# Adiabatic Hot-Air Balloon Rising Question

1. May 20, 2008

### TFM

[SOLVED] Adiabatic Hot-Air Balloon Rising Question

1. The problem statement, all variables and given/known data

A large research balloon containing 2.00 x 10^3 m^3 of helium gas at 1.00 atm and a temperature of 15.0 degrees Celsius rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 Atm. Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air.

2. Relevant equations

pv = nRT

3. The attempt at a solution

I am not sure what to do, I tried using

$$P_1*V_1 = P_2*V_2$$, but this doesn't work - I think it is becausethe temperature isn't quite constant. (The next question asks for the change in T)

Does anyoine have any ideas?

TFM

2. May 20, 2008

### astrorob

Hi again,

What exactly is the question?

3. May 20, 2008

### TFM

Opps, Sorry:

Calculate the volume of the gas at the higher altitude.

TFM

4. May 20, 2008

### alphysicist

Hi TFM,

You're right that PV=constant is true only for an isothermal process, which is not true here. What is the relationship between P and V for an adiabatic process?

5. May 21, 2008

### TFM

I cn't seem to find a formula for this in my text book. The nearest is:

$$W = \frac{C_v}{R}(p_1V_1 - p_2V_2)$$

But the work done is not given, nor is Cv?

TFM

6. May 21, 2008

### fredrick08

hi TFM, im quite a novice but im pretty sure for an adiabatic process that,

P2V2$$^{}\gamma$$=P1V1$$^{}\gamma$$

and T2V2^$$^{}(\gamma$$-1)=T1V1^$$^{}(\gamma$$-1)

also that helium is a monatomic gas, i got no idea how you can it thought without Cv, i could tell you that Cv for helium is 12.5 lol.... or that $$\gamma$$=5/3, then its just plug in numbers and solve, if there is another way then i got no idea, r u sure its just not isothermal??
i guess that saying that there is no heat exchanged, implies its adiabatic tho...

7. May 21, 2008

### TFM

I'm pretty sure it is Adiabatic, since partof the deifintion in my textbook:

...by carrying out the process so quickly that there is not enough time for appreicable heat-flow

Compared to the questions:

...the balloon's ascent is too rapid to permit much heat exchange with the surrounding air.

Thanks,

TFM

8. May 21, 2008

### fredrick08

ya sounds adiabatic, well if there is a way of finding V2 in an adiabatic process without knowing gamma, let me know lol, r u sure your not meant to get Cv or gamma from the book, since gamma=(R/Cv)+1, or Cp/Cv...

9. May 21, 2008

### fredrick08

im pretty sure that the solution to this prob is,
(101300*(2x10^3)^(1.67)/91170)^(.6)=2.16x10^3m^3

if its not let me know.....

10. May 21, 2008

### Vuldoraq

Hey TFM,

What Fredrick08 said looks good to me, that for an adiabatic process,

$$P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$$

To find $$\gamma$$ note that helium can be treated as an ideal monatomic gas and that $$C_{p}=C_{v}+R$$

For the temperature,

$$T_{1}V_{1}^{\gamma-1}=T_{2}V_{2}^{\gamma-1}$$

11. May 21, 2008

### TFM

$$\gamma$$ is gioven in the book as 1.67.

So

$$P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$$

$$\frac{P_1V_1^{\gamma}}{p_2} = v_2^{\gamma}$$

$$\sqrt[\gamma]{\frac{p_1V_1^{\gamma}}{p_2}} = v_2$$

Does this look right?

TFM

12. May 21, 2008

### Vuldoraq

Yeah that looks pretty good.

13. May 21, 2008

### TFM

I can't seem to put 1.67 in the root part of mastering physics?

(i.e: I can't do this:$$\sqrt[1.37]{G}$$)

TFM

14. May 21, 2008

### TFM

One other question, what is the pressure of the balloon after it has risen, since the 0.9atm is the pressure of the atmosphere oin the ballon, not the pressure of the helium in the balloon itself?

TFM

15. May 21, 2008

### alphysicist

You can rewrite it in terms of exponents:

$$\sqrt[1.67]{G} = G^{(1/1.67)} = G^{(0.5988)}$$

and get the final numerical value. Does that work?

I think some research balloons aren't as elastic to the extent that children's balloons are; they are more similar to large bags. It looks to me like that is what they want you to assume here, so we can treat the pressure inside the balloon as essentially equal to the outside air pressure.

16. May 21, 2008

### TFM

Thanks for all the help, I got an answer of 2130.

Thanks to you all,

TFM