Physics, 19.02.2020 06:00 TVASALLO31

A helium-filled toy balloon has a gauge pressure of 0.250 atm and a volume of 8.0 liters.
How much greater is the internal energy (in J) of the helium in the balloon than it would be at zero gauge pressure?

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Answer from: chessacs6540

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:

pV=nRT

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:

p=0.250 atm + 1.0 atm = 1.250 atm (the total pressure is sum of the gauge pressure + atmospheric pressure)

V=8.0 L is the volume

So we can write the temperature as

$T=\frac{pV}{nR}=\frac{(1.250)(8.0)}{nR}=\frac{10}{nR}$

Instead, when the helium balloon is at zero gauge pressure (=at atmospheric pressure), we have

p=1 atm

Therefore, the temperature would be

$T'=\frac{pV}{nR}=\frac{(1.0)(8.0)}{nR}=\frac{8}{nR}$

The ratio between the two temperatures is

$\frac{T}{T'}=\frac{10}{8}=\frac{5}{4}=1.25$

And we also know that the internal energy of a gas is directly proportional to the temperature:

$U\propto T$

Therefore, the internal energy of the gas is 1.25 times greater when the helium is at 0.250 atm gauge pressure.

Answer from: Quest

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Answer from: Quest

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