Fico1 Boyle Calculos Y Resultados (melquiades).docx

PROCEDIMIENTO EXPERIMENTAL Instalar el equipo de Boyle con baño a temperatura constante, anotar lecturas iniciales, luego tomar las mediciones a diferentes volúmenes del gas en la columna para obtener valores de la presión.

Presión (𝟏𝟎𝟓 Pa)

Volumen (mL)

1

3

2.1

2.1

2.35

2

2.4

1.9

2.6

1.8

2.9

1.7

3.1

1.6

3.4

1.5

3.7

1.4

4.3

1.3

4.6

1.2

5.3

1.1

6

1

6.9

0.9

8

0.8

9.6

0.7

11.8

0.6

14.7

0.5

19.8

0.4

29.1

0.3

37.1

0.25

a) Grafica P vs V

GRÁFICA P Vs V 40 35

PRESIÓN

30 25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

VOLUMEN

b) Grafica P vs 1/V

3

3.5

4

4.5

Grafico P vs 1/V 2 y = 6.7927x - 1.2267 R² = 0.9717

1.8

PRESIÓN (10^5PA)

1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

1/VOLUMEN

De los puntos obtenidos en la experiencia se escogen 10 de estos para resolver y encontrar K, hacemos uso de las otras ecuaciones: 

Gas ideal



Van der Waals



Redlich-kwong



Por compresibilidad de los gases



Virial

Los puntos escogidos convenientemente son: Volumen (𝐦𝐋)

Presión (𝟏𝟎𝟓 Pa)

3

1

2.5

1.5

2.1

2.1

1.8

2.6

1.5

3.4

1.1

5.3

0.8

8

0.6

11.8

0.4

19.8

0.25

37.1

a) Grafica P vs V

MUESTRA DE LA GRÁFICA P Vs V 40

35 30 25

Presión

20 15 10 5 0 0

0.5

1

1.5

2

Volumen

2.5

3

3.5

b) Grafico P vs 1/V Esti ma

Grafico P vs 1/V

mo

4

3.5

s K

y = 2.9032x + 0.4501 R² = 0.9438

con

Presión (10^5 Pa)

3

los

2.5

pun

2

tos 1.5

real

1

es:

0.5 0 0

0.2

0.4

0.6

0.8

1/Volumen

𝑘1 = 𝑃1 𝑉1 = 1𝑥105 𝑃𝑎. 3𝑐𝑚3 = 3𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘2 = 𝑃2 𝑉2 = 1,51𝑥105 𝑃𝑎. 2,5𝑐𝑚3 = 3,775𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘3 = 𝑃3 𝑉3 = 2,1𝑥105 𝑃𝑎. 2,1𝑐𝑚3 = 4,41𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘4 = 𝑃4 𝑉4 = 2,6𝑥105 𝑃𝑎. 1,8𝑐𝑚3 = 4,68𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘5 = 𝑃5 𝑉5 = 3,4𝑥105 𝑃𝑎. 1,5𝑐𝑚3 = 5,1𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘6 = 𝑃6 𝑉6 = 5,3𝑥105 𝑃𝑎. 1,1𝑐𝑚3 = 5,83𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘7 = 𝑃7 𝑉7 = 8𝑥105 𝑃𝑎. 0,8𝑐𝑚3 = 6,4𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘8 = 𝑃8 𝑉8 = 11,8𝑥105 𝑃𝑎. 0,6𝑐𝑚3 = 7,08𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘9 = 𝑃9 𝑉9 = 19,8𝑥105 𝑃𝑎. 0,4𝑐𝑚3 = 7,92𝑥105 𝑃𝑎. 𝑐𝑚3 𝑘10 = 𝑃10 𝑉10 = 37,1𝑥105 𝑃𝑎. 0,25𝑐𝑚3 = 9,275𝑥105 𝑃𝑎. 𝑐𝑚3

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 =

(3+3,775+4,41+4,68+5,1+5,83+6,4+7,08+7,92+9,275)𝑥105 𝑃𝑎.𝑐𝑚3 10

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 = 5,7465𝑃𝑎. 𝑐𝑚3

1

1.2

ECUACIÓN DE LOS GASES IDEALES

PVm  RT Vm n  V PV  k

𝑅 = 83,14𝑥105

𝑃𝑎. 𝑐𝑚3 𝑚𝑜𝑙. 𝑘

𝑇 = 298𝑘 Para el punto 1: 𝑃1 = 1𝑥105 𝑃𝑎

𝑉1 = 3𝑐𝑚3

𝑛 = 0,00012𝑚𝑜𝑙

𝑃1 𝑉𝑚1 = 𝑅𝑇 5

1𝑥10 𝑃𝑎𝑉𝑚1

𝑃𝑎. 𝑐𝑚3 = 83,14𝑥10 𝑥298𝑘 𝑚𝑜𝑙. 𝑘 5

𝑉𝑚1 = 24775,72

𝑐𝑚3 𝑚𝑜𝑙

𝑉1 = 𝑉𝑚1 𝑛 = 2,974𝑐𝑚3 𝑘1 = 𝑃1 𝑉1 = 1𝑥105 𝑃𝑎. 2,974𝑐𝑚3 = 297400𝑃𝑎. 𝑐𝑚3

Para el punto 2: 𝑃2 = 1,51𝑥105 𝑃𝑎

𝑉2 = 2.5𝑐𝑚3

𝑃2 𝑉𝑚2 = 𝑅𝑇 1,51𝑥105 𝑃𝑎𝑉𝑚2 = 83,14𝑥105 𝑉𝑚2 = 16407,76

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑛 = 0,00015𝑚𝑜𝑙

𝑉2 = 𝑉𝑚2 𝑛 = 2,461164𝑐𝑚3 𝑘2 = 𝑃2 𝑉2 = 1,51𝑥105 𝑃𝑎. 2,461164𝑐𝑚3 = 371635,16𝑃𝑎. 𝑐𝑚3

