# First Law of Thermodynamics

```Physics 201
Professor P. Q. Hung
311B, Physics Building
Physics 201 – p. 1/3
Laws of Thermodynamics
We will discuss the zeroth, first and 2nd laws of
thermodynamics in this lecture.
Physics 201 – p. 2/3
Laws of Thermodynamics
We will discuss the zeroth, first and 2nd laws of
thermodynamics in this lecture.
First Law: Statement about conservation of
energy.
Physics 201 – p. 2/3
Laws of Thermodynamics
We will discuss the zeroth, first and 2nd laws of
thermodynamics in this lecture.
First Law: Statement about conservation of
energy.
Second Law: Statement about how heat only
flows from hot to cold and not the reverse.
Physics 201 – p. 2/3
First Law of Thermodynamics
The net heat added to a system equals the
change in the internal energy of the system plus
the work done by the system.
Q = ∆U + W with ∆U = Uf − Ui
Q &gt; 0: heat added to the system
Physics 201 – p. 3/3
First Law of Thermodynamics
The net heat added to a system equals the
change in the internal energy of the system plus
the work done by the system.
Q = ∆U + W with ∆U = Uf − Ui
Q &gt; 0: heat added to the system
Q &lt; 0: heat removed from the system
Physics 201 – p. 3/3
First Law of Thermodynamics
The net heat added to a system equals the
change in the internal energy of the system plus
the work done by the system.
Q = ∆U + W with ∆U = Uf − Ui
Q &gt; 0: heat added to the system
Q &lt; 0: heat removed from the system
W &gt; 0: work done by the system
Physics 201 – p. 3/3
First Law of Thermodynamics
The net heat added to a system equals the
change in the internal energy of the system plus
the work done by the system.
Q = ∆U + W with ∆U = Uf − Ui
Q &gt; 0: heat added to the system
Q &lt; 0: heat removed from the system
W &gt; 0: work done by the system
W &lt; 0: work done on the system
Physics 201 – p. 3/3
First Law of Thermodynamics
Physics 201 – p. 4/3
First Law of Thermodynamics
Here W = 0; Q = Uf − Ui and Q &gt; 0 (Heat
added) ⇒ Uf &gt; Ui ⇒ Internal energy increases!
Physics 201 – p. 5/3
First Law of Thermodynamics
Here Q = 0; −W = Uf − Ui and W &gt; 0 (Work
done by the system) ⇒ Uf &lt; Ui ⇒ Internal
energy decreases!
Physics 201 – p. 6/3
First Law of Thermodynamics
A system consists of 3kg of water at 80 0 C. 25 kJ
of work is done on the system by stirring it with a
paddle wheel, while 15 kcal of heat is removed.
1) What is the change in the internal energy?; 2)
What is the final temperature?
(1)
Convert! 15 kcal = 15 &times; 4.18 kJ = 62.7 kJ.
Physics 201 – p. 7/3
First Law of Thermodynamics
A system consists of 3kg of water at 80 0 C. 25 kJ
of work is done on the system by stirring it with a
paddle wheel, while 15 kcal of heat is removed.
1) What is the change in the internal energy?; 2)
What is the final temperature?
(1)
Convert! 15 kcal = 15 &times; 4.18 kJ = 62.7 kJ.
Concept: Heat is removed ⇒ Q &lt; 0; Work is
done on ⇒ W &lt; 0
⇒ (−62.7 kJ) = ∆U + (−25 kJ)
Physics 201 – p. 7/3
First Law of Thermodynamics
A system consists of 3kg of water at 80 0 C. 25 kJ
of work is done on the system by stirring it with a
paddle wheel, while 15 kcal of heat is removed.
1) What is the change in the internal energy?; 2)
What is the final temperature?
(1)
Convert! 15 kcal = 15 &times; 4.18 kJ = 62.7 kJ.
Concept: Heat is removed ⇒ Q &lt; 0; Work is
done on ⇒ W &lt; 0
⇒ (−62.7 kJ) = ∆U + (−25 kJ)
⇒ ∆U = −62.7 kJ + 25 kJ = −37.7 kJ &lt; 0. It
is negative because more energy is removed
Physics 201 – p. 7/3
First Law of Thermodynamics
(2)
Concept: Use Q = mcW ∆T ; m = 3kg and
cW = 4.18 kJ/kg.0 C
Physics 201 – p. 8/3
First Law of Thermodynamics
(2)
Concept: Use Q = mcW ∆T ; m = 3kg and
cW = 4.18 kJ/kg.0 C
∆T =
−37.7 kJ
(4.18 kJ/kg.0 C)(3 kg)
= −3.01 0 C.
