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Second Law of Thermodynamic

  • Thread starter stark_1809
  • Start date
  • #1

Homework Statement


A balloon containing 2.00 x10^3 m3 of helium gas at 1.00 atm and a temperature of 15.0°C 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. For helium, γ = 1.67.

Homework Equations


(a) Calculate the volume of the gas at the higher altitude.
(b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?



The Attempt at a Solution

 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


A balloon containing 2.00 x10^3 m3 of helium gas at 1.00 atm and a temperature of 15.0°C 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. For helium, γ = 1.67.

Homework Equations


(a) Calculate the volume of the gas at the higher altitude.
(b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?



The Attempt at a Solution

You have to make an attempt first. What equation applies here?

AM
 
  • #3
Oh, I'm sorry. My mistake.
For (a) and (b), I use the equation for the adiabatic process.
But for (c): I don't know what equation to apply here.
 
  • #4
Well, I think it is an adiabatic process. Hence the internal energy will be:
E(int)= -W
but I don't know how to calculate this work here. The equation of Work is W= integral(pdV), right? But p changes?
 
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,583
346
Well, I think it is an adiabatic process. Hence the internal energy will be:
E(int)= -W
but I don't know how to calculate this work here. The equation of Work is W= integral(pdV), right? But p changes?
You can determine the work done by the gas, but that is doing it the hard way. What does property determines the internal energy of an ideal gas?

AM
 

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