Para el punto 3: 𝑃3 = 2,1𝑥105 𝑃𝑎

𝑉3 = 2,1𝑐𝑚3

𝑛 = 0,00017𝑚𝑜𝑙

𝑃3 𝑉𝑚3 = 𝑅𝑇 2,1𝑥105 𝑃𝑎𝑉𝑚3 = 83,14𝑥105 𝑉𝑚3 = 11797,96

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 𝑚𝑜𝑙

𝑉3 = 𝑉𝑚3 𝑛 = 2,00565𝑐𝑚3 𝑘3 = 𝑃3 𝑉3 = 2.1𝑥105 𝑃𝑎. 2,0065𝑐𝑚3 = 421186,5𝑃𝑎. 𝑐𝑚3

Para el punto 4: 𝑃4 = 2,6𝑥105 𝑃𝑎

𝑉4 = 1.8𝑐𝑚3

𝑛 = 0,00018𝑚𝑜𝑙

𝑃4 𝑉𝑚4 = 𝑅𝑇 2,6𝑥105 𝑃𝑎𝑉𝑚4 = 83,14𝑥105 𝑉𝑚4

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 = 9529,1231 𝑚𝑜𝑙

𝑉4 = 𝑉𝑚4 𝑛 = 1,71524𝑐𝑚3 𝑘4 = 𝑃4 𝑉4 = 2,6𝑥105 𝑃𝑎. 1,71524𝑐𝑚3 = 445962,4𝑃𝑎. 𝑐𝑚3

Para el punto 5: 𝑃5 = 3,4𝑥105 𝑃𝑎

𝑉5 = 1,5𝑐𝑚3

𝑃5 𝑉𝑚5 = 𝑅𝑇 3,4𝑥105 𝑃𝑎𝑉𝑚5 = 83,14𝑥105 𝑉𝑚5 = 7286,9764

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑛 = 0,00021𝑚𝑜𝑙

𝑉5 = 𝑉𝑚5 𝑛 = 1,53026𝑐𝑚3 𝑘5 = 𝑃5 𝑉5 = 3,4𝑥105 𝑃𝑎. 1,53026𝑐𝑚3 = 520322,4𝑃𝑎. 𝑐𝑚3

Para el punto 6: 𝑃6 = 5,3𝑥105 𝑃𝑎

𝑉6 = 1,1𝑐𝑚3

𝑛 = 0,000235𝑚𝑜𝑙

𝑃6 𝑉𝑚6 = 𝑅𝑇 5,3𝑥105 𝑃𝑎𝑉𝑚6 = 83,14𝑥105 𝑉𝑚6 = 4674,664

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 𝑚𝑜𝑙

𝑉6 = 𝑉𝑚6 𝑛 = 1,098𝑐𝑚3 𝑘6 = 𝑃6 𝑉6 = 5,3𝑥105 𝑃𝑎. 1,098𝑐𝑚3 = 581940𝑃𝑎. 𝑐𝑚3

Para el punto 7: 𝑃7 = 8𝑥105 𝑃𝑎

𝑉7 = 0,8𝑐𝑚3

𝑛 = 0,000258𝑚𝑜𝑙

𝑃7 𝑉𝑚7 = 𝑅𝑇 8𝑥105 𝑃𝑎𝑉𝑚7 = 83.14𝑥105 𝑉𝑚7

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 = 3096,965 𝑚𝑜𝑙

𝑉7 = 𝑉𝑚7 𝑛 = 0,799𝑐𝑚3 𝑘7 = 𝑃7 𝑉7 = 8𝑥105 𝑃𝑎. 0,799𝑐𝑚3 = 639200𝑃𝑎. 𝑐𝑚3

Para el punto 8: 𝑃8 = 11,8𝑥105 𝑃𝑎

𝑉8 = 0,6𝑐𝑚3

𝑃8 𝑉𝑚8 = 𝑅𝑇 11,8𝑥105 𝑃𝑎𝑉𝑚8 = 83,14𝑥105 𝑉𝑚8 = 2099,637

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑛 = 0,000285𝑚𝑜𝑙

𝑉8 = 𝑉𝑚8 𝑛 = 0,598𝑐𝑚3 𝑘8 = 𝑃8 𝑉8 = 11,8𝑥105 𝑃𝑎. 0,598𝑐𝑚3 = 705640𝑃𝑎. 𝑐𝑚3

Para el punto 9: 𝑃9 = 19,8𝑥105 𝑃𝑎

𝑉9 = 0,4𝑐𝑚3

𝑛 = 0,000319𝑚𝑜𝑙

𝑃9 𝑉𝑚9 = 𝑅𝑇 19,8𝑥105 𝑃𝑎𝑉𝑚9 = 83,14𝑥105 𝑉𝑚9 = 1251,298

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 𝑚𝑜𝑙

𝑉9 = 𝑉𝑚9 𝑛 = 0,399𝑐𝑚3 𝑘9 = 𝑃9 𝑉9 = 19,8𝑥105 𝑃𝑎. 0,399𝑐𝑚3 = 790020𝑃𝑎. 𝑐𝑚3

Para el punto 10: 𝑃10 = 37,1𝑥105 𝑃𝑎

𝑉10 = 0,25𝑐𝑚3

𝑛 = 0,000374𝑚𝑜𝑙

𝑃10 𝑉𝑚10 = 𝑅𝑇 37,1𝑥105 𝑃𝑎𝑉𝑚10 = 83,14𝑥105 𝑉𝑚10

𝑃𝑎. 𝑐𝑚3 𝑥298𝑘 𝑚𝑜𝑙. 𝑘

𝑐𝑚3 = 667,809 𝑚𝑜𝑙

𝑉10 = 𝑉𝑚10 𝑛 = 0,249𝑐𝑚3 𝑘10 = 𝑃10 𝑉10 = 37,1𝑥105 𝑃𝑎. 0,249𝑐𝑚3 = 923790𝑃𝑎. 𝑐𝑚3

Hallando el k promedio: 𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 =

𝑘1 + 𝑘2 + 𝑘3 + 𝑘4 + 𝑘5 + 𝑘6 + 𝑘7 + 𝑘8 + 𝑘9 + 𝑘10 10

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 (297400 + 371635,16 + 421186,5 + 445962,4 + 520322,4 + 581940 + 639200 + 705640 + 790020 + 923790) = 10

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 = 569709,646𝑃𝑎. 𝑐𝑚3 a) Gráfico P vs V

Gráfico P vs V 40 35 30

PRESIÓN

25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

VOLUMEN

b) Gráfico P vs 1/V

Gráfico P vs 1/V 40 35 30

Presión

25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

1/Volumen

3

3.5

4

4.5

ECUACIÓN DE VAN DER WAALS

𝑽𝟑𝒎 − (𝒃 +

𝑹𝑻 𝟐 𝒂 𝒂𝒃 ) 𝑽𝒎 + ( )𝑽𝒎 − =𝟎 𝑷 𝑷 𝑷

𝑃𝑎. 𝑐𝑚3 𝑚𝑜𝑙. 𝑘 𝑇 = 298𝑘

𝑅 = 83,14𝑥105

Cálculo del a y b para el aire, considerando que su composición es: 21% O 2 y 79% N2 𝑎̅ = ∑ 𝑥𝑖 . 𝑎𝑖 𝑖

𝑎̅ = 𝑥𝑁2 . 𝑎𝑁2 + 𝑥𝑂2 . 𝑎𝑂2

De datos obtenidos en tablas:

𝑎𝑁2 = 1,39

𝑎𝑡𝑚.𝑑𝑚6 𝑚𝑜𝑙2

𝑎𝑂2 = 1,36

𝑎𝑡𝑚.𝑑𝑚6 𝑚𝑜𝑙2

Reemplazando: 𝑎̅ = (0,79) × (1,39

𝑎𝑡𝑚. 𝑑𝑚6 𝑎𝑡𝑚. 𝑑𝑚6 (0,21) )+ × (1,36 ) 𝑚𝑜𝑙 2 𝑚𝑜𝑙 2 𝑎̅ = 1,38

𝑎𝑡𝑚. 𝑑𝑚6 𝑚𝑜𝑙 2

Realizamos el mismo procedimiento para hallar la otra constante:

𝑏̅ = ∑ 𝑥𝑖 . 𝑏𝑖 𝑖

𝑏̅ = 𝑥𝑁2 . 𝑏𝑁2 + 𝑥𝑂2 . 𝑏𝑂2

De datos obtenidos en tablas: 𝑏𝑁2 = 3,913 × 10−2

𝑑𝑚3

𝑏𝑂2 = 3,183 × 10−2

𝑚𝑜𝑙

𝑑𝑚3 𝑚𝑜𝑙

Reemplazando: 𝑏̅ = (0,79) × (3,913 × 10−2

𝑑𝑚3 𝑑𝑚3 ) + (0.21) × (3,183 × 10−2 ) 𝑚𝑜𝑙 𝑚𝑜𝑙

𝑏̅ = 0,0376

𝑑𝑚3 𝑚𝑜𝑙

Para un correcto trabajo, procedemos a hacer un cambio de unidades a las contantes a y b del aire: 𝑎̅ = 1.38

𝑎̅ = 1.38 × 101325 × 106

𝑎𝑡𝑚. 𝑑𝑚6 𝑃𝑎. 𝑐𝑚3 6 = 1.38 × 101325 × 10 𝑚𝑜𝑙 2 𝑚𝑜𝑙 2

𝑃𝑎.𝑐𝑚3 𝑚𝑜𝑙2

𝑏̅ = 0.0376

𝑑𝑚3 𝑐𝑚3 = 37.6 𝑚𝑜𝑙 𝑚𝑜𝑙

𝑏̅ = 37.6

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 1: 𝑃1 = 1𝑥105 𝑃𝑎 𝑉1 = 3𝑐𝑚3 𝑉𝑚3 − (8258,57

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (4,66095𝑥105 ) (17,525172𝑥10 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3

𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 1651714𝑉𝑚 + 466095 𝑉𝑚0 = 𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = 24775,72 𝑃 𝑚𝑜𝑙

𝐹(24775,72 ) 𝑐𝑚3 = 24775,72 − ′ = 24763,563 𝐹 (24775,72 ) 𝑚𝑜𝑙

𝑉𝑚2 = 24763,563 −

𝐹( 24763,563) 𝑐𝑚3 = 24757,513 𝐹 ′ (24763,563 ) 𝑚𝑜𝑙

𝑉𝑚3 = 24757,513 −

𝐹(24757,513 ) 𝑐𝑚3 = 24754.499 𝐹 ′ (24757,513) 𝑚𝑜𝑙

𝑉𝑚4 = 24754.499 −

𝐹( 24754.499) 𝑐𝑚3 = 24752.995 𝐹 ′ (24754.499 ) 𝑚𝑜𝑙

𝑉𝑚5 = 24752.995 −

𝐹( 24752.995) 𝑐𝑚3 = 24752.995 𝐹 ′ (24752.995) 𝑚𝑜𝑙

𝑉𝑚 = 24752.995

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,00012𝑚𝑜𝑙 𝑉1 = 𝑉𝑚 𝑥𝑛 = 24752.995

