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4.
Volume work
During chemical reactions, volume changes often occur
.
Caused by the volume change of the work environment system called volume work done, generally indicated by W is
.
As shown in Figure 2-2, the piston is used to seal the gas in the cylindrical cylinder, and the gas expansion opposes the external force F to push the piston from position I to position II, and the displacement is △ l
.
If the mass of the piston itself and the friction between the piston and the cylinder wall are ignored, the work done by the piston against the external force F is
W=F·△l
Suppose the cross-sectional area of the piston is S, then there is
Figure 2-2 Schematic diagram of work done by volume change
In the formula, it is the external pressure p resisted when the volume expands; (S△l) is the volume change △V
.
The system does work to the environment, and the work is negative, so
W=-p·△V
From the above definition of volume work, it can be seen that if the external pressure p=0 or the volume change △V=0, the volume work W=0
.
The systems and processes studied in this chapter do not do non-volume work, that is, the work done by the system change process is all volume work
.
From the unit of p, Pa, and the unit of △V, m 3 , the unit of volume work can be derived
Pa·m 3 =(N·m -2 )·m 3 =N·m=J
Therefore, the unit of volume work is J or kJ
.
[Example 2-1] Ideal gas expands from p 1 =16×10 5 Pa, V 1 =1dm 3 to p 2 =1×10 5 Pa, V 2 =16dm 3 under constant temperature conditions , the process is as shown in Figure 2-1 The three ways to complete, find the volume work of each way
.
Solve the ideal gas expansion, the system does work to the environment, and the volume work is a negative value
W=-p·△V
(1) Route F first resists external pressure 8×10 5 Pa expansion, and then resists external pressure 1×10 5 Pa expansion
W F =[-8×10 5 Pa×(2-1)×10 -3 m 2 ]+[-1×10 5 Pa×(16 -2 )×10 -3 m 3 ]
=-2200J
(2) Path G resists external pressure 1×10 5 Pa one time expansion
W G =-1×10 5 Pa×(16-1)×10 -3 mm=-1500J
(3) Route H first resists external pressure 4×10 5 Pa expansion, and then resists external pressure 1×10 5 Pa expansion
W H =[-4×10 5 Pa×(4-1)×10 -3 m 3 ]+[-1×10 5 Pa×(16 -4 )×10 -3 m 3 ]
=-2400J
It can be seen that volume work is a physical quantity related to a pathway, not a state function, and the value of work performed by different pathways may be different
.
Volume work can also be calculated using the pV diagram method
.
The external pressure p external is plotted against the volume V of the system, and the resulting curve is called the pV line
.
Figure 2-3 Use pV diagram to show volumetric work
Figure 2-3 shows the path F in Figure 2-1 as a pV line
.
Point A represents the initial state, V 1 =1×10 -3 m 3 , p 1 =16×10 5 Pa; Point E represents the final state, V 2 =16×10 -3 m 3 , p 2 =1×10 5 Pa
.
S=8×(2-1)+1×(16-2)=22 (unit area)
5.
Thermodynamic energy
Thermodynamic energy is also called internal energy, which is the sum of all energy in the system
.
Although the thermodynamic energy of the system cannot be obtained yet, the thermodynamic energy of the system is a fixed value when the state of the system is certain
.
Therefore, the thermodynamic energy U is the state function of the system
.
△U=U end-U beginning
Thermodynamic energy is a measuring property of the system, and it is additive
.
Ideal gas is the simplest system, and its thermodynamic energy is only a function of temperature
.
If the temperature does not change, the thermodynamic energy of the system does not change, that is, the amount of change in the thermodynamic energy of the system is zero (△T=0, then △U=0)
.
Related Links: Basic Concepts of Thermodynamics (1)