Question

50 PUNTOS !!! Resolver. Estado gaseoso

Answer 1

1. V1 = 550 mL / 1000 = 0.550 L

P1 = 758 mmHg / 760 = 0.997  1 atm

V2 = ¿?

P2 = (1/2 VECES MAYOR) = 0.997 + 0.5 = 1.497 = 1.5 atm

Aplicar Ley de Boyle =  (V1 x P1 = V2 x P2)

V2 = (0.550 L x 1 atm) / 1.5 atm

V2 = 0.3663 L x 100 = 366.67 mL

2. Aplicar ley de Charles = ( V1/T1 = V2/T2)

V1 = 360 mL / 1000 = 0.360 L

T1 = 27 ºC + 273 = 300 K

V2 = ¿?

T2 =  - 23 ºC + 273 = 250 K

V2 = (0.360 L x 250 K) / 300 K

V2 = 0.3 L x 1000 = 300 Ml

3. Aplicar ley de los gases ideales = V x P = n x R  x T

V = 11.5 pie3

     1 pie3 --- 28.316 L

11.5 pie3 ---      x

        X = 325.634 L

T = 72 ºF

K = (ºF – 32) / 1.8 + 273

K = (72 – 32) / 1.8 + 27

K = 295.9

P = 1 atm

R = 0.0821 (L atm / mol K)

Calcular n (moles)

n = (325.634 L  x   1 atm)/(0.0821 (L atm / mol K) x 295.9 K)

n = 13.404 moles

calcular Volumen

               1 mol  ---- 22.4 L

     13. 404 mol  ----     x

               X = 300.25 L

Calcular el volumen en pie3

V = 300.25 / 28.317 = 10.603  pie3

4. Aplicar Ley de los gases ideales = V x P = n x R  x T

V = 149 mL / 1000 = 0.149 L

T = 25 ºC + 273 = 298 K

 P = 755 mmHg / 760 = 0.99 atm

V gas CN= ¿?

R = 0.0821 (L atm / mol K)

Calcular n (moles)

n = ( 0.149 L  x  0.99 atm) /( R = 0.0821 (L atm / mol K) x 295 k)

n = 0.0061 moles

calcular volumen en CN

         1 mol ---- 22.4 L

0.0061 mol ----  x

            X = 0.136 L x 1000 = 136 mL

5. Aplicar ley general de los gases = ( V1 x P1 /T1 = V2 x P2/T2)

V1 = 333 mL / 1000 = 0.333 L

T1 = 25 ºC + 273 = 298 K

P1 = 750 mmHg / 760 = 0.986 atm

V2 = ¿?

T2 = - 11 ºC + 273 = 262 K

P2 = 730 mmHg / 760 = 0.961 atm

Calcular V2

V2 = (0.333 L  x 0.986 atm x 262 K) / (0.961 atm  x  298 K)

V2 = 0.300 L x 1000 = 300 mL

6. Aplicar  ley general de los gases = ( V1 x P1 /T1 = V2 x P2/T2)

T1 = 77 ºF

K = (ºF – 32)/1.8 + 273

K = (77 – 32)/ 1.8 + 273 = 298 K

P1 = 14.7 lb/pulg2

  1 lb/pulg2 ---- 0.06805 atm

                   14.7 lb/pulg2 ---    x

                                X = 1.00 atm

T2 = 70 ºF

K = (ºF – 32)/1.8 + 273

K = (70 – 32)/ 1.8 + 273 = 294 K

P2 = 2000 lb/pulg2

           1 lb/pulg2 ---- 0.06805 atm

   2000 lb/pulg2 ---    x

                                X = 136 atm

V2 = 10 cilindros capacidad c/u = 2 pie3 = 20 pie3 = 566.34 L

V1 = (566.34 L  x 136 atm x 298 K) / ( 1 atm x 294 K)

V1 = 78127.2 L / 28.317 = 2757 pie3

Answer 2

Respuesta:

1. V1 = 550 mL / 1000 = 0.550 L

P1 = 758 mmHg / 760 = 0.997  1 atm

V2 = ¿?