𝑐𝑚3 𝑥 0,00012𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉1 = 2.9703594𝑐𝑚3 𝐾1 = 𝑃1 𝑥𝑉1 = 1𝑥105 𝑃𝑎 𝑥 2.9703594𝑐𝑚3 𝐾1 = 297035.94𝑃𝑎. 𝑐𝑚3

 Para el punto 2: 𝑃2 = 1.5𝑥105 𝑃𝑎 𝑉2 = 2.5𝑐𝑚3 𝑉𝑚3 − (27721.94

𝑐𝑚3 2 𝑐𝑚9 ) 𝑉𝑚 + (9.32190𝑥105 𝑉𝑚 ) − (35.050344𝑥106 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 3

𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 55443.88𝑉𝑚 + 932190 𝑉𝑚0 = 𝑉𝑚1 = 16407.76 −

𝑅𝑇 𝑐𝑚3 = 16407.76 𝑃 𝑚𝑜𝑙

𝐹(16407.76 ) 𝑐𝑚3 = 16394.201 𝐹 ′ (16407.76 ) 𝑚𝑜𝑙

𝑉𝑚2 = 16394.201 −

𝐹( 16394.201) 𝑐𝑚3 = 16378.768 𝐹 ′ (16394.201 ) 𝑚𝑜𝑙

𝑉𝑚3 = 16378.768 − 𝑉𝑚4

𝐹(16378.768 ) 𝑐𝑚3 = 16374.53 𝐹 ′ (16378.768) 𝑚𝑜𝑙

𝐹( 16374.53) 𝑐𝑚3 = 16374.53 − ′ = 16373.287 𝐹 (16374.53 ) 𝑚𝑜𝑙

𝑉𝑚5 = 16373.287 −

𝐹( 16373.287) 𝑐𝑚3 = 16371.133 𝐹 ′ (16373.287) 𝑚𝑜𝑙

𝑐𝑚3 𝑉𝑚 = 16371.133 𝑚𝑜𝑙 𝑛 = 0,00015𝑚𝑜𝑙 𝑉2 = 𝑉𝑚 𝑥𝑛 = 16371.133

𝑐𝑚3 𝑥0,00015𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉2 = 2.456𝑐𝑚3 𝐾2 = 𝑃2 𝑥𝑉2 = 1.5𝑥105 𝑃𝑎 𝑥 2.456𝑐𝑚3 𝐾2 = 322800𝑃𝑎. 𝑐𝑚3

 Para el punto 3: 𝑃3 = 2.1𝑥105 𝑃𝑎 𝑉3 = 2.1𝑐𝑚3 𝑉𝑚3 − (11835.56

𝑐𝑚3 2 𝑐𝑚9 ) 𝑉𝑚 + (6.65850𝑥105 𝑉𝑚 ) − (25.03596𝑥106 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 23671.12𝑉𝑚 + 665850 𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 11797.96 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 11797.96 −

𝐹(11797.96 ) 𝑐𝑚3 = 11779.11 𝐹 ′ (11797.96 ) 𝑚𝑜𝑙

𝑉𝑚2 = 11779.11 −

𝐹( 11779.11) 𝑐𝑚3 = 11779.04 𝐹 ′ (11779.11) 𝑚𝑜𝑙

𝑉𝑚3 = 11779.04 −

𝐹(11779.04) 𝑐𝑚3 = 11779.04 𝐹 ′ (11779.04) 𝑚𝑜𝑙 𝑉𝑚 = 11779.04

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,00017𝑚𝑜𝑙

𝑉3 = 𝑉𝑚 𝑥𝑛 = 11779.04

𝑐𝑚3 𝑥0,00017𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉3 = 2.002𝑐𝑚3 𝐾3 = 𝑃3 𝑥𝑉3 = 2.1𝑥105 𝑃𝑎 𝑥 2.002𝑐𝑚3 𝐾3 = 420420𝑃𝑎. 𝑐𝑚3

 Para el punto 4: 𝑃4 = 2.6𝑥105 𝑃𝑎 𝑉4 = 1.8𝑐𝑚3 𝑉𝑚3 − (9566.72

𝑐𝑚3 2 𝑐𝑚9 ) 𝑉𝑚 + (5.378019𝑥105 𝑉𝑚 ) − (20.221352𝑥106 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 19133.44𝑉𝑚 + 537801.9 𝑉𝑚0

𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = = 9529.1231 𝑃 𝑚𝑜𝑙

𝐹(9529.1231 ) 𝑐𝑚3 = 9529.1231 − ′ = 9510.4682 𝐹 (9529.1231 ) 𝑚𝑜𝑙

𝑉𝑚2 = 9510.4682 −

𝐹(9510.4682) 𝑐𝑚3 = 9510.3947 𝐹 ′ (9510.4682 ) 𝑚𝑜𝑙

𝑉𝑚3 = 9510.3947 −

𝐹(9510.3947 ) 𝑐𝑚3 = 9510.3947 𝐹 ′ (9510.3947) 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 9510.3947 𝑚𝑜𝑙 𝑛 = 0,00018𝑚𝑜𝑙 𝑉4 = 𝑉𝑚 𝑥𝑛 = 9510.3947

𝑐𝑚3 𝑥0,00018𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉2 = 1.7119𝑐𝑚3 𝐾4 = 𝑃4 𝑥𝑉4 = 2.6𝑥105 𝑃𝑎 𝑥 1.7119𝑐𝑚3 𝐾4 = 445094𝑃𝑎. 𝑐𝑚3

 Para el punto 5: 𝑃5 = 3.4𝑥105 𝑃𝑎 𝑉5 = 1.5𝑐𝑚3 𝑉𝑚3 − (7286.97

𝑐𝑚3 2 𝑐𝑚9 ) 𝑉𝑚 + (4.11260𝑥105 𝑉𝑚 ) − (15.463387𝑥106 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 3

𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 14573.95𝑉𝑚 + 411260 𝑉𝑚0 = 𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = 7286.98 𝑃 𝑚𝑜𝑙

𝐹(7286.98 ) 𝑐𝑚3 = 7286.98 − ′ = 7231.255 𝐹 (7286.98 ) 𝑚𝑜𝑙

𝑉𝑚2 = 7231.255 −

𝐹( 7231.255) 𝑐𝑚3 = 7230.386 𝐹 ′ (7231.255 ) 𝑚𝑜𝑙

𝑉𝑚3 = 7230.386 −

𝐹(7230.386 ) 𝑐𝑚3 = 7230.386 𝐹 ′ (7230.386) 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 7230.386 𝑚𝑜𝑙 𝑛 = 0,00021𝑚𝑜𝑙 𝑉5 = 𝑉𝑚 𝑥𝑛 = 7230.386

𝑐𝑚3 𝑥0,00021𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉5 = 1.5185𝑐𝑚3 𝐾5 = 𝑃5 𝑥𝑉5 = 3.4𝑥105 𝑃𝑎 𝑥1.5185𝑐𝑚3 𝐾5 = 516290𝑃𝑎. 𝑐𝑚3

 Para el punto 6: 𝑃6 = 5,3𝑥105 𝑃𝑎 𝑉6 = 1,1𝑐𝑚3 𝑉𝑚3 − (4312,264

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (2,638273𝑥105 ) (9,9919𝑥10 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 8624,528𝑉𝑚 + 263827,3 𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 4674,6641 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 4674,6641 −

𝐹(4674,6641 ) 9142654657 𝑐𝑚3 = 4674,6641 − = 4316,1920 𝐹 ′ (4674,6641 ) 25504509,22 𝑚𝑜𝑙

𝑉𝑚2 = 4316,1920 −

𝐹( 4316,1920) 1201914356 𝑐𝑚3 = 4316,1920 − = 4252,6902 𝐹 ′ (4316,1920 ) 18927248,91 𝑚𝑜𝑙

𝑉𝑚3 = 4252,6902 −

𝐹(4252,6902 ) 34569423,34 𝑐𝑚3 = 4252,6902 − = 4250,7527 𝐹 ′ (4252,6902) 17842503,41 𝑚𝑜𝑙