Physics 201 – p. 8/3
First Law of Thermodynamics
(2)
Concept: Use Q = mcW ∆T ; m = 3kg and
cW = 4.18 kJ/kg.0 C
∆T =
−37.7 kJ
(4.18 kJ/kg.0 C)(3 kg)
= −3.01 0 C.
∆T = Tf − Ti ⇒ Tf = 80 0 − 3.01 0 C ≈ 77 0 C.
Physics 201 – p. 8/3
First Law of Thermodynamics
Thermal processes
Reversible process: Change in which thermal
equilibrium is maintained (slow process).
when external conditions are reversed.
Physics 201 – p. 9/3
First Law of Thermodynamics
Thermal processes
Reversible process: Change in which thermal
equilibrium is maintained (slow process).
when external conditions are reversed.
Irreversible process: Out-of-equilibrium
change.
Physics 201 – p. 9/3
First Law of Thermodynamics
Thermal processes
Take a cylinder fitted with a piston of
cross-sectional area A. Slowly push the piston ⇒
quasi-static ⇒ a succession of thermal
equilibrium.
Force: F = P A.
Physics 201 – p. 10/3
First Law of Thermodynamics
Thermal processes
Take a cylinder fitted with a piston of
cross-sectional area A. Slowly push the piston ⇒
quasi-static ⇒ a succession of thermal
equilibrium.
Force: F = P A.
Work done: ∆W = F ∆x = P A∆x = P ∆V .
Physics 201 – p. 10/3
First Law of Thermodynamics
Thermal processes
Take a cylinder fitted with a piston of
cross-sectional area A. Slowly push the piston ⇒
quasi-static ⇒ a succession of thermal
equilibrium.
Force: F = P A.
Work done: ∆W = F ∆x = P A∆x = P ∆V .
P
∆W ⇒ Area under the
Total work done:
P − V curve.
Physics 201 – p. 10/3
First Law of Thermodynamics
Thermal processes: Constant Pressure
Physics 201 – p. 11/3
First Law of Thermodynamics
Thermal processes: Constant Pressure
W = P0 (Vf − Vi )
Physics 201 – p. 12/3
First Law of Thermodynamics
Thermal processes: Constant Volume
Physics 201 – p. 13/3
First Law of Thermodynamics
Thermal processes: Constant Volume
∆V = 0 ⇒ W = 0
Physics 201 – p. 14/3
First Law of Thermodynamics
process which either occurs very rapidly or in
a system that is very well insulated that no
transfer of energy as heat occurs ⇒ Q = 0.
∆U = −W
Physics 201 – p. 15/3
First Law of Thermodynamics
process which either occurs very rapidly or in
a system that is very well insulated that no
transfer of energy as heat occurs ⇒ Q = 0.
∆U = −W
W &gt; 0: work done by the system ⇒ ∆U &lt; 0:
decrease in internal energy.
Physics 201 – p. 15/3
First Law of Thermodynamics
process which either occurs very rapidly or in
a system that is very well insulated that no
transfer of energy as heat occurs ⇒ Q = 0.
∆U = −W
W &gt; 0: work done by the system ⇒ ∆U &lt; 0:
decrease in internal energy.
W &lt; 0: work done on the system ⇒ ∆U &gt; 0:
increase in internal energy.
Physics 201 – p. 15/3
First Law of Thermodynamics
Physics 201 – p. 16/3
First Law of Thermodynamics
Cyclic process: Processes in which, after
interchanges of heat and work, the system is
restored to its initial state ⇒ No change in
internal energy ⇒ ∆U = 0.
Q=W
Physics 201 – p. 17/3
First Law of Thermodynamics
Cyclic process: Processes in which, after
interchanges of heat and work, the system is
restored to its initial state ⇒ No change in
internal energy ⇒ ∆U = 0.
Q=W
Free expansion: Adiabatic processes in which
Q = W = 0 ⇒ ∆U = 0.
Physics 201 – p. 17/3
First Law of Thermodynamics
Thermal processes: Isothermal process
Isothermal process: Processes which occur
at constant temperature.
Physics 201 – p. 18/3
First Law of Thermodynamics
Thermal processes: Isothermal process
Isothermal process: Processes which occur
at constant temperature.
