# lecture_4

```Lecture 5
First Law of Thermodynamics
First Law of Thermodynamics
You can’t get something for nothing.
We will discuss these statements later…
Thermodynamic System
Consider a closed system (e.g., a parcel of air).
It has internal energy (“u”) = energy due to
molecular kinetic and potential energies.
Suppose some energy (dq) is added to the
system.
Example: via radiation from the sun
What happens?
Thermodynamic System
Some of the energy goes into work done (dw)
by the system against its surroundings.
Example: expansion
What’s left is a change in internal energy.
du = dq – dw
Conservation of Energy principle
Thermodynamic System
Heat
Environment
System
System can exchange energy
with environment via heat flow.
Thermodynamic System
In addition, system can do work
on environment or vice versa.
Example: expansion
First Law, General Form
dU = dQ – dW
dU = change in internal energy of system
dQ = heat exchanged with environment

dQ &gt; 0  heat flowing into system
dW = work done by or on system

dW &gt; 0  system is doing work
Ideal Gas
Consider a system consisting of an ideal
gas in a cylinder.
Cylinder has a piston, which allows
volume to be changed.
Cylinder
weights
more weight, more
pressure
Walls of cylinder:
Gas
1) perfect heat
conductors
2) perfect insulators
Cases
Case 1: Walls of cylinder are perfect
conductors


 heat can freely flow between system and
environment
 in equilibrium, temperature of system must
equal temperature of environment
Case 2: Walls of cylinder are perfect
insulators


no heat flow between system and
environment
 temperatures of system and environment
need not be equal
Work (Qualitative)
System does work on environment
(dW &gt; 0) if gas expands

(Piston is pushed upward.)
Work is done on system (dW &lt; 0) if
gas contracts

(Piston is pushed downward.)
Work (Quantitative)
Suppose piston is pushed upward a distance dx.
Work (Quantitative)
Suppose piston is pushed upward a distance dx.
Area of piston = A
Pressure force = pA
dx
dW = pAdx
= pdV
First Law, Ideal Gas
dU  dQ  pdV
Usually, we are interested in energy
per unit mass
Divide both sides by m
Define u = U/m; q = Q/m;  = V/m
First Law, Ideal Gas (per unit mass)
du  dq  pd
Internal Energy
For an ideal gas, u is a function of T only

u is an increasing function of T
Change in internal energy depends only
on change in temperature.



Doesn’t depend on the way in which that
temperature change is accomplished.
du = kdT, where k is a constant
(Value of k will be determined shortly.)
Heat Capacity, C
dQ
C
dT
SI units: JK-1
Amount of heat required to change a “body’s” temperature
by a given amount
For a gas, C depends on the
particular process.
Cv = heat capacity at constant volume (dV = 0)
Cp = heat capacity at constant pressure (dp = 0)
Specific Heat
Specific heat is heat capacity per unit mass
cv (lower case) = specific heat at constant
volume
cp = specific heat at constant pressure
Specific heat is the heat energy needed to
raise the temperature of a unit mass of a
substance by one degree.
Specific Heat
Suppose we add some thermal energy (dq) to a unit
mass of a substance like air, water, soil.
We expect T(substance) to increase
How much?
dq
We can define Specific Heat as
dT
Temp change
 dq 
cv  

 dT v
 dq 
cp  

 dT  p
Constant volume
Constant pressure
Specific Heat of Dry Air
cv = 717 Jkg-1K-1
cp = 1004 Jkg-1K-1
Note: cp – cv = 287 Jkg-1K-1



Look familiar?
cp – cv = Rd
(Not a coincidence!)
Constant Volume Processes
du  dq  pd
du  dq  cv d T
0
Internal Energy Change
The following is always true:
du  cv d T
(3.40) W&amp;H
First Law for Ideal Gas, Re-written
cv dT  dq  pd
(1)
Constant-Pressure Processes
Go back to ideal-gas law:
p  RT
Take differential of both sides:
d ( p )  RdT
But,…
d ( p )  pd  dp

pd  RdT   dp
(2)
Substitute (2) into (1)
pd  RdT   dp
cv dT  dq  pd
 cv dT  dq  RdT   dp
Simplify
cv  RdT  dq   dp (3)
Constant-pressure process 
dp  0
 dq  cv  R  dT
But, …
For a constant-pressure process,
dq  c p dT
Compare
dq  cv  R dT
and
dq  c p dT
Result:
c p  cv  R
Rewrite (3)
c p dT  dq   dp
We will use this in the
next lecture.
(4)
```