𝑉𝑚4 = 4250,7527 −

𝐹( 4250,7527) 31270.48721 𝑐𝑚3 = 4250,7527 − = 4250,7527 𝐹 ′ (4250,7527 ) 17809787.17 𝑚𝑜𝑙

𝑉𝑚 = 4250,7527

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,000235𝑚𝑜𝑙 𝑐𝑚3 𝑉6 = 𝑉𝑚 𝑥𝑛 = 4250,7527 𝑥 0,000235𝑚𝑜𝑙 𝑚𝑜𝑙 𝑉6 = 0,99893𝑐𝑚3 𝐾6 = 𝑃6 𝑥𝑉6 = 5,3𝑥105 𝑃𝑎 𝑥 0,99893𝑐𝑚3 𝐾6 = 529432,9𝑃𝑎. 𝑐𝑚3

 Para el punto 7: 𝑃7 = 8𝑥105 𝑃𝑎 𝑉7 = 0,8𝑐𝑚3 𝑉𝑚3 − (3134,565

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (1,747856𝑥105 ) (6,5719𝑥10 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 6269,13𝑉𝑚 + 174785,6 𝑉𝑚0 =

𝑉𝑚1 = 3096,965 −

𝐹( 3096,965) 𝐹 ′ (3096,965 )

𝑉𝑚2 = 3078,7019 − 𝑉𝑚3

𝑅𝑇 𝑐𝑚3 = 3096,965 𝑃 𝑚𝑜𝑙

= 3096,965 −

174104158,6 𝑐𝑚3 = 3078,7019 9533086,043 𝑚𝑜𝑙

𝐹(3078,7019 ) 2047750,723 𝑐𝑚3 = 3078,7019 − = 3078,4819 𝐹 ′ (3078,7019 ) 9309219,325 𝑚𝑜𝑙

𝐹(3078,4819 ) 17,77524 𝑐𝑚3 = 3078,4819 − ′ = 3078,4819 − = 3078,4819 𝐹 (3078,4819 ) 9306534,792 𝑚𝑜𝑙

𝑉𝑚4 = 3078,4818 −

𝐹(3078,4818 ) −912,87812 𝑐𝑚3 = 3078,4818 − = 3078,4819 𝐹 ′ ( 3078,4818) 9306533,572 𝑚𝑜𝑙 𝑉𝑚 = 3078,4819

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,000258𝑚𝑜𝑙 𝑐𝑚3 𝑉7 = 𝑉𝑚 𝑥𝑛 = 3078,4819 𝑥 0,000258𝑚𝑜𝑙 𝑚𝑜𝑙 𝑉7 = 0,79425𝑐𝑚3 𝐾7 = 𝑃7 𝑥𝑉7 = 8𝑥105 𝑃𝑎 𝑥 0,79425𝑐𝑚3 𝐾7 = 635400𝑃𝑎. 𝑐𝑚3

 Para el punto 8: 𝑃8 = 11,8𝑥105 𝑃𝑎 𝑉8 = 0,6𝑐𝑚3 𝑉𝑚3 − (2137,237

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (1,184987𝑥105 ) (4,4555𝑥10 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 4274,474𝑉𝑚 + 118498,7 𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 2099,6372 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 2099,6372 −

𝐹(2099,6372 ) 78590948,79 𝑐𝑚3 = 2099,6372 − = 2081,6492 𝐹 ′ ( 2099,6372) 4369083,194 𝑚𝑜𝑙

𝑉𝑚2 = 2081,6492 −

𝐹(2081,6492 ) 1340645,282 𝑐𝑚3 = 2081,6492 − = 2081,3315 𝐹 ′ (2081,6492) 4220333,493 𝑚𝑜𝑙

𝑉𝑚3 = 2081,3315 −

𝐹(2081,3315 ) 259,9039 𝑐𝑚3 = 2081,3315 − = 2081,3314 𝐹 ′ (2081,3315 ) 4217723.757 𝑚𝑜𝑙

𝑉𝑚4 = 2081,3314 −

𝐹( 2081,3314) −161.86843 𝑐𝑚3 = 2081,3314 − = 2081,3314 𝐹 ′ ( 2081,3314) 4217722.935 𝑚𝑜𝑙 𝑉𝑚 = 2081,3314

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,000285𝑚𝑜𝑙 𝑉8 = 𝑉𝑚 𝑥𝑛 = 2081,3314

𝑐𝑚3 𝑥 0,000285𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉8 = 0,59318𝑐𝑚3 𝐾8 = 𝑃8 𝑥𝑉8 = 11,8𝑥105 𝑃𝑎 𝑥 0,59318𝑐𝑚3 𝐾8 = 699952,4𝑃𝑎. 𝑐𝑚3

 Para el punto 9: 𝑃9 = 19,8𝑥105 𝑃𝑎 𝑉9 = 0,4𝑐𝑚3 𝑉𝑚3 − (1288,898

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (0,706204𝑥105 ) (2,6553𝑥10 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 2577,796𝑉𝑚 + 70620,4

𝑅𝑇 𝑐𝑚3 = 1251,2989 𝑃 𝑚𝑜𝑙

𝑉𝑚0 = 𝑉𝑚1 = 1251,2989 −

𝐹(1251,2989) 26841177,98 𝑐𝑚3 = 1251,2989 − = 1233,8952 𝐹 ′ (1251,2989) 1542273,912 𝑚𝑜𝑙

𝑉𝑚2 = 1233,8952 −

𝐹(1233,8952 ) 741890,6834 𝑐𝑚3 = 1233,8952 − = 1233,3861 𝐹 ′ (1233,8952 ) 1457382,383 𝑚𝑜𝑙

𝑉𝑚3 = 1233,3861 −

𝐹(1233,3861 ) −73,61255 𝑐𝑚3 = 1233,3861 − = 1233,3862 𝐹 ′ (1233,3861 ) 1454926,46 𝑚𝑜𝑙

𝑉𝑚4 = 1233,3862 −

𝐹(1233,3862) 71,88011 𝑐𝑚3 = 1233,3862 − = 1233,3862 𝐹 ′ (1233,3862) 1454926,942 𝑚𝑜𝑙 𝑉𝑚 = 1233,3862

𝑐𝑚3 𝑚𝑜𝑙

𝑛 = 0,000319𝑚𝑜𝑙 𝑉9 = 𝑉𝑚 𝑥𝑛 = 1233,3862

𝑐𝑚3 𝑥 0,000319𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉9 = 0,39345𝑐𝑚3 𝐾9 = 𝑃9 𝑥𝑉9 = 19,8𝑥105 𝑃𝑎 𝑥 0,39345𝑐𝑚3 𝐾9 = 779031𝑃𝑎. 𝑐𝑚3

 Para el punto 10: 𝑃10 = 37,1𝑥105 𝑃𝑎 𝑉10 = 0,25𝑐𝑚3 𝑉𝑚3 − (705,409

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 6 𝑉 − ) 𝑉𝑚 + (0,376896𝑥105 ) (1,4171𝑥10 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 1410,818𝑉𝑚 + 37689,6 𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 667,8091 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 667,8091 −

𝐹( 667,8091) 6983968,276 𝑐𝑚3 = 667,8091 − = 651,6962 𝐹 ′ ( 667,8091) 433439,4833 𝑚𝑜𝑙

𝑉𝑚2 = 651,6962 −

𝐹(651,6962 ) 332816,616 𝑐𝑚3 = 651,6962 − = 650,8480 𝐹 ′ (651,6962 ) 392388,6818 𝑚𝑜𝑙

𝑉𝑚3 = 650,8480 −

𝐹(650,8480) 890,999367 𝑐𝑚3 = 650,8480 − = 650,8457 𝐹 ′ (650,8480) 390270,8836 𝑚𝑜𝑙

𝑉𝑚4 = 650,8457 −

𝐹(650,8457) 6,617068 𝑐𝑚3 = 650,8457 − = 650,8457 𝐹 ′ (650,8457) 390265,1468 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 650,8457 𝑚𝑜𝑙 𝑛 = 0,000374𝑚𝑜𝑙 𝑉10 = 𝑉𝑚 𝑥𝑛 = 650,8457

𝑐𝑚3 𝑥 0,000374𝑚𝑜𝑙 𝑚𝑜𝑙

𝑉10 = 0,24341𝑐𝑚3 𝐾10 = 𝑃10 𝑥𝑉10 = 37,1𝑥105 𝑃𝑎 𝑥 0,24341 𝑐𝑚3 𝐾10 = 903051,1𝑃𝑎. 𝑐𝑚3

