Video: Determining the Molar Internal Energy Change of a Gas in an Isochoric Process

The temperature of an ideal monatomic gas rises by 8.0 K while the volume of the gas remains constant. What is the change in the internal energy per mole of the gas?

03:39

Video Transcript

The temperature of an ideal monatomic gas rises by 8.0 Kelvin, while the volume of the gas remains constant. What is the change in the internal energy per mole of the gas?

So in this question, we’ve got an ideal monatomic gas. And its temperature rises by 8.0 Kelvin, while the volume of the gas remains constant. We need to find the change in the internal energy per mole of the gas.

Let’s start by recalling the first law of thermodynamics. The change in internal energy, Δ𝑒, is equal to 𝑄, the heat supplied, plus π‘Š, the work done. And we can apply this to our gas. The change in internal energy of the gas is equal to the heat supplied to the gas plus the work done on the gas.

However, when heat is supplied at a constant volume, all of the heat supplied to a gas goes towards increasing the temperature. In other words, the work done is equal to zero. And the change in internal energy of the gas is equal to the heat supplied to the gas.

So basically, what this is telling us is that we don’t need to worry about any work being done on the gas. All we need to worry about is the change in internal energy due to the heat supplied to the gas. And that is the relevance of the constant volume bit.

Now we can recall that the internal energy of an ideal monatomic gas is given by 𝐸 sub int, the internal energy, is equal to three by two times 𝑛𝑅𝑇. The lowercase 𝑛 is the number of moles of gas that we have. 𝑅 is simply the molar gas constant. And 𝑇 is the temperature of the gas.

In this question, we’re trying to find the change in internal energy per mole of the gas. The internal energy per mole is given by the total internal energy of the gas, 𝐸 sub int, divided by 𝑛, the number of moles of gas that we have. And so that’s just three by two 𝑛𝑅𝑇 divided by 𝑛. The 𝑛s cancel out, leaving us with three by two 𝑅𝑇.

But we’re not just trying to find out the internal energy per mole of the gas. We’re trying to find out the change in the internal energy per mole of the gas when the temperature rises by 8.0 Kelvin. This change in internal energy per mole is given by 𝐸 sub int over 𝑛 sub final, the final internal energy after the temperature of the gas has risen by 8.0 Kelvin, minus 𝐸 sub int over 𝑛 sub initial, the initial internal energy per mole of the gas before the temperature has risen.

And on the right-hand side, we can substitute in three by two 𝑅𝑇 in each case. However, we know that three by two and 𝑅 are just constant. They do not change over time. So we can write the right-hand side of the expression like this. In other words, we’ve pulled out three by two 𝑅 in each case because they don’t change. And at this point, we can actually factorize, which gives us that the right-hand side is equal to three by two 𝑅 multiplied by 𝑇 sub final minus 𝑇 sub initial.

In other words, this bracket is equal to the change in temperature. And we know that that change in temperature, Δ𝑇, is 8.0 kelvin because we’ve been told this in the question. So our final mathematical expression for the change in internal energy per mole of the gas is three by two 𝑅 multiplied by Δ𝑇. At this point, we can substitute in the values, which gives us a value of 99.72 dot dot dot, so on and so forth, joules.

Now the value that we’ve been given in the question and used in our calculation of 8.0 kelvin has been given to us to two significant figures. So we need to give our final answer to two significant figures as well. And so our final answer is that Δ𝐸 sub int over 𝑛, the change in internal energy per mole of a gas, is 100 joules to two significant figures.

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