Surface energy

  • Thread starter vissh
  • Start date
  • #1
vissh
82
0
hiii :)

Homework Statement


(Q)Air is pushed into a soap bubble of radius r to double its radius.If the surface tension of the soap solution is S, the work done in the process is
<a>8(pie)(r)2(S) <b>12(pie)(r)2(S) <c>16(pie)(r)2(S)
<d>24(pie)(r)2(S)

Homework Equations


Surface energy of a liquid = (S)(A)
where S is surface tension and A is the area of the surface of liquid.

The Attempt at a Solution


Taking the bubble as system.
Isn't it like this:-
>Work done in process i.e. work done by external agent = change in Potential energy = change in surface energy(Uf - Ui)
> Ui = 4(pie)(r)2(S)
> Uf = 4(pie)(2r)2(S) = 16(pie)(r)2(S)]
> Uf - Ui = 12(pie)(r)2(S) = Work done we require
But the answer is <d>
Can any1 help me where i am getting wrong :)
Thanks for reading ^.^
 

Answers and Replies

  • #2
zorro
1,384
0
For a bubble surface tension is on its two sides i.e. 2S :smile:
 
  • #3
vissh
82
0
lol . I missed that xD Thanks abdul :) Got it ^.^
 

Suggested for: Surface energy

Replies
4
Views
262
  • Last Post
Replies
1
Views
192
  • Last Post
Replies
13
Views
617
Replies
2
Views
214
  • Last Post
Replies
5
Views
288
Replies
1
Views
326
  • Last Post
Replies
5
Views
235
Replies
7
Views
278
Replies
9
Views
403
Replies
5
Views
300
Top