# Homework Help: Balloon in a vacuum

1. Mar 20, 2017

### charlie05

1. The problem statement, all variables and given/known data
Balloon in vacuum, radius R0 = 5,0 cm, surface tension σ = 25 N/m.

a/ the pressure in the balloon p in,0 by radius R0………. pin,0 = ?

b/ relation to the pressure inside the balloon pin, depending on its radius R…?

c/relation to the pressure inside the balloon pin , depending on the pressure outside the balloon pout and on its radius R, surface tension is σ…….?

2. Relevant equations
a/ pin,0=4σ/R0 = 4.25/0,05 = 2kPa

b/ pin = 4σ/R

c/ pin = pout + 4σ/R

3. The attempt at a solution

2. Mar 20, 2017

### Staff: Mentor

Looks right.

3. Mar 20, 2017

### charlie05

in B/ I have to use pressure p in 0.....

p in = 4σ/R .....R = R0 + ΔR = R0 + ( R-R0)....p in = 4σ/R0 + ( R-R0)

4. Mar 20, 2017

### Staff: Mentor

If you add proper brackets to the denominators, you get the same result as in post 1.

5. Mar 20, 2017

### charlie05

so:?

p in = p in 0 + 4σ/ ( R- R0)

6. Mar 20, 2017

### Staff: Mentor

That looks wrong, no matter how I interpret the notation, and I don't see how you got it. In particular, it is undefined for the easiest case of R=R0.

Why do you want to change a correct answer?

7. Mar 20, 2017

### charlie05

Because I have to find dependence p in on p in 0 : Find the equation for the pressure p in inside the balloon, depending on the radius R, where
at radius R 0 is pressure p in 0

8. Mar 20, 2017

### Staff: Mentor

That's not what you write in post 1.

9. Mar 20, 2017

### charlie05

Yes, it's true, I did not read the assignment to the end :-( I am sorrry.....

10. Mar 20, 2017

### haruspex

Your notation is hard to interpret, but I think you are saying:
With zero external pressure, the radius is R0 and the internal pressure is pin0.
With some unknown external pressure the radius is R; what is the internal pressure, pin, now?

If that is right, write out the equation for each of the two circumstances. Please use subscripts, as I have, to make the notation clearer. Use the X2 button above the text entry box.

Last edited: Mar 21, 2017
11. Mar 21, 2017

### charlie05

b/ Find the equation for the pressure p in inside the balloon, depending on the radius R, where
at radius R 0 is pressure p in 0

12. Mar 21, 2017

### haruspex

Yes, that agrees with my interpretation. So what two equations can you write, one for the zero external pressure case and one for the general case?

13. Mar 21, 2017

### charlie05

in vacuum : ......pin = 4σ/R.....for R0 .....pin0 = 4σ/R0
p in = 4σ/R .....R = R0 + ΔR = R0 + ( R-R0)....p in = 4σ/ (R0 + ( R-R0))
- But here now I can not express pin0

for the general case: pin = pout + 4σ/R

14. Mar 21, 2017

### haruspex

It's all the same balloon and contents. What does not vary as the external pressure changes?

15. Mar 21, 2017

### charlie05

surface tension?

16. Mar 21, 2017

### haruspex

What else?

17. Mar 21, 2017

### charlie05

the amount of air in the balloon ?

18. Mar 21, 2017

### haruspex

Right. What law can you use?

19. Mar 21, 2017

### charlie05

p1V1 =p2V2........?

20. Mar 21, 2017

### haruspex

Yes. (You can assume the temperature is constant.)

21. Mar 21, 2017

### charlie05

pin0V0=pinV
pin = pin0*V0 / V = pin0 * (V0 / V ) = pin0 * (R03 / R3 ) .........?

22. Mar 21, 2017

### haruspex

Yes.

23. Mar 21, 2017

### charlie05

super..thanks very much :-)

24. Mar 21, 2017

### haruspex

Ok, but what is your answer to the question, i.e. as an equation for pin in terms of pout, R and σ?

25. Mar 21, 2017

### charlie05

oh yes.........pin = pout + 4σ/R...?