The Thermodynamic Cycle

Heat engines and refrigerators operate on thermodynamic cycles where a gas is carried from an initial state through a number of intermediate states and ultimately returned to its initial state.

During a thermodynamic cycle:

Heat Engines convert a portion of the heat input to work

Refrigerators convert input work to the transfer of heat from low temperature to high temperature

Examples:Steam engineInternal gas combustion engineDiesel engine

Examples:RefrigeratorHeat Pump

4.5 moles of O2 are carried through a thermodynamic cycle composed of four steps: (i → 1→ 2→ 3→ i)

The molar specific heats for oxygen are:

CP =29.18 Jmole⋅K

CV =20.83 Jmole⋅K

i → 1: (Isochoric Vi =0.06m3) increase in

2.5pressure from ×105Pa 3.5to ×105Pa

2 → 3 : Isothermal expansion returning

2.5the gas to its initial pressure of×105Pa

1→ 2 : 0.10Isobaric expansion to m3

3 → i : Isobaric compression returning

the gas to its initial state

3.5×105

2.5 ×105

0.06 0.10 V3

i

1 2

3

isochoric

isobaric

isothermal

isobaric

Pressure, Pa

Volume ,m3

(4.5,0.06, 3.5×105 ,T1 )

(4.5,0.06, 2.5×105 ,Ti )

(4.5,0.10, 3.5×105 ,T2 )

(4.5,V3 , 2.5×105 ,T3 )

What is the gas temperature at point "i"?

PiVi =nRTi

Ti =PiVinR

(4 .5 , 0.06 , 2 .5 ×10 5 ,T i )

Ti =2.5 ×105 Pa( ) .06m3( )

4.5moles( ) 8.31 Jmole• K( )

Ti =401K

What is the gas temperature at point "1"?

(4 .5 , 0.06 , 3 .5 ×10 5 ,T 1)

P1V1 =nRT1

T1 =P1V1nR

T1 =3.5×105 Pa( ) .06m3( )

4.5moles( ) 8.31 Jmole• K( )

T1 =562K

What is the gas temperature at point "2"?

(4 .5 , 0 .10 , 3 .5 ×10 5 ,T 2 )

P2V2 =nRT2

T2 =P2V2nR

T2 =3.5×105 Pa( ) .10m3( )

4.5moles( ) 8.31 Jmole• K( )

T2 =936K

What is the gas volume at point "3"?

(4 .5 , V3 , 2 .5 ×10 5 ,T 3 )

Since the step 2 → 3 ,is isothermal 3 the temperature at point is the same

2. as the temperature at point T3 =T2 =936KP3V3 =nRT3

V3 =nRT3

P3

V3 =(4.5moles)(8.31 J

mole⋅K )(936K )

2.5×105Pa

V3 =.14m3

3.5×105

2.5 ×105i

1 2

3

isochoric

isobaric

isothermal

isobaric

0.06 0.10 V3

Pressure, Pa

Volume ,m3

0.06 0.10 0.14

(4.5,0.06, 3.5×105 , 562) (4.5,0.10,3.5×105 , 936)

(4.5, 0.14, 2.5×105 ,936 )5(4.5,0.06, 2.5×10 ,401 )

What is the magnitude and direction

of the heat flow in the step "i → 1 "?

Qi→ 1 =nCVΔTi→ 1 =nCV T1 −Ti( )

Qi→ 1 = 4.5moles( ) 20.83 Jmole• K( ) 562K −401K( )

Qi→ 1 =1.509 ×104 J K into the gas

What is the magnitude and direction

of the work done in the step "i → 1 "?

ΔVi→1 = 0∴ Wi→1 = 0

What is the magnitude and direction

of the change in thermal energy in the

step "i → 1 "?ΔUi→1 = Q i→1 + Wi→1

ΔUi→1 = 1.509 × 104 J + 0

ΔUi→1 = 1.509 × 104 JK increase

What is the magnitude and direction

of the heat flow in the step "1 → 2 "?

Q1→ 2 =nCPΔT1→ 2 =nCP T2 −T1( )

Q1→ 2 = 4.5moles( ) 29.18 Jmole• K( ) 936K −562K( )

Q1→ 2 =4.911×104 J K into the gas

What is the magnitude and direction

of the work done in the step "1→ 2 "?W1→ 2 =−P1ΔV1→ 2 =−P1 V2 −V1( )W1→ 2 =−3.5×105 Pa .1m3 −.06m3( )

W1→ 2 =−1.4 ×104 J K by the gas

What is the magnitude and direction

of the change in thermal energy

in the step "1→ 2 "?

ΔU1→2 = Q1→2 + W1→2

ΔU1→2 = 4.911 × 104 J + −1.4 × 104 J( )

ΔU1→2 = 3.511 × 104 JK increase

What is the magnitude and direction

of the work done in the step "2 → 3 "?

