Answer:
The internal energy of the gas is 1.25 times greater when the helium is at 0.250 atm gauge pressure.
Explanation:
Since the helium in the balloon is an ideal gas, we can use the equation of state for ideal gases:
[tex]pV=nRT[/tex]
where
p is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For the helium gas in the balloon, we have:
[tex]p=0.250 atm + 1.0 atm = 1.250 atm[/tex] (the total pressure is sum of the gauge pressure + atmospheric pressure)
[tex]V=8.0 L[/tex] is the volume
So we can write the temperature as
[tex]T=\frac{pV}{nR}=\frac{(1.250)(8.0)}{nR}=\frac{10}{nR}[/tex]
Instead, when the helium balloon is at zero gauge pressure (=at atmospheric pressure), we have
[tex]p=1 atm[/tex]
Therefore, the temperature would be
[tex]T'=\frac{pV}{nR}=\frac{(1.0)(8.0)}{nR}=\frac{8}{nR}[/tex]
The ratio between the two temperatures is
[tex]\frac{T}{T'}=\frac{10}{8}=\frac{5}{4}=1.25[/tex]
And we also know that the internal energy of a gas is directly proportional to the temperature:
[tex]U\propto T[/tex]
Therefore, the internal energy of the gas is 1.25 times greater when the helium is at 0.250 atm gauge pressure.
