 0
***there is no graph plotted joining the
initial and final point because there is
no work done (area under graph).
***Note that Free expansion is the
subset of Adiabatic process.
Cyclical process ∆𝑈 = 0
***the initial and final point is the same.
𝑇𝑖 = 𝑇𝑓
𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓
𝑃
𝑉
𝑖
𝑓
𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙
𝑃
𝑉
𝑖
𝑓
𝑊𝑖→𝑓→𝑖 > 0

7. ∆𝑈 = 𝑄 − 𝑊
Process Condition
Cyclical process ∆𝑈 = 0
𝑊 > 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑊 < 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑎𝑛𝑡𝑖 − 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Since 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 𝑎𝑛𝑑 𝑑𝑈 = 0,
𝑑𝑄 = 𝑑𝑊
2nd Law of Thermodynamics
Reversible and Irreversible process
Most of the processes in
Thermodynamics are reversible.
2 mains irreversible processes
(i). Transfer of heat energy between 2
systems with different temperatures.
Such as; isobaric and isochoric
(ii). Free expansion (not quasi-static)
***isothermal and adiabatic are
reversible
Entropy 𝑺 & Entropy change ∆𝑺
;Entropy shows the property of the
state which is disorder.
For example,
𝑆 𝑔𝑎𝑠 > 𝑆𝑙𝑖𝑞𝑢𝑖𝑑 > 𝑆𝑠𝑜𝑙𝑖𝑑
For Entropy change,
∆𝑆 = 𝑆𝑓 − 𝑆𝑖
∆𝑆 = 0 → 𝑆𝑖 = 𝑆𝑓 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Suppose there is an initial state,
𝑑𝑆 =
𝑑𝑄
𝑇
***A tiny change in entropy is equal to
the tiny input energy over a period of
time.
***Entropy is a state function, meaning
that Entropy change does not depend
on the path it takes.
Specifically for isothermal,
𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
∴ ∆𝑆 =
1
𝑇
𝑑𝑄
𝑓
𝑖
=
𝑄
𝑇
From the 1st Law of Thermodynamics,
∆𝑈 = 𝑄 − 𝑊
For isothermal ∆𝑈 = 0,
𝑄 = 𝑊
∴ ∆𝑆 =
𝑄
𝑇
=
𝑊
𝑇
Since 𝑊 = 𝑁𝑘𝑇 𝑙𝑛
𝑉 𝑓
𝑉 𝑖
,
∆𝑆 =
𝑁𝑘𝑇 𝑙𝑛
𝑉𝑓
𝑉𝑖
𝑇
∆𝑆 = 𝑆𝑓 − 𝑆𝑖 =
𝑑𝑄
𝑇
𝑓
𝑖
∆𝑆 = 𝑁𝑘 𝑙𝑛
𝑉𝑓
𝑉𝑖

8. 2nd Law of Thermodynamics
For any closed system,
1. If 𝑠 of the system increases ∆𝑆 > 0 ,
it is irreversible process.
2. If 𝑠 of the system is constant
∆𝑆 = 0 , it is reversible process.
In Thermodynamics, a closed system
means that there is no transfer of heat
energy between the system and
surroundings.
***Only Adiabatic is a closed system
because 𝑄 = 0.
However, to increase the boundary of
the application of the 2nd Law, we must
consider,
Closed system = system + surroundings
so that there is no transfer of heat
energy in and out of our closed system.
For Reversible processes, ∆𝑺 = 𝟎
; isothermal and adiabatic
Isothermal, ∆𝑈 = 0, 𝑄 = 𝑊
∆𝑆𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚≥ 0
∆𝑆 =
𝑑𝑄
𝑇
𝑓
𝑖
Since 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
∆𝑆 =
𝑄
𝑇
=
𝑊
𝑇
From 𝑖 → 𝑓, 𝑊 < 0
(the gas is compressed)
∴ ∆𝑆 𝑔𝑎𝑠=
𝑊𝑖→𝑓
𝑇
= −
𝑄
𝑡
< 0
***This is not a closed system since
𝑄 ≠ 0.
Consider the surrounding.
Since 𝑄 𝑔𝑎𝑠 < 0, heat energy is lost to
the surrounding.
𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = −𝑄 𝑔𝑎𝑠
∴ ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠=
𝑄
𝑇
∆𝑆𝑡𝑜𝑡𝑎𝑙=
𝑄
𝑇
−
𝑄
𝑇
= 0 (𝑝𝑟𝑜𝑣𝑒𝑛)
For Reversible processes, ∆𝑺 > 𝟎
; Free expansion, isochoric, isobaric
For Free expansion, ∆𝑈 = 𝑄 = 𝑊 = 0
Only gas itself is already a closed system
since 𝑄 = 0
Since, ∆𝑠 is a state function
∆𝑆 𝐹𝑟𝑒𝑒 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛= ∆𝑠𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙
∆𝑆 =
𝑄
𝑇
> 0 (𝑝𝑟𝑜𝑣𝑒𝑛)
𝑃
𝑉
∆𝑈 = 0, Q = W
𝑉
𝑖
𝑓
𝑖
∆𝑆𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚≥ 0

9. Engine
𝑄 𝐻 → 𝑊𝑜𝑢𝑡 𝑝𝑢𝑡 + 𝑄 𝐿
𝐻 = 𝐻𝑖𝑔ℎ, 𝐿 = 𝐿𝑜𝑤
Engine converts heat energy to work.
However, not all heat energy will be
entirely converted to work.
(𝑄 𝐻 → 𝑊𝑜𝑢𝑡 𝑝𝑢𝑡 + 𝑄 𝐿)
𝑄 𝐻 > 𝑄 𝐿
Hot reservoir is the source heat energy
𝑄 𝐻 , has the temperature of 𝑇 𝐻.
Cold reservoir is where it receives 𝑄 𝐿,
has the temperature of 𝑇𝐿.
***𝑇 𝐻 = 𝑇𝐿; we assume that the hot and
cold reservoir are very large. Small 𝑄 𝐻
and 𝑄 𝐿 do not make any noticeable
change of their temperature.
Working substance
; A substance, such as a fluid, used to
effect a thermodynamic or other change
in a system.
***The lost or gain of heat energy
involves only the working substances
and not the engine itself.
Carnot Engine
; It is an ideal engine.
; We assume that all processes in Carnot
Engine are reversible.
; All work is done by the working
substance only and not the frictional
force or viscosity.
Carnot Cycle
There are 4 steps.
𝑎 → 𝑏 → 𝑐 → 𝑑 → 𝑎
***
𝑎 → 𝑏, 𝑐 → 𝑑 ∶ 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙
𝑏 → 𝑐, 𝑑 → 𝑎 ∶ 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐
𝒂 → 𝒃 ∶ 𝑸 𝑯 𝒊𝒏
𝒄 → 𝒅: 𝑸 𝑳 𝒐𝒖𝒕
Since it is a clockwise cycle,
𝑊𝑎→𝑏→𝑐→𝑑→𝑎 > 0
Efficiency, η
η =
𝑊𝑜𝑢𝑡
𝐸𝑖𝑛
=
𝑊
𝑄 𝐻
=
𝑄 𝐻 − 𝑄 𝐿
𝑄 𝐻
= 1 −
𝑄 𝐿
𝑄 𝐻
ℎ𝑜𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
𝐸𝑛𝑒𝑟𝑔𝑖𝑛𝑒
𝑐𝑜𝑙𝑑 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
𝑄 𝐻
𝑄 𝐿
𝑇 𝐻
𝑇𝐿
𝑊
𝑃
𝑉
𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒
𝑎
𝑏
𝑑
𝑐
𝑇 𝐻
𝑇𝐿
𝑄 𝐿
𝑄 𝐻
=
𝑇𝐿
𝑇 𝐻

