What Factors Affect the Internal Energy of a Gas?

How much greater is the internal energy of the helium in the ball than it would be at zero gauge pressure?

The internal energy of the helium in the ball is not greater than it would be at zero gauge pressure. The difference in internal energy is 0.

Answer:

The internal energy of a gas depends on its temperature, which in turn is related to its pressure and volume. In this case, we are comparing the internal energy of helium in a ball at a gauge pressure of 0.200 atm to the internal energy at zero gauge pressure.

To calculate the difference in internal energy, we need to consider the change in pressure. The gauge pressure represents the pressure above atmospheric pressure, so at zero gauge pressure, the absolute pressure would be atmospheric pressure, which is approximately 1 atm.

We are given that the volume of the helium-filled ball is 10.0 L. To calculate the difference in internal energy, we can use the formula:

ΔU = Δn * R * ΔT

where:

ΔU is the change in internal energy

Δn is the change in the number of moles of gas

R is the gas constant

ΔT is the change in temperature

Since the volume remains constant, the number of moles of gas remains constant as well. Therefore, Δn = 0.

Now, let's calculate the change in temperature (ΔT). We can use the ideal gas law:

PV = nRT

at a gauge pressure of 0.200 atm, the absolute pressure is 1 atm, so we can calculate the initial temperature (T1):

T1 = (1 atm * 10.0 L) / (n * R)

Next, at zero gauge pressure, the absolute pressure is 1 atm, so we can calculate the final temperature (T2):

T2 = (1 atm * 10.0 L) / (n * R)

Since the number of moles of gas remains constant, we can simplify the equation to:

T2 = T1

Therefore, the change in temperature (ΔT) is 0.

Substituting the values into the equation for ΔU, we have:

ΔU = Δn * R * ΔT = 0 * R * 0 = 0

So, the internal energy of the helium in the ball is not greater than it would be at zero gauge pressure. The difference in internal energy is 0.

← Exploring data on peach sales and prices Understanding oxidation numbers in compounds →