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Work, Energy & Heat. The First Law: Some Terminology. System: Well defined part of the universe Surrounding: Universe outside the boundary of the system Heat (q) may flow between system and surroundings. Closed system: No exchange of matter with surroundings - PowerPoint PPT Presentation
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Work, Energy & HeatThe First Law: Some Terminology
System: Well defined part of the universe
Surrounding: Universe outside the boundary of the system
Heat (q) may flow between system and surroundings
Closed system: No exchange of matter with surroundings
Isolated System: No exchange of q, or matter with surroundings
Isothermal process: Temperature of the system stays the same
Adiabatic: No heat (q) exchanged between system and surroundings
THE CONCEPT OF REVERSABILTY
Irreversible processes:
Hot
Cold
Warm
WarmTemperature equilibration
Mixing of two gases
Expansion into a vacuumP = 0
Evaporation into a vacuumP = 0
THE CONCEPT OF REVERSABILTY
Reversible processes:
Po
Po + P
Po
Tiny weight
Condensation(pressure minimally increases by adding tiny weight)
Evaporation(pressure minimally decreases by removing tiny weight)
P
VV1 V2
Pext
w = -Pext(V2 – V1)
P, V1 P, V2
IRREVERSIBLE EXPANSION
Pext
Pext
V1
V2
THE CONCEPT OF REVERSABILTY
REMOVE PINS
Irreversible Expansion
pins
P = 2 atm
Pext = 1 atm
(1)
(2)
P = 2 atm
Pext = 2 atm
P = 1.999 atm
Pext = 1.999 atm
Step 1
P = 1.998 atm
Pext = 1.998 atm
Step 2Infinite number of steps
Reversible ExpansionP = 1 atm
Pext = 1 atm
P = 1 atm
Pext = 1 atm
REVERSIBLE EXPANSION
Pext = Pressure of gas. If the gas is ideal, then Pext = nRT/V
How does the pressure of an ideal gas vary with volume?
P
V
This is the reversible path. The pressure at each point along curve is equal to the external pressure.
1
2ln12
1
2
1
2
1 V
VnRTdV
VnRTdV
V
nRTPdVdVPw
V
V
V
V
V
Vext
REVERSIBLE EXPANSION
P
VVi Vf
Pi
Pf
A
B
The reversible path
i
f
V
VnRTw ln
The shaded area is
IRREVERSIBLE EXPANSION
P
VVi Vf
Pi
Pf
A
B
fif VVPw
The shaded area is
Pext = Pf
Reversible expansion gives the maximum work
REVERSIBLE COMPRESSION
P
VVf Vi
Pf
Pi
A
BThe reversible path
i
f
V
VnRTw ln
The shaded area is
P
Reversible compression gives the minimum work
IRREVERSIBLE COMPRESSION
VVf Vi
Pf
Pi
A
B
fif VVPw
The shaded area is
Pext = Pf
A system from a state 1 (or 2) to a new state 2 (or 1). Regions A, B, C, D, and E correspond to the areas of the 5 segments in the diagram.
1. If the process is isothermal reversible expansion from state 1 to state 2, the total work done by the system is equal to A. Area C + Area EB. Area C onlyC. Area E onlyD. Area A + Area C + Area EE. Area A only
3. If the process in question 1 was carried out irreversibly against a constant pressure of 2 atm, the total work done by the system is equal toA. Area C + Area EB. Area C onlyC. Area A + Area C + Area ED. Area E + Area DE. Area E only
2. If state 2 undergoes irreversible compression to state 1 against an external pressure of 5 atm, the work done by the surroundings on the system is equal toA. Area A + Area C + Area EB. Area C onlyC. Area A + Area CE. Area E onlyF. Area A only
6. Which one of the following is not true?A. There is no heat flow between system and surrounding for a reversible adiabatic processB. Work done by the gas in a reversible expansion is a maximumC. Work done by the gas in a reversible expansion is a minimumD. Work done by the gas in a reversible compression is a minimumE. Work done by the gas in a reversible expansion is not the same as the work done against a constant external pressure
4. For a reversible adiabatic expansion of a gas, which one of the following is correct?A. Heat flows to maintain constant temperatureB. The gas suffers a maximum drop in temperatureC. The gas suffers a minimum drop in temperatureD. The work done is a positive quantityE. There is zero change in internal energy
5. The heat capacity (Cp) for a solid at low temperatures is approximately represented by Cp = AT3, where A is a constant. Using the equation for Cp, the change in entropy (S) for heating a solid from 0K to 1K is
A. A/4 C. A/2B. A/3 D. A
P/atm
1.0 10 Volume/L
2
5
A B
C
D E
State 1
State 2
Consider four molecules in two compartments:
1 way 4 ways 6 ways 4 ways 1 way
Total number of microstates = 16
The most probable(the “even split”)
If N the “even split” becomes overwhelmingly probable
Statistical Entropy
Boltzmann S = kB lnW
Consider spin (or dipole restricted to two orientations)
or W = 2, and S = kB ln21 particle
2 particles
3 particles
1 mole
W = 4, and S = kB ln4, , ,
W = 8, and S = kB ln8
W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln
