Adiabatic Hot-Air Balloon Rising Question

[TFM]

[SOLVED] Adiabatic Hot-Air Balloon Rising Question

Homework Statement

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.

Homework Equations

pv = nRT

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

 

[astrorob]

Hi again,

What exactly is the question?

 

[TFM]

Opps, Sorry:

Calculate the volume of the gas at the higher altitude.

TFM

 

[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?

 

[TFM]

I cn't seem to find a formula for this in my textbook. 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

 

[fredrick08]

hi TFM, I am quite a novice but I am 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...

 

[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

 

[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...

 

[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...

 

[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}

 

[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

 

[Vuldoraq]

Yeah that looks pretty good.

 

[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

 

[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

 

[alphysicist]

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

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?

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

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.

 

[TFM]

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

Thanks to you all,

TFM