Question

Alguien me podria ayudar con los siguientes problemas

Con fórmulas por favor :)

Answer 1

1. APLICAR LA lEY DE LOS GASES: LEY DE BOYLE

V1 · P1 = V2 · P2

V1 = 638 mL
P1 = 69.9 PKa
V2 = 208 mL

DESPEJAR P2 = ?

P2 = 69.6 Pka x 638 mL
        ```````````````````````````
              208 mL

P2 = 213.48 Pka

2.  Aplicar la Ley de Charles

V1  =  V2
`````   ``````
 T1     T2

V1 =1.50 L
T1 = 22 ºC + 273 = 295 K
V2 = ?
T2 = 450 ºC + 273 = 723 K
despejar V2 = ?

V2 = 1.50 L  x  723 K
        ``````````````````````
               295 K

V2 = 3.676 L

3. Aplicar LEY DE BOYLE

V1 = 2.45 L
P1 = 0.950 atm
V2 = 5.38 L
P2 = ?

P1 x V1 = V2 x P2

P2 = 2.45 L  x 0.950 atm
        ```````````````````````````
              5.38 L

P2 = 0.4326  atm

4. Ley de Charles

V1 = V2
````   `````
 T1    T2

T1 = 15 ºC + 273 = 288 K
V1 = 282 mL/1000 = 0.282 L
T2 = ?
V2 = 82.0 mL/1000 = 0.082 L

T2 = 0.082 L x 288 K
        ```````````````````````
            0.282 l

T2 = 83.74 K

5. Aplicar Ley de los gases ideales

diámetro = 2 cm

radio = diametro/2

r = 2.00 cm/2 = 1 cm
altura = 4 m x 100 = 400  cm

Calcular volumen
V cilindro = π r² h
V = π ( 1 cm)² (400 cm)
V cilindro = 1257 cm³

2 calcular n (moles) P V = n R T

R = 0.082 (L atm / mol K)
T = 35 ºC + 273 = 308 K
Mm Ne= 20.17 g/mol
P = 200 Pa

    1 Pa ----- 0.00010 atm
200 Pa -----       x

x = 0.002 atm

n =     0.002 atm x 1257 L
      ``````````````````````````````````
     0.082 (L atm/mol K) x 308 K

n = 0.00995 moles

g = 0.00995 moles  x   20.17 g
                                  ``````````````
                                     1 mol

g = 2.0077 de neón