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1st Law of Thermodynamics
The change in the internal energy of a closed system is equal to
the amount of heat supplied to the system, minus the amount of
work performed by the system on its surroundings. (Version of
conservation of energy applied to thermodynamical systems.
Thermodynamic Processes: Isothermal
Using Boyle’s Law –
To keep the temperature
constant both the pressure
and volume change to
compensate. (Volume goes
up, pressure goes down)
Change in internal energy is a
function of ΔT.
If ΔT = 0, then ΔU = 0
Then Q = W
W = nRTln(Vf/Vi)
ln is the natural log function
Thermodynamic Processes: Isobaric
No change in pressure:
ΔP = 0
Heat is added to the gas which
increases the Internal Energy
(U). Work is done by the gas as it
changes in volume.
On a PV diagram, the isobaric
process is a horizontal line called
an isobar.
∆U = Q - W can be used since
the WORK is POSITIVE in this
case
∆U = Q – W
where W = PΔV
Thermodynamic Processes: Isovolumetric
No change in volume:
ΔV = 0
On a PV diagram, an
isovolumetric process is a
vertical line. No area
under the curve, and no
work done.
ΔU = Q
Thermodynamic Processes: Adiabatic
System exchanges no heat
with its surroundings
Q=0
PV diagram is a steep
hyperbola.
ΔU = - W
Example Problem 13.17
Two moles of an ideal gas expands isothermically at 27°C to three times its
initial volume.
(a) Find the work done by the gas
(b) The heat flow into the system
Example Problem 13.22
A 1-kg block of aluminum is heated at atmospheric pressure such that its
temperature increases from 22°C to 40°C. Find:
(a) The work done by the aluminum
(a) The heat added to the aluminum
(b) The change in internal energy of the aluminum
Example Problem 13.54
A gas follows the path 123 on the PV diagram below and 418 J of heat flows into
the system. Also, 167 J of work is done.
Pressure, P
2
1
3
4
Volume, V
Find the internal energy of the system:
Example Problem 13.54
A gas follows the path 123 on the PV diagram below and 418 J of heat flows into
the system. Also, 167 J of work is done.
Pressure, P
2
1
3
4
Volume, V
How much heat flows into the
system if the process follows the
path 143? The work done by the
gas along this path is 63 J:
Example Problem 13.54
A gas follows the path 123 on the PV diagram below and 418 J of heat flows into
the system. Also, 167 J of work is done.
Pressure, P
2
1
3
4
Volume, V
What net work would be done on
or by the system if the system
followed the path 12341 ?