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 297035.94 + 322800 + 420420 + 445094 + 516290 + 529432,9 + 635400 + 699952,4 + 779031 + 903051,1 = 10

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 = 554850.694𝑃𝑎. 𝑐𝑚3

a) Gráfica P vs V

Gráfica P vs V 40 35

PRESIÓN

30 25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

VOLUMEN

b) Grafica P vs 1/V

Gráfica P vs 1/V 40

Presión

E C U A C I Ó N

35 30 25 20 15 10

5 0

D E

0

0.5

1

1.5

2

2.5

1/Volumen

REDLICH-KWONG

3

3.5

4

4.5

Vm3 

RT 2 bRT a ab Vm  (b2   )Vm  0 P P T .P T .P 𝑎𝑎𝑖𝑟𝑒 = 16,07𝑥106 𝑥 101325 𝑏𝑎𝑖𝑟𝑒 = 0,026𝑥103

𝑐𝑚6 𝑚𝑜𝑙 2 𝑘 −0,5

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 𝑅 = 83,14𝑥10 𝑚𝑜𝑙. 𝑘 𝑇 = 298𝑘 5

 Para el punto 1: 𝑃1 = 3𝑥105 𝑃𝑎 𝑉1 = 1𝑐𝑚3 𝑉𝑚3 − (24775,72

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 ) 𝑉𝑚 + (15339.01147 ) 𝑉 − (314414.881 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 49551.44𝑉𝑚 + 15339.01147

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0

𝑉𝑚1 = 24775,72 −

𝑅𝑇 𝑐𝑚3 = = 24775,72 𝑃 𝑚𝑜𝑙

𝐹(24775,72 ) 379720638,4 𝑐𝑚3 = 24775,72 − = 24775.1014 𝐹 ′ (24775,72 ) 613851640.5 𝑚𝑜𝑙

𝑉𝑚2 = 24775.1014 −

𝐹(24775.1014) 18960,56 𝑐𝑚3 = 24775.1014 − = 24775.1013 𝐹 ′ (24775.1014) 613790337,9 𝑚𝑜𝑙

𝑉𝑚3 = 24775.1013 −

𝐹(24775.1013) 0.089744 𝑐𝑚3 = 24775.1013 − = 24775.1013 𝐹 ′ (24775.1013) 613790334 𝑚𝑜𝑙

𝑉𝑚 = 24775.1013

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 2: 𝑃2 = 1.5𝑥105 𝑃𝑎 𝑉2 = 2.5𝑐𝑚3 𝑉𝑚3

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 − (16407,76 ) 𝑉 + (1054755.536 ) 𝑉 − (16349574.18 )=0 𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 32815.52𝑉𝑚 + 1054755.536

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0 =

𝑉𝑚1 = 16407,76 −

𝑅𝑇 𝑐𝑚3 = 16407,76 𝑃 𝑚𝑜𝑙

𝐹(16407,76 ) 17289826120 𝑐𝑚3 = 16407,76 − = 16343.7874 𝐹 ′ (16407,76 ) 270269343.8 𝑚𝑜𝑙

𝐹(16343.7874) 134035395.8 𝑐𝑚3 = 16343.7874 − ′ = 16343.7874 − = 16343.2836 𝐹 (16343.7874) 266083034.5 𝑚𝑜𝑙

𝑉𝑚2

𝑉𝑚3 = 16343.2836 −

𝐹(16343.2836) 8278.1066 𝑐𝑚3 = 16343.2836 − = 16343.2836 𝐹 ′ (16343.2836) 266050168 𝑚𝑜𝑙

𝑉𝑚4 = 16343.2836 −

𝐹(16343.2836) −0.012 𝑐𝑚3 = 16343.2836 − = 16343.2836 𝐹 ′ (16343.2836) 266050168 𝑚𝑜𝑙

𝑉𝑚 = 16343.2836

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 3: 𝑃3 = 2.1𝑥105 𝑃𝑎 𝑉3 = 2.1𝑐𝑚3 𝑉𝑚3

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 − (11797,96 ) 𝑉 + (755235.0858 ) 𝑉 − (11678267.27 )=0 𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 23595.92𝑉𝑚 + 755235.0858

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 11797,96 𝑃 𝑚𝑜𝑙

𝑉𝑚1

𝐹(11797,96 ) 8898555066 𝑐𝑚3 = 11797,96 − ′ = 11797,96 − = 11734.37 𝐹 (11797,96 ) 139947095.2 𝑚𝑜𝑙

𝑉𝑚2

𝐹(11734.37) 95142865.41 𝑐𝑚3 = 11734.37 − ′ = 11734.37 − = 11733.68 𝐹 (11734.37) 136958524.9 𝑚𝑜𝑙

𝑉𝑚3

𝐹(11733.68) 11294.71039 𝑐𝑚3 = 11733.68 − ′ = 11733.68 − = 11733.68 𝐹 (11733.68) 136926008 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 11733.68 𝑚𝑜𝑙

 Para el punto 4: 𝑃4 = 2.6𝑥105 𝑃𝑎 𝑉4 = 1.8𝑐𝑚3 𝑉𝑚3 − (9529,1231

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 ) 𝑉𝑚 + (609867.6099 ) 𝑉 − (9432446.642 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 19058.2462𝑉𝑚 + 609867.6099

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0

𝑉𝑚1 = 9529,1231 −

𝑅𝑇 𝑐𝑚3 = = 9529,1231 𝑃 𝑚𝑜𝑙

𝐹(9529,1231 ) 5802071083 𝑐𝑚3 = 9529,1231 − = 9465.65287 𝐹 ′ (9529,1231 ) 91414054.66 𝑚𝑜𝑙

𝑉𝑚2

𝐹(9465.65287) 76519887.2 𝑐𝑚3 = 9465.65287 − ′ = 9465.65287 − = 9464.7931 𝐹 (9465.65287) 89006877.53 𝑚𝑜𝑙

𝑉𝑚3

𝐹(9464.7931) 13944.5223 𝑐𝑚3 = 9464.7931 − ′ = 9464.7931 − = 9464.7930 𝐹 (9464.7931) 88974438.11 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 9464.7930 𝑚𝑜𝑙

 Para el punto 5: 𝑃5 = 3.4𝑥105 𝑃𝑎 𝑉5 = 1.5𝑐𝑚3

𝑉𝑚3

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 − (7286,9764 ) 𝑉 + (466210.2876 ) 𝑉 − (7213047.432 )=0 𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 14573.9528𝑉𝑚 + 466210.2876

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0

𝑅𝑇 𝑐𝑚3 = = 7286,9764 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 7286,9764 −

𝐹(7286,9764 ) 3390050316 𝑐𝑚3 = 7286,9764 − = 7223.6893 𝐹 ′ (7286,9764 ) 53566235.34 𝑚𝑜𝑙

𝑉𝑚2 = 7223.6893 −

𝐹(7223.6893) 58118898 𝑐𝑚3 = 7223.6893 − = 7222.5658 𝐹 ′ (7223.6893) 51733565.44 𝑚𝑜𝑙

𝑉𝑚3 = 7222.5658 −

𝐹(7222.5658) 18152.58547 𝑐𝑚3 = 7222.5658 − = 7222.5655 𝐹 ′ (7222.5658) 51701250.26 𝑚𝑜𝑙 𝑉𝑚 = 7222.5655

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 6: 𝑃6 = 5,3𝑥105 𝑃𝑎 𝑉6 = 1,1𝑐𝑚3 𝑉𝑚3 − (4674,6642

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 𝑉 − ) 𝑉𝑚 + (55753,4 ) (4627237,98 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3

𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 9349,3284𝑉𝑚 + 55753,4

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0 =

𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = 4674,6642 𝑃 𝑚𝑜𝑙

𝐹(4674,6642 ) 256001185 𝑐𝑚3 = 4674,6642 − ′ = 4674,6642 − = 4662,9790 𝐹 (4674,6642 ) 21908238,78 𝑚𝑜𝑙

𝑉𝑚2 = 4662,9790 −

𝐹(4662,9790) 1274031,415 𝑐𝑚3 = 4662,9790 − = 4662,9203 𝐹 ′ (4662,9790) 21690150,87 𝑚𝑜𝑙

𝑉𝑚3 = 4662,9203 −

𝐹(4662,9203) 851,65202 𝑐𝑚3 = 4662,9203 − = 4662,9203 𝐹 ′ (4662,9203) 21689057,38 𝑚𝑜𝑙

