1. First Law of Thermodynamics
Eint = Q + W
Eint is a state variable, so only depends on condition (P, V, T,…) of system.
Therefore, Eint only depends on initial and final states of the system.
Q and W, however, depend on the process.
other state variables }

2. Eint = Q + W
We will mostly be concerned with (nonmagnetic, nonelectric) liquids and gases, for which W = - P dV (if the pressure stays well defined during the process: quasi-static)
Types of Processes:
Isothermal: System is in contact with an (“infinite”) energy reservoir (also called a “heat bath”) that keeps its temperature constant.
2) Isobaric: P = constant: W = -P V
3) Isovolumetric: V = constant: W = 0, Eint = Q.
4) Adiabatic Process: Q = 0. Either:
a) in “perfectly” insulated container
b) occurs rapidly (and perhaps not quasi-static) so heat does not have enough time to flow
in/out. (e.g. this is what happens during power and compression strokes in engines)
Example: Adiabatic Free Expansion: Gas rushing to fill a volume. Not quasi-static and
pressure not defined during the process. Q = 0, W = 0  Eint = 0.
Since Q depends on process, heat capacity C  dQ/dT also depends on process:
Molar specific heats: CP  (1/n) dQ/dT when P = constant ;
CV  (1/n) dQ/dT when V = constant.
[For liquids and solids, CP  CV , but they can be quite different for gases.]

3. Ethanol: c = 2400 J/kgoC [or CV  CP = (2400 J/kg oC) (0.046 kg/mol)]
Problem: Ethanol , with an initial volume of V0 = m3, is in a cylinder with a piston at a constant pressure of 2 atm. The ethanol is heated from 20 oC to oC. What is the change in its internal energy?
Ethanol: c = 2400 J/kgoC [or CV  CP = (2400 J/kg oC) (0.046 kg/mol)]
 = 1.12 x 10-4 / oC
 = 790 kg/m3
Q = mc T [= nCP T]
m = m3  790 kg/m3 = 3.16 kg [or n=3.16 kg/(0.046 kg/mol)]
Q = (3.16 kg)(2400 J/kgoC) (15 oC)
Q = 1.14 x 105 J
W = -P V = -P V0 T
W = - (2 x 1.01 x 105 N/m2) (1.12 x 10-4 /oC) (0.004 m3) (15 oC)
W = J
[W<0 because ethanol expands, doing work on surroundings]
Eint = Q + W  Q = 1.14 x 105 J
[|Q| >> |W|, as is usually case for liquids and solids,
since their volumes are almost constant –
i.e. they are very incompressible]
F = PA

4. Cycle
System returns to initial T,V,P,Eint (so Eint = 0), but there may be net Q and net W.
For cycle with steps “j”: Eint(cycle) = Eint(j) = Qj + Wj = 0, but may have Qj = -Wj  0.
For clockwise (quasi-static) cycle, net W = Wj = - area inside.
For counterclockwise (quasi-static) cycle, net W = Wj = + area inside.

5. Some Properties of an Ideal Gas
Equation of State: P = nRT/V (definition of Kelvin temperature scale)
Eint  n (i.e. it is “extensive” – proportional to amount of gas)
Eint / n only depends on T [= (PV)/(Rn)], so if T is constant, Eint is also constant.
(i.e. Eint does not separately depend on V or p.) – discovered by Joule
CP – CV = R (Derived in Ch. 21.)
5) For a monoatomic ideal gas (e.g. He, Ne, Ar, but not N2, O2, H2, CO2, …),
CV = 3/2 R, CP = 5/2 R (Derived in Ch. 21)
Adiabatic Free Expansion: Since Eint = 0  T = 0.
Quasi-Static Isothermal Process:
P = nRT/V, W = - nRTdV/V = -nRTdV/V = -nRT ln(Vf/Vi)
Since T = constant, Eint = 0, so Q = - W = +nRT ln(Vf/Vi)

6. n moles of helium (monoatomic ideal gas)
Problem: Find Q and W for each step and check that Eint = Q + W = 0.

