Rate of effusion = = = A m o u n t o f g a s t i m e Amount of gas \over time t im e A m o u n t o f g a s
Combine with Grahams law;
R a t e o f e f f u s i o n o f h y d r o g e n R a t e o f e f f u s i o n o f O x y g e n Rate of effusion of hydrogen\over Rate of effusion of Oxygen R a t eo f e ff u s i o n o f O x y g e n R a t eo f e ff u s i o n o f h y d ro g e n = = = M O M H \sqrt MO \over \sqrt MH M H M O
To get
A m o u n t o f H t r a n s f e r r e d t i m e o f H A m o u n t o f O t r a n s f a r e d t i m e o f O {Amount of H transferred \over time of H} \over{Amount of O transfared\over time of O} t im eo f O A m o u n t o f Ot r an s f a re d t im eo f H A m o u n t o f H t r an s f erre d = = = M O M H \sqrt MO \over \sqrt MH M H M O
∴ \therefore ∴ T i m e o f H T i m e o f 0 Time of H\over Time of 0 T im eo f 0 T im eo f H = = = M O M H \sqrt MO\over \sqrt MH M H M O = = = M O M H \sqrt MO\over \sqrt MH M H M O
T i m e o f O 4 s Time of O\over 4s 4 s T im eo f O = = = 3 2 g / m o l 2 m o l / L \sqrt 32g/mol \over 2mol/L 2 m o l / L 3 2 g / m o l = = = 1 6 1 \sqrt 16 \over \sqrt 1 1 1 6
T i m e o f O 4 s Time of O \over 4s 4 s T im eo f O = = = 4 1 4\over 1 1 4
= 4 ∗ 4 = 16 =4*4=16 = 4 ∗ 4 = 16
Oxygen takes 16 16 16 seconds to travel through the same distance.
Comments
Amazing