P2 = (1/2 VECES MAYOR) = 0.997 + 0.5 = 1.497 = 1.5 atm

Aplicar Ley de Boyle =  (V1 x P1 = V2 x P2)

V2 = (0.550 L x 1 atm) / 1.5 atm

V2 = 0.3663 L x 100 = 366.67 mL

2. Aplicar ley de Charles = ( V1/T1 = V2/T2)

V1 = 360 mL / 1000 = 0.360 L

T1 = 27 ºC + 273 = 300 K

V2 = ¿?

T2 =  - 23 ºC + 273 = 250 K

V2 = (0.360 L x 250 K) / 300 K

V2 = 0.3 L x 1000 = 300 Ml

3. Aplicar ley de los gases ideales = V x P = n x R  x T

V = 11.5 pie3

    1 pie3 --- 28.316 L

11.5 pie3 ---      x

       X = 325.634 L

T = 72 ºF

K = (ºF – 32) / 1.8 + 273

K = (72 – 32) / 1.8 + 27

K = 295.9

P = 1 atm

R = 0.0821 (L atm / mol K)

Calcular n (moles)

n = (325.634 L  x   1 atm)/(0.0821 (L atm / mol K) x 295.9 K)

n = 13.404 moles

calcular Volumen

              1 mol  ---- 22.4 L

    13. 404 mol  ----     x

              X = 300.25 L

Calcular el volumen en pie3

V = 300.25 / 28.317 = 10.603  pie3

4. Aplicar Ley de los gases ideales = V x P = n x R  x T

V = 149 mL / 1000 = 0.149 L

T = 25 ºC + 273 = 298 K

P = 755 mmHg / 760 = 0.99 atm

V gas CN= ¿?

R = 0.0821 (L atm / mol K)

Calcular n (moles)

n = ( 0.149 L  x  0.99 atm) /( R = 0.0821 (L atm / mol K) x 295 k)

n = 0.0061 moles

calcular volumen en CN

        1 mol ---- 22.4 L

0.0061 mol ----  x

           X = 0.136 L x 1000 = 136 mL

5. Aplicar ley general de los gases = ( V1 x P1 /T1 = V2 x P2/T2)

V1 = 333 mL / 1000 = 0.333 L

T1 = 25 ºC + 273 = 298 K

P1 = 750 mmHg / 760 = 0.986 atm

V2 = ¿?

T2 = - 11 ºC + 273 = 262 K

P2 = 730 mmHg / 760 = 0.961 atm

Calcular V2

V2 = (0.333 L  x 0.986 atm x 262 K) / (0.961 atm  x  298 K)

V2 = 0.300 L x 1000 = 300 mL

6. Aplicar  ley general de los gases = ( V1 x P1 /T1 = V2 x P2/T2)

T1 = 77 ºF

K = (ºF – 32)/1.8 + 273

K = (77 – 32)/ 1.8 + 273 = 298 K

P1 = 14.7 lb/pulg2

 1 lb/pulg2 ---- 0.06805 atm

                  14.7 lb/pulg2 ---    x

                               X = 1.00 atm

T2 = 70 ºF

K = (ºF – 32)/1.8 + 273

K = (70 – 32)/ 1.8 + 273 = 294 K

P2 = 2000 lb/pulg2

          1 lb/pulg2 ---- 0.06805 atm

  2000 lb/pulg2 ---    x

                               X = 136 atm

V2 = 10 cilindros capacidad c/u = 2 pie3 = 20 pie3 = 566.34 L

V1 = (566.34 L  x 136 atm x 298 K) / ( 1 atm x 294 K)

V1 = 78127.2 L / 28.317 = 2757 pie3

Explicación:

SOY DEL SACO