7. } n moles of helium (monoatomic ideal gas) WAB = 0 (because V = 0)
QAB = nCV(TB-TA) = 3/2 nR (3Pi Vi /nR – Pi Vi/nR) = 3PiVi
WBC = -3Pi (3Vi-Vi) = -6PiVi
QBC = nCP (TC-TB) = 5/2 nR (9PiVi-3PiVi)/nR = 15 PiVi
WCD = 0
QCD = nCV(TD-TC) = 3/2 nR (3Pi Vi /nR – 9Pi Vi/nR) = -9PiVi
WDA = Pi (3Vi-Vi) = 2PiVi
QDA = nCP (TA-TD) = 5/2 nR (PiVi-3PiVi)/nR = -5 PiVi
W = 0-6PiVi+0+2PiVi
W = -4 PiVi
[W<0: net work done by gas]
Q = ( )PiVi
Q = +4 PiVi
[Q > 0: net heat into gas]
Eint(cycle) = Q + W = 0

9. } W = 0 - 2P0V0 +3P0V0 ln(3) W = P0V0 [3ln(3) -2]
[W>0: net work done on gas]
Q = ( ln(3))P0V0
Q = [2 - 3 ln(3)] P0V0
[Q < 0: net heat out of gas]
Eint(cycle) = Q + W = 0
WAB = 0
QAB = nCV (TB – TA) = 3/2 nR (P0V0 – 3P0V0)/nR = -3P0V0
WBC = -P0 (3V0 – V0) = -2 P0V0
QBC = nCP (TC – TB) = 5/2 nR (3P0V0 – P0V0) = 5 P0V0
CA: isotherm
WCA = -nRT ln[(V0/3V0] = 3P0V0 ln(3)
But Eint(CA) = 0 since isothermal
 QCA = -WCA = -3P0V0 ln(3)

10. Problem: Find Q and W for each step.

11. n moles of helium
3P0 P0 V0
3V0
WBC = 0
QBC = nCV (TC-TB) = 3/2 nR (3P0V0 - 9P0V0)/nR = - 9P0V0
WCA = -P0(V0-3V0) = + 2P0V0
QCA = nCP(TA-TC) = 5/2 nR (P0V0 – 3P0V0) / nR = -5P0V0
WAB = -pdV = - area under AB = -[½ (2P0*2V0 ) + P0*2V0]
= - 4P0V0
QAB = ??

12. n moles of helium
3P0 P0 V0
3V0
To find QAB, use Eint(cycle) = Q + W = 0
QAB = – (QBC + QCA + WAB + WBC + WCA)
QAB = - ( ) P0V0
QAB = 16 P0V0

13. 8 moles of Argon (a monoatomic ideal gas) undergo the cycle below, in which step CD is an isotherm and AB is adiabatic. How much work is done in step AB?

14. 8 moles of Argon (a monoatomic ideal gas) undergo the cycle below, in which step CD is an isotherm and AB is adiabatic. How much work is done in step AB?
Eint = 0 = WAB + QAB + WBC + QBC + WCD + QCD + WDA + QDA
[WCD = - QCD]
0 = WAB + 0 – PB (VC – VB)+ n CP (TC-TB) – PA (VA – VD) + nCP (TA – TD)
[at any point: nCPT = 5/2 nR (PV/nR) = 5/2 PV]
WAB = [(3.03 x 105) (0.31) -5/2 (3.03 x 105) (0.31) – (1.01 x 105)(1.00)
+ 5/2 (1.01 x 105) (1.00)] (N/m2) (m3)
WAB = 3/2 ( x x 105) J = x 104 J

15. T0 C, T
Suppose an object, with heat capacity C is at temperature T, in thermal contact with surroundings, which are at temperature T0.
Q will flow into object (Q >0) if T < T0 and Q will flow out (Q< 0) if T > T0.
dQ = C dT
 dQ/dt = C dT/dt

16. Mechanisms of Heat Flow
dQ/dt = C dT/dt
[Note: dQ/dt > 0 if T < T0 and dQ/dt <0 if T > T0.]
Mechanisms of Heat Flow
Conduction
Radiation
Convection

17. Convection: Heating by matter transfer
Suppose there is a temperature gradient in a gas or liquid in a gravitational field. If the density varies with temperature [e.g. ideal gas: n/V = P/RT, so density decreases with increasing T if P = constant ], then the more dense (usually the colder) material will “sink” (move in direction of gravitational force) and the less dense material will flow opposite the gravitational field. That is, the warmer material will flow to the colder regions and the colder material to the warmer regions
cool
air
warm
air g
Earth

18. Water
There will also be convection in water. For T > 4 oC, warm water is less dense and will rise. However, for 0 oC < T < 4 oC, cold water is less dense and will rise  lakes freeze from the top.