W2→ 3 =−nRT2 lnV3V2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

W2→ 3 =−4.5moles( ) 8.31 Jmole• K( ) 936K( ) ln .14m3

.10m3 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

W2→ 3 =−1.178 ×104 J K by the gas

What is the magnitude and direction

of the heat flow in the step "2 → 3 "?

Q2→ 3 =−W2→ 3 =1.178 ×104 J K into the gas

What is the magnitude and direction

of the change in thermal energy in the

step "2 → 3 "?

ΔU2→3 = 0

What is the magnitude and direction

of the heat flow in the step "3 → "i ?Q3→ i =nCPΔT3→ i =nCP Ti −T3( )

Q3→ i = 4.5moles( ) 29.18 Jmole• K( ) 401K −936K( )

Q3→ i =−7.025 ×104 J K out of the gas

What is the magnitude and direction

of the work done in the step "3 → "i ?W3→ i =−P3ΔV3→ i =−P3 Vi −V3( )

W3→ i =−2.5×105 Pa .06m3 −.14m3( )

W3→ i =2.0 ×104 J K on the gas

What is the magnitude and direction

of the change in thermal energy in the

step "3 → "i ?ΔU3→i = Q3→i + W3→i

ΔU3→i = −7.025 × 104 J + 2.0 × 104 J

ΔU3→i = −5.025 × 104 JK decrease

3.5×105

2.5 ×105i

1 2

3

isochoric

isobaric

isothermal

isobaric

Pressure, Pa

Volume ,m30.06 0.10 0.14

Qi→ 1 =1.509×104 J

Q1→ 2 =4.911×104 J

W1→ 2 =−1.4×104 J

Q2→ 3 =1.178×104 JW2→ 3 =−1.178×104 J

Q3→ i =−7.025×104 J W3→ i =2.0×104 J

Summary

Net for Cycle

Step Process Work-W J

Heat Flow-Q J

Change in Thermal

Energy-Δ UJ

i→1 Isochoric

1→2 Isobaric

2→3 Isothermal

3→ i Isobaric

0 1.509x104J 1.509x104J≈0

-1.400x104J 4.911x104J 3.511x104J

-1.178x104J 1.178x104J 0

2.000x104J -7.025x104J -5.025x104J

-5.780x103J 5.730x103J -.05x103J ≈0

When an ideal gas is carried through a complete cycle returning it to its initial state the net change in thermal energy is zero.

What is the efficiency of this heat engine?

net work out = 5.78 ×103 J

total heat in = 1.509×104 J +4.911×104 J +1.178×104 Jtotal heat in = Q i→ 1 +Q1→ 2 +Q2→ 3

total heat in = 7.598 ×104 J

efficiency =η= net work out total heat in×100%

η= 5.78 × 103J7.598 × 104 J

× 100%

η=7.6%

Why is the molar specific heat at constant pressure, CP, always greater than the molar specific heat at constant volume, CV?

Recall: The specific heat of a substance is the amount of heat required to change the temperature of 1kg of the substance by 1C°,(or 1K). The equation relating heat flow and change in temperature is: Q=mcΔT, or Q=nCΔT. From the second equation an expression for molar specific heat can be obtained:

C = QnΔT

CP = QnΔTP

CV = QnΔTV

Suppose we take two containers each holding 1 mole of the same gas.

We will take the first contained gas through an isobaric expansion process where heat Q=1000J flows into the gas and the gas does 200J of work.

We will take the second contained gas through an isochoric expansion process where heat Q=1000J flows into the gas and the gas does no work.

For an ideal gas the thermal energy, U only depends on the temperature, T.

Let’s say that for every 100J change in thermal energy there is a corresponding 1C° change in the temperature.

Isobaric

Qin =1000J

Wout =200J

ΔU = 800J

ΔT = 8K

CP = QnΔTP

CP = 1000J1mole⋅8K

CP =125 Jmole⋅K

Isochoric

Qin =1000J ΔU = 1000J

ΔT = 10K

CV = QnΔTV

CV = 1000J1mole⋅10K

CV =100 Jmole⋅K

CP >CV

0

25

50

75

100

125

150

175

200

0 25 50 75 100 125 150 175 200

Cp, J/mol · K

Cv, J/mol · KCp versus Cv

Cv =Cp −8.31

Cv =Cp −R

Gas Cp, J/mol · K Cv, J/mol · K

Air 29.11 20.80Argon 20.78 12.47Butane 99.76 91.45Carbon Dioxide 37.23 28.91Carbon Monoxide 29.13 20.84Ethane 53.11 44.80Ethylene 43.43 35.12Helium 20.79 12.47Hydrogen 28.84 20.53Methane 36.16 27.84Neon 20.79 12.47Nitrogen 29.11 20.81Octane 195.48 187.17Oxygen 29.38 21.06Propane 74.06 65.74Steam 33.73 25.42