𝑉𝑚 = 4662,9203

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 7: 𝑃7 = 8𝑥105 𝑃𝑎 𝑉7 = 0,8𝑐𝑚3 𝑉𝑚3 − (3096,965

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 𝑉 − ) 𝑉𝑚 + (36708,493 ) (3065545,16 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 6193,93𝑉𝑚 + 36708,493

-

Hallamos el Vm según Redlich-Kwong:

𝑉𝑚0 =

𝑅𝑇 𝑐𝑚3 = 3096,965 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 3096,965 −

𝐹(3096,965) 110619372,9 𝑐𝑚3 = 3096,965 − = 3085,476 𝐹 ′ (3096,965) 9627900,704 𝑚𝑜𝑙

𝑉𝑚2 = 3085,476 −

𝐹(3085,476) 820486,0857 𝑐𝑚3 = 3085,476 − = 3085,389 𝐹 ′ (3085,476) 9485972,572 𝑚𝑜𝑙

𝑉𝑚3 = 3085,389 −

𝐹(3085,389) 3097506,745 𝑐𝑚3 = 3085,389 − = 3085,062 𝐹 ′ (3085,389) 9484900.848 𝑚𝑜𝑙

𝑉𝑚 = 3085,062

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 8: 𝑃8 = 11,8𝑥105 𝑃𝑎 𝑉8 = 0,6𝑐𝑚3 𝑉𝑚3

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 − (2099,6373 ) 𝑉 + (24669,419 ) 𝑉 − (2078335,701 )=0 𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 4199,2746𝑉𝑚 + 24669,419

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0

𝑅𝑇 𝑐𝑚3 = = 2099,6373 𝑃 𝑚𝑜𝑙

𝑉𝑚1 = 2099,637 −

𝐹(2099,637) 49718496,6 𝑐𝑚3 = 2099,637 − = 2088,422 𝐹 ′ (2099,637) 4433146,211 𝑚𝑜𝑙

𝐹(2088,422) 526218,3762 𝑐𝑚3 = 2088,422 − ′ = 2088,422 − = 2088,301 𝐹 (2088,422) 4339331,311 𝑚𝑜𝑙

𝑉𝑚2

𝑉𝑚3 = 2088,301 −

𝐹(2088,301) 1220,274809 𝑐𝑚3 = 2088,301 − = 2088,301 𝐹 ′ (2088,301) 4338323,272 𝑚𝑜𝑙 𝑉𝑚 = 2088,301

𝑐𝑚3 𝑚𝑜𝑙

 Para el punto 9: 𝑃9 = 19,8𝑥105 𝑃𝑎 𝑉9 = 0,4𝑐𝑚3 𝑉𝑚3 − (1251,2989

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 𝑉 − ) 𝑉𝑚 + (14428,8457 ) (1238604,105 )=0 𝑚 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚𝑜𝑙 3 𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 2502,5978𝑉𝑚 + 14428,8457

-

Hallamos el Vm según Redlich-Kwong: 𝑉𝑚0 =

𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = 1251,298 𝑃 𝑚𝑜𝑙

𝐹(1251,298) 16816194,65 𝑐𝑚3 = 1251,298 − ′ = 1251,298 − = 1240,657 𝐹 (1251,298) 1580177,783 𝑚𝑜𝑙

𝑉𝑚2 = 1240,657 −

𝐹(1240,657) 282314,7949 𝑐𝑚3 = 1240,657 − = 1240,472 𝐹 ′ (1240,657) 1527252,742 𝑚𝑜𝑙

𝑉𝑚3 = 1240,472 −

𝐹(1240,472) −142,4099496 𝑐𝑚3 = 1240,472 − = 1240,472 𝐹 ′ (1240,472) 1526338,696 𝑚𝑜𝑙 𝑐𝑚3 𝑉𝑚 = 1240,472 𝑚𝑜𝑙

 Para el punto 10: 𝑃10 = 37,1𝑥105 𝑃𝑎 𝑉10 = 0,25𝑐𝑚3 𝑉𝑚3 − (667,8092

𝑐𝑚3 2 𝑐𝑚6 𝑐𝑚9 𝑉 − ) 𝑉𝑚 + (7454,0608 ) (661033,9965 )=0 𝑚𝑜𝑙 𝑚𝑜𝑙 2 𝑚 𝑚𝑜𝑙 3

𝐹 ′ (𝑉𝑚 ) = 3𝑉𝑚2 − 1335,6184𝑉𝑚 + 7454,0608

-

Hallamos el Vm según Redlich-Kwong:

𝑉𝑚0 = 𝑉𝑚1

𝑅𝑇 𝑐𝑚3 = 667,8092 𝑃 𝑚𝑜𝑙

𝐹(667,809) 4316856,383 𝑐𝑚3 = 667,809 − ′ = 667,809 − = 658,289 𝐹 (667,809) 453423,1884 𝑚𝑜𝑙

𝑉𝑚2 = 658,289 −

𝐹(658,289) 120366,805 𝑐𝑚3 = 658,289 − = 658,001 𝐹 ′ (658,289) 428264,3824 𝑚𝑜𝑙

𝑉𝑚3 = 658,001 −

𝐹(658,001) −2864,94844 𝑐𝑚3 = 658,001 − = 658,001 𝐹 ′ (658,001) 427511,766 𝑚𝑜𝑙

𝑉𝑚 = 658,001

𝑐𝑚3 𝑚𝑜𝑙

El volumen molar fue hallado por el método de Newton- Raphson: 𝑉𝑛 = 𝑉𝑛+1 −

𝐹(𝑉𝑛+1 ) 𝐹′(𝑉𝑛+1 )

 Hallando el k para Redlick- Kwong −𝑉1 = 𝑉𝑚1 𝑛 = 24775,1013

𝑐𝑚3 𝑥 0,00012𝑚𝑜𝑙 = 2,97305𝑐𝑚3 𝑚𝑜𝑙

𝑘1 = 𝑃1 𝑉1 = 1𝑥105 𝑃𝑎. 2.9730𝑐𝑚3 = 297305𝑃𝑎. 𝑐𝑚3 𝑐𝑚3 −𝑉2 = 𝑉𝑚2 𝑛 = 16432.836 𝑥0.00015 𝑚𝑜𝑙 = 2,4649𝑐𝑚3 𝑚𝑜𝑙 𝑘2 = 𝑃2 𝑉2 = 1.5𝑥105 𝑃𝑎. 2,4649𝑐𝑚3 = 369746𝑃𝑎. 𝑐𝑚3 𝑐𝑚3 −𝑉3 = 𝑉𝑚3 𝑛 = 11733,68 𝑥 0.00017𝑚𝑜𝑙 = 1,9947𝑐𝑚3 𝑚𝑜𝑙 𝑘3 = 𝑃3 𝑉3 = 2.1𝑥105 𝑃𝑎. 1,9947𝑐𝑚3 = 418894𝑃𝑎. 𝑐𝑚3 −𝑉4 = 𝑉𝑚4 𝑛 = 9464,7930

𝑐𝑚3 𝑥 0.00018𝑚𝑜𝑙 = 1,7037𝑐𝑚3 𝑚𝑜𝑙

𝑘4 = 𝑃4 𝑉4 = 2.6𝑥105 𝑃𝑎. 1,7037𝑐𝑚3 = 442953𝑃𝑎. 𝑐𝑚3

𝑐𝑚3 −𝑉5 = 𝑉𝑚5 𝑛 = 7222,5655 𝑥 0,00021𝑚𝑜𝑙 = 1,5167𝑐𝑚3 𝑚𝑜𝑙 𝑘5 = 𝑃5 𝑉5 = 3,4𝑥105 𝑃𝑎. 1,5167𝑐𝑚3 = 515691𝑃𝑎. 𝑐𝑚3 −𝑉6 = 𝑉𝑚6 𝑛 = 4662,9203

