1. The First Law of Thermodynamics
Chapter 19
The First Law of Thermodynamics
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2. Learning Goals for Chapter 19
Calculate work done by a system when its volume changes.
Interpret & use 1st Law of Thermodynamics.
Four important kinds of thermodynamic processes:
Isothermal
Isochoric
Isobaric
Adiabatic
Why internal energy of ideal gas depends on T only.
Difference between molar heat capacities at constant volume Cv & at constant pressure Cp.

3. Key Ideas to Concentrate Upon!
PV diagrams as a way to see cyclic processes
Pressure (atm, kPA, barr, lb/in2)
Volume (liters, m3, cm3)

4. Key Ideas to Concentrate Upon!
PV diagrams as a way to see cyclic processes
Pressure (atm, kPA, barr, lb/in2)
Volume (liters, m3, cm3)

5. Key Ideas to Concentrate Upon!
PV diagrams as a way to see cyclic processes
Pressure (atm, kPA, barr, lb/in2)
Isothermal
isochoric
Adiabatic
Isobaric
Volume (liters, m3, cm3)

6. Introduction Steam locomotive operates using laws of thermodynamics
So car engines…
And air conditioners…
And refrigerators!
Revisit conservation of energy in form of 1st law of thermodynamics.
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7. Thermodynamics systems
A thermodynamic system is any collection of objects that may exchange energy with its surroundings.
Popcorn in a pot can be a ‘thermodynamic system’.
In the thermodynamic process shown here, heat is added to the system,
Qin is positive!
The system does work on its surroundings to lift pot lid.
Wdone is positive!
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8. Thermodynamics systems
In a thermodynamic process, changes occur in the state of the system.
Be careful with signs!
Q is positive when heat flows into a system.
W is work done BY system, positive for expansion.
Both can appear NEGATIVE
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9. Thermodynamics systems
Q + heat flows into a system.
+500J of heat added
Q - heat flows out from a system.
Cool gas with ice – 500 J
W + done BY system as it expands.
Piston expands +10 cm
Gas does J of work
W – done ON system as it is compressed.
500 J work done on gas

10. Work done during volume changes
Consider a single molecule in a gas.
Collides with a surface moving to the right, which moves, so volume occupied by gas increases,
Molecule does positive work on the piston.
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11. Work done during volume changes
If piston moves left, so volume occupied by gas decreases; positive work is done ON the gas molecule during collision.
Gas molecules do “negative work” on piston.
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12. Work done during volume changes
Infinitesimal work done by system during small expansion dx is dW = pA dx
In finite change of volume from V1 to V2:
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13. Calculating Work Done BY Gas
The work done in moving a piston by an infinitesimal displacement is:
Easy: if V doesn’t change or P constant over volume change!
Harder: If P & V BOTH change
Figure The work done by a gas when its volume increases by dV = A dl is dW = P dV.

14. System undergoing an expansion with constant pressure.
Work on a pV-diagram
Work done equals area under curve on pV-diagram
If Pressure is constant, easiest!
W = pDV = N/m2 x m3 = Nm
W = p(V2 – V1)
System undergoing an expansion with constant pressure.
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15. System undergoing an expansion with varying pressure.
Work on a pV-diagram
Work done equals area under curve on pV-diagram
If pressure varies, easy!??
W =  p dV
Integral depends on the pressure function p(Volume)
System undergoing an expansion with varying pressure.
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16. Work on a pV-diagram
System undergoing a compression with varying pressure.
Work is done ON the gas, not BY the gas.
Work is negative
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17. Work depends on path chosen: Slide 1 of 4
Consider three different paths on a pV-diagram for getting from state 1 to state 2.
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18. Work depends on path chosen: Slide 2 of 4
System does a LARGE amount of work on path
WHY?

19. Work depends on path chosen!
Gas does LARGE amount of work under path
Why? HIGH Pressure expansion (1 => 3)
How? ADD heat!
Going from 3 => 2 no DV
No work done!
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20. Work depends on path chosen: Slide 3 of 4
System does a small amount of work on path
WHY??

21. Work depends on path chosen!
Gas does small amount of work under path
Why? LOW Pressure expansion (4 => 2)
How? ADD less heat!
Going from 1 => 4 no DV
No work done!
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22. Work depends on path chosen: Slide 4 of 4
Along smooth curve from 1 to 2, work done is different from that for other paths.
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23. First law of thermodynamics
Change in internal energy U of system equals heat added minus work done by system!
1st Law of Thermodynamics just a generalization of principle of conservation of energy.
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24. The First Law of Thermodynamics
DEinternal = +Q in – W out
Types of +Qin:
Chemical Energy (Gasoline, Diesel) through combustion
Conduction of Heat from Hot reservoir (steam from a boiler or nuclear power plant)
Absorption of heat from solar energy

25. The First Law of Thermodynamics
DEinternal = +Q in – W out
Types of –W out:
Expansion of Pistons (Internal combustion & steam engines)
Turning of a rotor (solar powered fan)