10. Carnot Cycle (step-by-step)
1). Isothermal expansion, 𝑎 → 𝑏
2). Adiabatic expansion, 𝑏 → 𝑐
3). Isothermal compression, 𝑐 → 𝑑
∆𝑈 = 0, 𝑄 = 𝑊
𝑊𝑐→𝑑 = −𝑄 𝐿 = 𝑁𝑘𝑇𝐿 ln
𝑉𝑐
𝑉𝑑
< 0
4). Adiabatic compression, 𝑑 → 𝑎
𝑄 = 0, ∆𝑈 = −𝑊𝑑→𝑎
∴ 𝑊𝑑→𝑎 = −∆𝑈 𝑑→𝑎< 0
***−∆𝑈 𝑑→𝑎= −∆𝑈 𝑎𝑑= − 𝑈 𝑎 − 𝑈 𝑑
Carnot Cycle (step-by-step)
1). Isothermal expansion, 𝑎 → 𝑏
∆𝑈 = 0, 𝑄 = 𝑊
𝑊𝑎→𝑏 = 𝑄 𝐻 = 𝑁𝑘𝑇 𝐻 ln
𝑉𝑏
𝑉𝑎
> 0
2). Adiabatic expansion, 𝑏 → 𝑐
𝑄 = 0, ∆𝑈 = −𝑊𝑏→𝑐
∴ 𝑊𝑏→𝑐 = −∆𝑈 𝑏→𝑐> 0
***−∆𝑈 𝑏→𝑐= −∆𝑈𝑐𝑏= − 𝑈𝑐 − 𝑈 𝑏
3). Isothermal compression, 𝑐 → 𝑑
4). Adiabatic compression, 𝑑 → 𝑎
𝑃
𝑉
𝑎
𝑏
𝑑
𝑐
𝑇 𝐻
𝑇𝐿
𝑃
𝑉
𝑎
𝑏
𝑑
𝑐
𝑃
𝑉
𝑎
𝑏
𝑑
𝑐
𝑃
𝑉
𝑎
𝑏
𝑑
𝑐

11. Summary
1). Isothermal expansion, 𝑎 → 𝑏
𝑊𝑎→𝑏 = 𝑄 𝐻
2). Adiabatic expansion, 𝑏 → 𝑐
𝑊𝑏→𝑐 = −∆𝑈 𝑏→𝑐= −𝑈𝑐 + 𝑈 𝑏
3). Isothermal compression, 𝑐 → 𝑑
𝑊𝑐→𝑑 = −𝑄 𝐿
4). Adiabatic compression, 𝑑 → 𝑎
𝑊𝑑→𝑎 = −∆𝑈 𝑑→𝑎= −𝑈 𝑎 + 𝑈 𝑑
∴ 𝑊𝑎→𝑏→𝑐→𝑑→𝑎 = 𝑄 𝐻 −𝑈𝑐 + 𝑈𝑏 −𝑄 𝐿 −𝑈 𝑎 + 𝑈 𝑑
Since for isothermal, ∆𝑈 = 0
𝑎 → 𝑏 ∶ 𝑈 𝑎 = 𝑈 𝑏
𝑐 → 𝑑 ∶ 𝑈𝑐 = 𝑈 𝑑
Efficiency, η
η = 1 −
𝑄 𝐿
𝑄 𝐻
Consider ∆𝑠 𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒
∆𝑆 𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒= ∆𝑆 𝑎→𝑏→𝑐→𝑑→𝑎= 0
∆𝑆 =
𝑑𝑄
𝑇
𝑏
𝑎
+
𝑑𝑄
𝑇
𝑐
𝑏
+
𝑑𝑄
𝑇
𝑑
𝑐
+
𝑑𝑄
𝑇
𝑎
𝑑
∆𝑆 =
𝑑𝑄
𝑇
𝑏
𝑎
+ 0 +
𝑑𝑄
𝑇
𝑑
𝑐
+ 0
∆𝑆 =
𝑑𝑄
𝑇
𝑏
𝑎
+ 0 +
𝑑𝑄
𝑇
𝑑
𝑐
+ 0
∆𝑆 =
𝑄 𝑎→𝑏
𝑇 𝐻
+
𝑄 𝑐→𝑑
𝑇𝐿
= 0
𝑄 𝐻
𝑇 𝐻
+
−𝑄 𝐿
𝑇𝐿
= 0
𝑊 = 𝑄 𝐻 − 𝑄 𝐿
𝑄 𝐻
𝑇 𝐻
+
−𝑄 𝐿
𝑇𝐿
= 0
𝑄 𝐻
𝑇 𝐻
=
𝑄 𝐿
𝑇𝐿
𝑇𝐿
𝑇 𝐻
=
𝑄 𝐿
𝑄 𝐻
(𝑝𝑟𝑜𝑣𝑒𝑛)
***For engine, it is impossible for
𝑄 𝐿 = 0 𝑜𝑟 η = 100%
η = 1 −
𝑄 𝐿
𝑄 𝐻
= 1 −
𝑇𝐿
𝑇 𝐻