1. How much wood… ? U -W Q

2. Evaluating heat engines with PV diagrams

3. Process Meaning Implications Example Isobaric P=0 +Q  V
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙=𝑄−𝑃𝑉
Isovolumetric
V=0 W
Q  P  T
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑄
Isothermal
T=0
PV = constant
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =0
𝑄=−𝑊
Adiabatic
Q=0
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =−𝑃𝑉

4. Process Implications Example In English Isobaric +Q  V
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙=𝑄−𝑃𝑉
To expand (contract) a gas at constant pressure, heat must be added (taken away).
Isovolumetric W
Q  P  T
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑄
Adding (subtracting) heat to a gas at constant volume increases (decreases) pressure.
Isothermal
PV = constant
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =0
𝑄=−𝑃𝑉
When energy is added to a gas at constant temperature, it will expand and its pressure will drop.
Adiabatic
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑃𝑉
Doing work on a gas decreases its volume, and increases its pressure and temperature.

5. How much work is done to gas. How much does internal energy change
How much work is done to gas? How much does internal energy change? How much heat is added?
Conventions:
W+ = work done TO gas by environment
i.e., gas is compressed, final volume is smaller
W- = work done BY gas on environment
i.e., gas expands, final volume is larger
Q+ = heat flows INTO gas
Q- = heat flows FROM gas

6. Isobaric process: work
𝑊=−𝑃𝑉
Final
Initial
W= - (2 Pa) (10 m3 – 1 m3) = -18 J

7. Isobaric process: change in internal energy
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇
𝑇= 𝑃 𝑉𝑓−𝑉𝑖 𝑛𝑅
Final
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑃(𝑉𝑓−𝑉𝑖)
Initial
Where T = P(Vf-Vi)/nR
So, U = 3/2 nR P (Vf-Vi) / nR
Or, U = 3/2 P (Vf-Vi)
Here, U = 3/2 (2 Pa) (9 m3) = 27 J

8. Isobaric process: heat
𝑈=𝑄+𝑊
𝑄=𝑈−𝑊
Final
Initial
Take-away:
Isobaric processes do work on the environment, but require that heat be added.
Q= 27 J – 18 J = 9 J

9. Isovolumetric process: work
𝑊=−𝑃𝑉
Final
V = 0 m3, so W = 0 J
Initial

10. Isovolumetric process: change in internal energy
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇
Final
𝑇= 𝑉 𝑃𝑓−𝑃𝑖 𝑛𝑅
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑉(𝑃𝑓−𝑃𝑖)
Where T = V(Pf-Pi)/nR
So, U = 3/2 nR V (Pf-Pi) / nR
Or, U = 3/2 V (Pf-Pi)
Here, U = 3/2 (1 m3) (20 Pa – 1 Pa) = 28.5 J
Initial

11. Isovolumetric process: heat
𝑈=𝑄+𝑊
Final
𝑄=𝑈−𝑊
Take-away:
Isovolumetric processes do no work on the environment; additional pressure requires heat be added to the system.
Q= 28.5 J – 0 J = 28.5 J
Initial

12. Isothermal process: change in internal energy
𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇
Final
Initial
T = 0 K, so U = 0 J

13. Isothermal process: work and heat
𝑈=𝑄+𝑊
Initial
@𝑈=0, 𝑄=−𝑊
Take-away:
Heat added to isothermal process does work on environment with no change in temperature.
Final

14. The process makes a difference!
Final
Initial
Initial
Final
Work in isovolumetric / isobaric process < work in isothermal process

15. Rank processes from those that do most work to those that do least work.
Initial
Final B C D
Area under the curve does most work

16. Efficiency
Ratio of what you get out by what you put in: 𝑒= 𝑊 𝑜𝑢𝑡 𝑊 𝑖𝑛 In the case of heat engines, 𝑊 𝑖𝑛 = 𝑄 𝐻 (energy input from, e.g., gasoline) So, 𝑒= 𝑊 𝑜𝑢𝑡 𝑄 𝐻 where, since energy is conserved, 𝑊 𝑜𝑢𝑡 = 𝑄 𝐻 − 𝑄 𝐿 So, 𝑒= 𝑄 𝐻 − 𝑄 𝐿 𝑄 𝐻 =1− QL QH

17. Example
An automobile engine has an efficiency of 20% and produces an average of 23,000 J of mechanical work per second.
How much heat input is required?
How much is discharged as waste?
a) QH = W / e = J / 0.20 = 115 kJ
b) e= 1 – QL / QH, so QL = QH(1-e) = 115 kJ (1-0.2) = 92 kJ

18. Ideal efficiency
In an ideal engine (i.e., perfectly reversible process),
QH  TH and QL  TL
So,
𝑒𝑖𝑑𝑒𝑎𝑙=1− TL TH
Note: this equation is sometimes called Carnot efficiency in honor of its discoverer, a French physicist Sadi Carnot.

19. Coefficient of performance
For cooling devices, 𝐶𝑂𝑃≡ 𝑄 𝐿 𝑊 = 𝑄 𝐿 (𝑄 𝐻 − 𝑄 𝐿 )
For heating devices,
𝐶𝑂𝑃≡ 𝑄 𝐻 𝑊
Since we care about TL for cooling devices, COP defined as a function of TL for cooling devices.

20. Why heat pumps?
How much thermal energy is delivered by 1500-W electric heater?
How much thermal energy is delivered by 1500-W heat pump with a COP of 3.0?
1500 W
COP = QH/W, so QH = (COP) (W) = 4500 W… NOT a violation of conservation of energy; extra heat coming from outside