26. First law of thermodynamics
DU = Qin - Wout
Both Q (heat in) & W (work done by) depend on path chosen between states, but DU is independent of the path.
WHY???
If changes are infinitesimal, write 1st law as dU = dQ – dW.
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27. Internal energy
The internal energy of a cup of coffee depends on just its thermodynamic state.
how much water & ground coffee it contains… and…
its temperature.
It does not depend on the history of how the coffee was prepared—that is, the thermodynamic path that led to its current state.
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28. First law of thermodynamics
In a thermodynamic process, internal energy U of a system may increase.
Example:
More heat is added to system than system does work.
Internal energy of system increases.
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29. First law of thermodynamics
In a thermodynamic process, internal energy U of a system may decrease.
Example:
More heat flows out of the system than work is done ON the system (note sign of work!!)
Internal energy of the system decreases.
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30. First law of thermodynamics
In a thermodynamic process, internal energy U of a system may remain the same.
Example:
Heat added to system equals work done BY system.
Internal energy of system is unchanged.
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31. Example Using the first law.
2500 J of heat is added to a gas under a piston in a closed cylinder, and 1800 J of work is done by the system as its piston expands. What is the change in internal energy of the system?
Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.

32. Example 19-7: Using the first law.
2500 J of heat is added to a system, and 1800 J of work is done by the system. What is the change in internal energy of the system?
DE = J – (1800J) = +700 J
Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.
DEinternal = +Q in – W out

33. Example 19-7: Using the first law.
2500 J of heat is added to a gas under a piston in a closed cylinder, and 1800 J of work is done on the system as the piston is pushed back down. What is the change in internal energy of the system?
Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.

34. Example 19-7: Using the first law.
2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in internal energy of the system?
DE = J – (-1800J) = J
Solution: Both the heat added and the work done ON the system add to the system’s internal energy, which increases by 4300 J.
DEinternal = +Q in – W out

35. First law of exercise thermodynamics
Your body is a thermodynamic system.
When you do a push-up, your body does work, so W > 0.
Your body also warms up during exercise; but by perspiration & other means the body rids itself of this heat, so Q < 0.
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36. First law of exercise thermodynamics
Since Q is negative & W is positive, U = Q − W < and your body’s internal energy decreases.
That’s why exercise helps you lose weight! It uses up some of internal energy stored in your body in the form of fat.
Yay! Studying Physics helps you maintain your shape!
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37. First law of exercise thermodynamics
Since Q is negative & W is positive, U = Q − W < and your body’s internal energy decreases.
The same holds for studying! You do work lifting your book and calculator and pencil!
Yay! Thermodynamics proves that studying Physics helps you maintain your shape!
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38. Thermodynamic process for a human day
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39. Cyclic thermodynamic process: Qin +
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40. Cyclic thermodynamic process
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41. Four kinds of thermodynamic processes
Isochoric: Volume remains constant
Isobaric: Pressure remains constant
Isothermal: Temperature remains constant
Adiabatic: No heat is transferred in or out of system
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42. The four processes on a pV-diagram
Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a.
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43. The four processes on a pV-diagram
Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a.
Adiabatic – no heat in or out
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44. The four processes on a pV-diagram
Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a.
Isochoric – no work done
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45. The four processes on a pV-diagram
Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a.
Isobaric – constant pressure
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46. The four processes on a pV-diagram
Paths on pV-diagram for all four different processes for constant amount of ideal gas, all starting at state a.
Isothermal – constant Temp
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47. Four kinds of thermodynamic processes
Four specific kinds of thermodynamic processes that occur often in practical situations:
Adiabatic: No heat is transferred into or out of the system, so Qin = 0.
Therefore: DU = U2 – U1 = Qin –Wout = -W
Since no heat allowed in or out, change in internal energy is opposite of work.
If work is done BY system, internal energy decreases.
If work is done ON system, internal energy increases.
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48. Adiabatic process
When cork is popped on a bottle of champagne, pressurized gases inside the bottle expand rapidly & do positive work on outside air.
VERY little time for gases to exchange heat with their surroundings, so expansion is nearly adiabatic.
Hence the internal energy of expanding gases decreases (ΔU = –W ) and temperature drops.
This makes water vapor condense & form a miniature cloud.
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49. Exhaling adiabatically
Put your hand a few centimeters in front of your mouth, open your mouth wide, and exhale.
Your breath will feel warm on your hand, because the exhaled gases emerge at roughly the temperature of your body’s interior.
Now purse your lips as though you were going to whistle, and again blow on your hand.
The exhaled gases will feel much cooler.
In this case the gases undergo a rapid, essentially adiabatic expansion as they emerge from between your lips, so the temperature of the exhaled gases decreases.
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50. Four kinds of thermodynamic processes
Four specific kinds of thermodynamic processes that occur often in practical situations:
Adiabatic
Isochoric: The volume remains constant, so W = 0.
Therefore: DU = U2 – U1 = Qin – 0 = Qin
Since no work is done by, nor on the system, change in internal energy is just how much heat energy enters or leaves.
If the system is cooled, internal energy decreases.
If heat is added, internal energy increases.
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51. Four kinds of thermodynamic processes
Four specific kinds of thermodynamic processes that occur often in practical situations:
Adiabatic
Isochoric
Isobaric: The pressure remains constant, so W = p(V2 – V1).
Therefore DU = U2 – U1 = Qin – pDV
If dV is negative (compress the system!), internal energy could still decrease if you pull enough heat out! Qin< 0
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52. Isobaric process Most cooking involves isobaric processes.
Air pressure above a saucepan or frying pan, or inside a microwave oven, remains essentially constant while food is being heated.
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53. Four kinds of thermodynamic processes
Four specific kinds of thermodynamic processes that occur often in practical situations:
Adiabatic:
Isochoric:
Isobaric:
Isothermal: The temperature remains constant.
How??
PV = nRT means PV is constant! Use this in W =  p dV
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54. Calculating Work – Isothermal Processes
An isothermal process is one in which the temperature does not change.
Figure PV diagram for an ideal gas undergoing isothermal processes at two different temperatures.
Figure An ideal gas in a cylinder fitted with a movable piston.

55. Calculating Work – Isothermal Processes
An isothermal process is one in which the temperature does not change.
For isothermal process to take place, assume system is in contact with a heat “reservoir”.
Assume that system remains in equilibrium throughout all steps.
Figure PV diagram for an ideal gas undergoing isothermal processes at two different temperatures.
Figure An ideal gas in a cylinder fitted with a movable piston.

56. Isothermal processes are S L O W
To keep T constant, SLOWLY add a bit of heat, let gas expand a bit, repeat..
Use a large reservoir of surrounding material to keep T constant.
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57. Calculating Work: Isothermal Process
For isothermal process, P = nRT/V.
Integrate to find work done in taking gas from A to B:
Figure Work done by an ideal gas in an isothermal process equals the area under the PV curve. Shaded area equals the work done by the gas when it expands from VA to VB.

58. Adiabatic vs Isothermal processes
In adiabatic process, no heat is transferred in or out of gas, so Q = 0.
As gas expands, it gas does positive work W on its environment, so its internal energy decreases.
Gas temperature drops!
Note that adiabatic curve at any point is always steeper than an isotherm at that point.
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59. Internal energy of an ideal gas
The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.
Test this with “free expansion” (Joule expansion)
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60. Free Expansion
A “free expansion” is when gas does no work expanding into a larger volume. Wout = 0
Assume insulated so Qin = 0
By first law, DU = Qin – Wout = 0
Since U is a function of T only, T won’t change in this process!
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61. Free Expansion
In reality, most gases DO cool down as they expand into a vacuum.
Change kinetic energy into internal intermolecular potential energy
He gas above 40K and H gas above 200 K both get *warmer*.
Change intermolecular PE into KE!
The real key:
Free expansion is an IRREVERSIBLE process!
Hints at “ENTROPY”
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62. Heat capacities of an ideal gas
Consider a constant volume process.
Add heat & raise temperature of ideal gas in rigid container with constant volume (ignoring container thermal expansion).
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63. Heat capacities of an ideal gas
For a Constant volume process
DU = Qin - Wout
Wout = 0 for Cv process
DU = Qin = nCvDT
CV = molar heat capacity at Constant Volume.
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64. Heat capacities of an ideal gas
CV = molar heat capacity at Constant Volume.
From the First Law of Thermodynamics
DU = Qin = nCvDT
So between ANY two temperatures in an ideal gas, no matter what the process is, you know change in U:
DU = nCvDT
Change in internal energy of ideal gas no matter what process!
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65. Heat capacities of an ideal gas
Now consider a constant pressure process to same T…
Add heat, let gas expand just enough to keep pressure constant as temperature rises, to the same final T.
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66. Heat capacities of an ideal gas
First Law:
DU = Qin - Wout
How much heat is added?
Define Cp = molar heat capacity at Constant Pressure
Qin = nCpDT
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67. Heat capacities of an ideal gas at Constant P
First Law:
DU = Qin - Wout
But now, work IS done!
DU = nCpDT – pDV
From ideal gas law
pDV = nRDT
DU = nCpDT – nRDT
But wait…. There’s more!! Didn’t we just say we know DU no matter what process is involved??
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68. Relating Cp and CV for an ideal gas
DU = nCvDT
Change in internal energy of ideal gas no matter what process!
If is same (ending at same T!)
Then DU = nCpDT – nRDT
becomes
n CvDT = n CpDT – nRDT
or… n CvDT + nRDT= n CpDT
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69. Relating Cp and CV for an ideal gas
n CvDT + nRDT = n CpDT
So… Cp = Cv + R
Cp > CV
To produce same temperature change, more heat is required at Constant Pressure than at Constant Volume
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70. Relating Cp and CV for an ideal gas
To produce same temperature change, more heat is required at constant pressure than at constant volume since is the same in both cases.
Cp > CV
Cp = CV + R
R is the gas constant R = J/mol ∙ K.
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71. The ratio of heat capacities
The ratio of heat capacities is:
For monatomic ideal gases,
For diatomic ideal gases,
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