P V = N kT = nRT ⇒ (Using ∆W = P ∆V
and calculus)
Vf
Vf
W = N kT ln( Vi ) = nRT ln( Vi )
Physics 201 – p. 18/3
First Law of Thermodynamics
Thermal processes: Isothermal process
V
W = nRT ln( Vfi )
Physics 201 – p. 19/3
First Law of Thermodynamics
Summary of thermodynamic processes
Physics 201 – p. 20/3
First Law of Thermodynamics
Thermal processes: Example
1 kg of water at 100 0 C is contained in a cylinder
fitted with a piston. It is then converted into
steam at 100 0 C by boiling it at standard
atmospheric pressure (which is at 1 atm or
1.01 &times; 105 P a). The volume of water changes
from an initial value of 1.00 &times; 10−3 m3 as a liquid
to 1.671m3 as steam. (a) How much work is done
by the system?; (b) How much energy is
transferred as heat?; (c) What is the change in
the system’s internal energy?
Physics 201 – p. 21/3
First Law of Thermodynamics
Thermal processes: Example
Change at constant pressure ⇒
W = P ∆V = (1.01 &times; 105 P a)(1.671m3 − 1.00 &times;
10−3 m3 ) = 169kJ.
Physics 201 – p. 22/3
First Law of Thermodynamics
Thermal processes: Example
Change at constant pressure ⇒
W = P ∆V = (1.01 &times; 105 P a)(1.671m3 − 1.00 &times;
10−3 m3 ) = 169kJ.
Liquid to steam ⇒
Q = LV m = (2256kJ/kg)(1kg) = 2256kJ.
Physics 201 – p. 22/3
First Law of Thermodynamics
Thermal processes: Example
Change at constant pressure ⇒
W = P ∆V = (1.01 &times; 105 P a)(1.671m3 − 1.00 &times;
10−3 m3 ) = 169kJ.
Liquid to steam ⇒
Q = LV m = (2256kJ/kg)(1kg) = 2256kJ.
∆U = Q − W = 2256kJ − 169kJ = 2087kJ.
Physics 201 – p. 22/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas
Molar specific heat at constant volume:
Heat is added at constant volume ⇒
QV = nCV ∆T = ∆U ⇒ nCV = ∆U
∆T .
Physics 201 – p. 23/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas
Molar specific heat at constant volume:
Heat is added at constant volume ⇒
QV = nCV ∆T = ∆U ⇒ nCV = ∆U
∆T .
Since the internal energy for a monatomic
gas is U = 23 nRT , it follows that
CV = 23 R
Molar specific heat at constant volume.
Physics 201 – p. 23/3
First Law of Thermodynamics
Molar specific heat at constant volume
Physics 201 – p. 24/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas
Molar specific heat at constant pressure:
QP = nCP ∆T = ∆U + W =
nCV ∆T + P ∆V = nCV ∆T + P nR
P ∆T
⇒
CP = CV + R
Physics 201 – p. 25/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas
Molar specific heat at constant pressure:
QP = nCP ∆T = ∆U + W =
nCV ∆T + P ∆V = nCV ∆T + P nR
P ∆T
⇒
CP = CV + R
Monatomic gas: CP = 25 R ⇒ CP − CV = R
Physics 201 – p. 25/3
First Law of Thermodynamics
Physics 201 – p. 26/3
First Law of Thermodynamics
Molar specific heat at constant pressure
Physics 201 – p. 27/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
No heat in or out of the system:
Physics 201 – p. 28/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
No heat in or out of the system:
P V γ = constant
CP
γ = CV
Physics 201 – p. 28/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
No heat in or out of the system:
P V γ = constant
CP
γ = CV
For monatomic gas, γ = 5/3. Different for
diatomic or triatomic gases, etc...
Physics 201 – p. 28/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
Physics 201 – p. 29/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
A quantity of air (γ = 1.4) expands adiabatically
from 2 atm and 2 L at T = 20 0 C to twice its
original volume. Final pressure?
γ
P1 V1
=
γ
P2 V2
Physics 201 – p. 30/3
First Law of Thermodynamics
Molar Specific Heat of an Ideal Gas: Adiabatic
process
A quantity of air (γ = 1.4) expands adiabatically
from 2 atm and 2 L at T = 20 0 C to twice its
original volume. Final pressure?
γ
P1 V1
=
γ
P2 V2
2L 1.4
P2 = P1 ( VV21 )γ = (2atm)( 4L
) = 0.758atm.
Physics 201 – p. 30/3
Second Law of Thermodynamics
When objects of different temperatures are
brought into thermal contact, the spontaneous
flow of heat that results is always from the high
temperature object to the low temperature object
but never the reverse.
Physics 201 – p. 31/3
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