Question

6. A large balloon containing 2.00×103 m3 of helium gas at 1.00 atm and a temperature 15.0◦ C rises rapidly from ground level to an altitude where the atmospheric pressure is only 0.900 atm. Assume that the helium behaves as an ideal gas and that the balloon rises so rapidly that no heat is exchanged with the surrounding atmosphere. a. Calculate the volume of the helium gas at the higher altitude. b. Calculate the temperature of the gas at the higher altitude. c. What is the change in the internal energy of the helium as the balloon rises to the higher altitude?

Answer 1

2130 m³, 3.11 °C, 1.25×10⁷ J

Step-by-step explanation:

No heat is exchanged, so this is an adiabatic process. For an ideal gas, this means:

PV^((f+2)/f) = constant,

where P is pressure, V is volume, and f is degrees of freedom.

For a monatomic gas, f=3. For a diatomic gas, f=5. Since helium is monatomic, f=3.

Therefore:

PV^(5/3) = PV^(5/3)

(1.00 atm) (2.00×10³ m³)^(5/3) = (0.900 atm) V^(5/3)

V = 2130 m³

For an ideal gas in an adiabatic process, we can also say:

VT^(f/2) = constant

Therefore:

(2.00×10³ m³) (15.0 + 273.15 K)^(3/2) = (2130 m³) T^(3/2)

T = 276.26 K

T = 3.11 °C

Finally, the change in internal energy is:

ΔU = (f/2) nRΔT

We need to find the number of moles, n, using ideal gas law:

PV = nRT

n = PV/(RT)

n = (1.00 atm) (2.00×10³ m³) / ((8.21×10⁻⁵ atm m³ / mol / K) (15.0 + 273.15 K))

n = 84,500 mol

So the change in internal energy is:

ΔU = (3/2) (84,500 mol) (8.314 J/mol/K) (15.0 - 3.11) K

ΔU = 1.25×10⁷ J