𝑐𝑚3 𝑥 0,000235𝑚𝑜𝑙 = 1,09579𝑐𝑚3 𝑚𝑜𝑙

𝑘6 = 𝑃6 𝑉6 = 5,3𝑥105 𝑃𝑎. 1,09579𝑐𝑚3 = 580768,7 𝑃𝑎. 𝑐𝑚3 −𝑉7 = 𝑉𝑚7 𝑛 = 3085,062

𝑐𝑚3 𝑥 0,000258𝑚𝑜𝑙 = 0,79595𝑐𝑚3 𝑚𝑜𝑙

𝑘7 = 𝑃7 𝑉7 = 8𝑥105 𝑃𝑎. 0,79595𝑐𝑚3 = 636760 𝑃𝑎. 𝑐𝑚3 𝑐𝑚3 −𝑉8 = 𝑉𝑚8 𝑛 = 2088,301 𝑥 0,000285𝑚𝑜𝑙 = 0,59517𝑐𝑚3 𝑚𝑜𝑙 𝑘8 = 𝑃8 𝑉8 = 11,8𝑥105 𝑃𝑎. 0,59517𝑐𝑚3 = 702300,6 𝑃𝑎. 𝑐𝑚3 −𝑉9 = 𝑉𝑚9 𝑛 = 1240,472

𝑐𝑚3 𝑥 0,000319𝑚𝑜𝑙 = 0,39571𝑐𝑚3 𝑚𝑜𝑙

𝑘9 = 𝑃9 𝑉9 = 19,8𝑥105 𝑃𝑎. 0,39571𝑐𝑚3 = 783505,8 𝑃𝑎. 𝑐𝑚3 −𝑉10 = 𝑉𝑚10 𝑛 = 658,001

𝑐𝑚3 𝑥 0,000374𝑚𝑜𝑙 = 0,24609𝑐𝑚3 𝑚𝑜𝑙

𝑘10 = 𝑃10 𝑉10 = 37,1𝑥105 𝑃𝑎. 0,24609𝑐𝑚3 = 912993,9 𝑃𝑎. 𝑐𝑚3

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 = (297305 + 369746 + 418894 + 442953 + 515691 + 580768,7 + 636760 + 702300,6 + 783505,8 + 912993,9) 𝑃𝑎. 𝑐𝑚3 10

𝑘𝑝𝑟𝑜𝑚𝑒𝑑𝑖𝑜 = 566091.8𝑃𝑎. 𝑐𝑚3

a) Gráfico P vs V b) ráfica P vs 1/V

Gráfico P vs V 40

G

35

Presión

30

Gráfico P vs 1/V

25 20

40

15

35

10

30

0 0

PRESIÓN

5 0.5

25 20

1

1.5

2

2.5

3

3.5

Volumen

15 10 5 0 0

0.5

1

1.5

2

2.5

1/VOLUMEN

ECUACI ÓN DE VIRIAL (BEATTIE-BRIDGEMAN)

𝑅𝑇 𝐶𝑃 𝑉𝑚 = + 𝐵 + 𝑃 𝑅𝑇 𝑃𝑎. 𝑐𝑚3 𝑎̅ = 1.38 × 101325 × 10 𝑚𝑜𝑙 2 6

𝑏̅ = 37.6

𝑐𝑚3 𝑚𝑜𝑙

3

3.5

4

4.5

Reemplazando: 𝐵=𝑏−

𝑎 𝑐𝑚3 = 𝑏̅ = 37.6 𝑅𝑇 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 3 𝑚𝑜𝑙2 = −18.8377 𝑐𝑚 − 𝑚𝑜𝑙 𝑃𝑎. 𝑐𝑚3 83,14 × 105 × 298𝐾 𝑚𝑜𝑙. 𝐾 1.38 × 101325 × 106

2

𝑐𝑚3 𝑐𝑚6 𝐶 = 𝑏 = (37.6 ) = 1413.76 𝑚𝑜𝑙 𝑚𝑜𝑙 2 2

Trabajando con los puntos escogidos:

 Para el punto 1: 𝑃1 = 1x105 𝑃𝑎 𝑉1 = 3𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 1×105 𝑃𝑎

83,14×105 ×

𝑍= 

− 18.8377

𝑃𝑉𝑚 = 𝑅𝑇

𝑐𝑚3 𝑚𝑜𝑙

+

𝑐𝑚6 ×1×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

1𝑥105 𝑃𝑎 × 24756.939

83.14 × 105 ×

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × 298𝐾 𝑚𝑜𝑙. 𝐾

= 0.9992

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾1

𝐾1 =

𝑐𝑚3

=24756.939 𝑚𝑜𝑙

𝑃. 𝑉 105 𝑃𝑎(3𝑐𝑚3 ) = = 300240.19𝑃𝑎. 𝑐𝑚3 𝑍 0.9992

 Para el punto 2: 𝑃2 = 1.51𝑥105 𝑃𝑎 𝑉2 = 2.5𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 1.51×105 𝑃𝑎

83,14×105 ×

− 18.8377

𝑐𝑚3 𝑚𝑜𝑙

+

𝑐𝑚6 ×1.51×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

𝑐𝑚3

=16389.0100𝑚𝑜𝑙

𝑐𝑚3 1.51𝑥105 𝑃𝑎 × 16389.0100 𝑃𝑉𝑚 𝑚𝑜𝑙 = 0.9988 𝑍= = 3 𝑃𝑎. 𝑐𝑚 𝑅𝑇 83,14 × 105 × × 298𝐾 𝑚𝑜𝑙. 𝐾 

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾2

𝐾2 =

𝑃. 𝑉 1.51𝑥105 𝑃𝑎 × 2.5𝑐𝑚3 = = 377953.5442𝑃𝑎. 𝑐𝑚3 𝑍 0.9988

 Para el punto 3: 𝑃3 = 2.1𝑥105 𝑃𝑎 𝑉3 = 2.1𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 3.4×105 𝑃𝑎

83,14×105 ×

− 18.8377

𝑃𝑉𝑚 𝑍= = 𝑅𝑇 

𝑐𝑚3 𝑚𝑜𝑙

+

𝑐𝑚6 ×2.1×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

2.1𝑥105 𝑃𝑎 × 7268.33

83,14 × 105 ×

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × 298𝐾 𝑚𝑜𝑙. 𝐾

𝑐𝑚3

=7268.33𝑚𝑜𝑙

= 0.9974

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾5

𝑃. 𝑉 2.1𝑥105 𝑃𝑎 × 2.1𝑐𝑚3 𝐾3 = = = 441706.73𝑃𝑎. 𝑐𝑚3 𝑍 0.9974

 Para el punto 4: 𝑃4 = 2.6𝑥105 𝑃𝑎 𝑉4 = 1.8𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 2.6×105 𝑃𝑎

83,14×105 ×

− 18.8377

𝑃𝑉𝑚 𝑍= = 𝑅𝑇 

𝑐𝑚3 𝑚𝑜𝑙

+

𝑐𝑚6 ×2.6×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

2.6𝑥105 𝑃𝑎 × 9510.4337

83,14 × 105 ×

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × 298𝐾 𝑚𝑜𝑙. 𝐾

𝑐𝑚3

=9510.4337𝑚𝑜𝑙

= 0.9980

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾5

𝐾4 =

𝑃. 𝑉 2.6𝑥105 𝑃𝑎 × 1.8𝑐𝑚3 = = 511329.46𝑃𝑎. 𝑐𝑚3 𝑍 0.9980

 Para el punto 5: 𝑃5 = 3.4𝑥105 𝑃𝑎 𝑉5 = 1.5𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 3.4×105 𝑃𝑎

83,14×105 ×

𝑍= 

− 18.8377

𝑃𝑉𝑚 = 𝑅𝑇

𝑐𝑚3 𝑚𝑜𝑙

+

𝑐𝑚6 ×3.4×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

3.4𝑥105 𝑃𝑎 × 7268.33

83,14 ×

105

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × × 298𝐾 𝑚𝑜𝑙. 𝐾

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾5

𝑐𝑚3

=7268.33𝑚𝑜𝑙

= 0.9974

𝑃. 𝑉 3.4𝑥105 𝑃𝑎 × 1.5𝑐𝑚3 𝐾5 = = = 511329.46𝑃𝑎. 𝑐𝑚3 𝑍 0.9974  Para el punto 6: 𝑃6 = 5,3𝑥105 𝑃𝑎 𝑉6 = 1,1𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 5.3×105 𝑃𝑎

83,14×105 ×

𝑐𝑚3

− 18.8377 𝑚𝑜𝑙 +

𝑐𝑚6 ×5.3×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

𝑐𝑚3

=4656.1288 𝑚𝑜𝑙

𝑐𝑚3 𝑃𝑉𝑚 𝑚𝑜𝑙 = 0.9960 𝑍= = 3 𝑃𝑎. 𝑐𝑚 𝑅𝑇 83,14 × 105 × × 298𝐾 𝑚𝑜𝑙. 𝐾 5.3𝑥105 𝑃𝑎 × 4656.1288



Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾6

𝐾6 =

𝑃. 𝑉 5.3𝑥105 𝑃𝑎 × 1.1𝑐𝑚3 = = 585341.37𝑃𝑎. 𝑐𝑚3 𝑍 0.9960

 Para el punto 7: 𝑃7 = 8𝑥105 𝑃𝑎 𝑉7 = 0,8𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 5 8×10 𝑃𝑎

83,14×105 ×

𝑍=

𝑐𝑚3

− 18.8377 𝑚𝑜𝑙 +

𝑃𝑉𝑚 = 𝑅𝑇

𝑐𝑚6 ×8×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

8𝑥105 𝑃𝑎 × 3078.58 83,14 ×

105

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × × 298𝐾 𝑚𝑜𝑙. 𝐾

𝑐𝑚3

=3078.58 𝑚𝑜𝑙

= 0.9941



Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾7

𝐾7 =

𝑃. 𝑉 8𝑥105 𝑃𝑎 × 0.8 𝑐𝑚3 = = 643822.02𝑃𝑎. 𝑐𝑚3 𝑍 0.9941

 Para el punto 8: 𝑃8 = 11,8𝑥105 𝑃𝑎 𝑉8 = 0,6𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 5 11.8×10 𝑃𝑎

83,14×105 ×

𝑐𝑚3

− 18.8377 𝑚𝑜𝑙 +

𝑃𝑉𝑚 𝑍= = 𝑅𝑇 

𝑐𝑚6 ×11.8×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

11.8𝑥105 𝑃𝑎 × 2081.47 83,14 ×

105

𝑐𝑚3 𝑚𝑜𝑙

𝑃𝑎. 𝑐𝑚3 × × 298𝐾 𝑚𝑜𝑙. 𝐾

𝑐𝑚3

=2081.47 𝑚𝑜𝑙

= 0.9913

Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾8

𝐾8 =

𝑃. 𝑉 11.8𝑥105 𝑃𝑎 × 0.6 𝑐𝑚3 = = 714213.66𝑃𝑎. 𝑐𝑚3 𝑍 0.9913

 Para el punto 9: 𝑃8 = 19,8𝑥105 𝑃𝑎 𝑉8 = 0,4𝑐𝑚3

𝑉𝑚 =

𝑃𝑎.𝑐𝑚3 ×298𝐾 𝑚𝑜𝑙.𝐾 19.8×105 𝑃𝑎

83,14×105 ×

𝑐𝑚3

− 18.8377 𝑚𝑜𝑙 +

𝑐𝑚6 ×19.8×105 𝑃𝑎 𝑚𝑜𝑙2 𝑃𝑎.𝑐𝑚3 83,14×105 × ×298𝐾 𝑚𝑜𝑙.𝐾

1413.76

𝑐𝑚3

=1233.5911 𝑚𝑜𝑙

𝑐𝑚3 𝑃𝑉𝑚 𝑚𝑜𝑙 = 0.9858 𝑍= = 3 𝑃𝑎. 𝑐𝑚 𝑅𝑇 83,14 × 105 × × 298𝐾 𝑚𝑜𝑙. 𝐾 19.8𝑥105 𝑃𝑎 × 1233.5911



Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾8

𝐾8 =

𝑃. 𝑉 19.8𝑥105 𝑃𝑎 × 0.4 𝑐𝑚3 = = 803408.39𝑃𝑎. 𝑐𝑚3 𝑍 0.9858

 Para el punto 10: 𝑃10 = 37,1𝑥105 𝑃𝑎 𝑉10 = 0,25𝑐𝑚3

𝑉𝑚 =

𝑃𝑎. 𝑐𝑚3 × 298𝐾 𝑐𝑚3 𝑚𝑜𝑙. 𝐾 − 18.8377 37.1 × 105 𝑃𝑎 𝑚𝑜𝑙

83,14 × 105 ×

= 651.0884

𝑐𝑚3 𝑚𝑜𝑙

𝑐𝑚6 5 2 × 37.1 × 10 𝑃𝑎 𝑚𝑜𝑙 + 𝑃𝑎. 𝑐𝑚3 83,14 × 105 × × 298𝐾 𝑚𝑜𝑙. 𝐾 1413.76

𝑐𝑚3 𝑃𝑉𝑚 𝑚𝑜𝑙 = 0.9749 𝑍= = 3 𝑃𝑎. 𝑐𝑚 𝑅𝑇 83,14 × 105 × × 298𝐾 𝑚𝑜𝑙. 𝐾 37.1𝑥105 𝑃𝑎 × 651.0884



Dónde: 𝑃. 𝑉 = 𝑍. 𝑛. 𝑅. 𝑇 𝑛. 𝑅. 𝑇 = 𝐾8

𝐾10

𝑃. 𝑉 37.1𝑥105 𝑃𝑎 × 0.25 𝑐𝑚3 = = = 714213.66𝑃𝑎. 𝑐𝑚3 𝑍 0.9749

K prom (Virial) = 579823.5296

a) Gráfica P vs Vm

Gráfico P vs V 14 12

P 10 R E 8 S I 6 O 4 N 2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

Volumen molar

b) Gráfica P vs 1/Vm

Gráfica P vs 1/Vm 14 12 P 10 r e 8 s i 6 ó 4 n 2 0 0

10

20

30 1/Volumen molar

40

50

60

RESULTADOS: P

V

(𝟏𝟎𝟓 𝑷𝒂) (𝒄𝒎𝟑 )

𝑽𝒎 Gas Ideal 𝒄𝒎𝟑 ( ) 𝒎𝒐𝒍

𝑽𝒎 Van der Waals

𝑽𝒎 de Redlich-Kwon

𝒄𝒎𝟑 ( ) 𝒎𝒐𝒍

𝒄𝒎𝟑 ( ) 𝒎𝒐𝒍

𝑽𝒎 de la ecuación de Virial 𝒄𝒎𝟑 ( ) 𝒎𝒐𝒍

Factor de compres ibilidad

𝑷𝟏

1

3

24775,72

24752.995

24775.1013

24756.939

𝑷𝟐

1.51

2.5

16407,76

16371.133

16343.2836

16389.0100

0.9988

𝑷𝟑

2.1

2.1

11797,96

11779.04

11733.68

11779.2440

0.9984

𝑷𝟒

2.6

1.8

9529,1231

9510.3947

9464.7930

9510.4337

0.9980

𝑷𝟓

3.4

1.5

7286,9764

7230.386

7222.5655

7268.33

0.9974

𝑷𝟔

5.3

1.1

4674,664

4250,7527

4662,9203

4656.1288

0.9960

𝑷𝟕

8

0.8

3096,965

3078,4819

3085,062

3078.58

0.9941

𝑷𝟖

11.8

0.6

2099,637

2081,3314

2088,301

2081.47

0.9913

𝑷𝟗

19,8𝑥105

0.4

1251,298

1233,3862

1240,472

1233.5911

0.9858

𝑷𝟏𝟎 37,1𝑥105

0.25

667,809

650,8457

658,001

651.0884

0.9750

0.9992

P

V

(𝟏𝟎𝟓 𝑷𝒂) (𝒄𝒎𝟑 )

𝑲 Gas Ideal 𝑷𝒂. 𝒄𝒎𝟑

𝑲 Van der Waals

𝑲 de RedlichKwon

𝑷𝒂. 𝒄𝒎𝟑

𝑷𝒂. 𝒄𝒎𝟑

𝑲 de la ecuación de Virial 𝑷𝒂. 𝒄𝒎𝟑

𝑷𝟏

1

3

297400

297035.94

297305

300240.19

𝑷𝟐

1,51

2.5

371635,16

322800

369746

377953.54

𝑷𝟑

2.1

2.1

421186,5

420420

418894

441706.73

𝑷𝟒

2.6

1.8

445962,4

445094

442953

468937.8757

𝑷𝟓

3.4

1.5

520322,4

516290

515691

511329.46

𝑷𝟔

5.3

1.1

581940

529432.9

580768.7

585341.37

𝑷𝟕

8

0.8

639200

635400

636760

643822.02

𝑷𝟖

11.8

0.6

705640

699952.4

702300.6

714213.66

𝑷𝟗

19,8

0.4

790020

779031

783505,8

803408.40

𝑷𝟏𝟎

37,1

0.25

92379

903051,1

912993,9

951282.05