Question Video: Calculating the Thermodynamic State Parameters in an Adiabatic Quasi-static Process

An ideal diatomic gas at a temperature of 78 K is slowly compressed adiabatically to 0.20 times its original value. What is the gas’s final temperature?

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Video Transcript

An ideal diatomic gas at a temperature of 78 Kelvin is slowly compressed adiabatically to 0.20 times its original value. What is the gas’s final temperature?

So in this question, we’ve got an ideal diatomic gas at a temperature of 78 Kelvin, and it’s being slowly compressed adiabatically to 0.20 times its original volume. We need to find the gas’s final temperature.

Let’s start by considering the fact that the gas is slowly compressed adiabatically. The slow compression just means that the gas will be in equilibrium, even whilst the compression is happening. So let’s imagine that we’ve got a piston compressing a gas. There’s a piston in orange compressing the pink gas particles.

What we’re saying is that the piston moves slowly enough so that the entirety of the gas is in equilibrium. Basically, the piston collides with the particles nearest to it, and that results in some form of change in the behavior of the particles. This change in behavior should be allowed to be transmitted from the particles in this region to the particles in this region. And hence, the piston must move really, really slowly, because it takes some time for these particles here to collide into these particles here, and so on and so forth, and thus affect how the particles around here are behaving. So that’s what the slow compression is about.

Now let’s talk about the fact that it’s an adiabatic compression. In thermodynamics, an adiabatic process is when there is no heat transfer or mass transfer between the system and its surroundings. So basically, our gas is not giving out or taking in any heat, and neither is it giving out or taking in any particles or any mass.

When we’ve got an adiabatic process for an ideal gas, it can be shown that this relationship holds. The pressure of the gas multiplied by the volume of the gas to the power of 𝛾 is equal to a constant. 𝛾 is known as the specific heat ratio. It’s the ratio between the specific heat capacity at a constant pressure, 𝐶 sub 𝑃, and the specific heat capacity at a constant volume, 𝐶 sub 𝑉.

Now we also know that we’ve got an ideal diatomic gas. It can be shown that, for an ideal diatomic gas, the value of 𝛾 is seven over five. This has to do with the number of degrees of freedom within a diatomic gas. For example, a diatomic molecule — let’s say this is our diatomic molecule — will have translational degrees of freedom so it can move left and right, up or down, or in and out of the screen, as well as rotational degrees of freedom and vibrational degrees of freedom, where the bond in the molecule vibrates. This is directly linked to the value of 𝛾. And as we said earlier, for an ideal diatomic gas, the value of 𝛾 is seven over five.

So putting this together with the first equation in green, we find that 𝑃 multiplied by 𝑉 to the power of seven over five is equal to a constant. However, in this question, we’re not working with pressures and volumes. We’re working with temperatures and volumes. We’ve been given the initial temperature of the gas. We’ll call this 𝑇 sub one, and it’s 78 Kelvin. That’s the temperature before the gas is compressed. And we also know the relationship between the initial volume of the gas and the final volume of the gas.

We know that the gas is compressed as 0.20 times the original volume. So we can say that 𝑉 sub two — that’s the final volume — is equal to 0.20 times 𝑉 sub one — that’s the original or initial volume. So basically, what we need to do here is to convert this relationship into one that contains temperature and volume. We can do this by remembering that we’re dealing with an ideal gas. Therefore, the ideal gas equation must apply.

The pressure multiplied by the volume is equal to lowercase 𝑛, which stands for the number of moles of gas that we have, multiplied by 𝑅, which is the molar gas constant, multiplied by the temperature, 𝑇. We can rearrange this to give us 𝑃 is equal to 𝑛𝑅𝑇 over 𝑉 and sub this into our equation. This gives us 𝑛𝑅𝑇 over 𝑉 multiplied by 𝑉 to the power of seven-fifths is equal to a constant.

What we can do is to divide both sides of the equation by 𝑛𝑅. So the 𝑛𝑅s on the left-hand side cancel. And what happens on the right-hand side? Well, since the number of moles of gas is not changing in this compression, because we’ve still got the same amount of gas, the value of 𝑛 must be a constant. And we already know that the value of 𝑅 is a constant. So what we have on the right-hand side is a constant divided by a constant. And a constant divided by a constant is another constant. So we can just say that the right-hand side is equal to a constant, and the left-hand side simplifies a little bit.

As we saw earlier, the 𝑛𝑅 canceled on the left-hand side. So we were left with 𝑇 over 𝑉 multiplied by 𝑉 to the power of seven-fifths. But we can simplify this even further because 𝑉 to the power of seven-fifths divided by 𝑉 is equal to 𝑉 to the power of two-fifths. And at this point, we’ve got the relationship we’re looking for. 𝑇 multiplied by 𝑉 to the power of two-fifths is equal to a constant.

In other words, the product between the initial temperature and the initial volume to the power of two-fifths must be equal to the product of the final temperature and the final volume to the power of two-fifths must be equal to the same constant. What that constant is actually doesn’t really matter.

And so what we’re left with is the relationship that will help us find our answer. And so let’s get rid of everything else and substitute in the values we’ve been given. We know that 𝑇 one is equal to 78 Kelvin. Multiply this by the initial volume to the power of two-fifths and we get that this must be equal to 𝑇 two multiplied by 𝑉 two to the power of two-fifths.

But we said earlier that 𝑉 two is equal to 0.20 times the original volume, which is 𝑉 one. Distributing the power, we get that the right-hand side of the equation is equal to 𝑇 two multiplied by 0.20 to the power of two-fifths times 𝑉 one to the power of two-fifths. And at this point, we can divide both sides of the equation by 𝑉 one to the power of two-fifths, which causes these to cancel. And so we’re left with 78 Kelvin is equal to 𝑇 two multiplied by 0.20 to the power of two-fifths. At this point, we can divide both sides of the equation by 0.20 to the power of two-fifths, which gives us 78 Kelvin divided by 0.20 to the power of two-fifths is equal to 𝑇 two. Evaluating this fraction, we find that 𝑇 two is equal to 148.485 dot dot dot, so on and so forth, Kelvin.

However, the values that we’ve been given in the question of 78 Kelvin and 0.20 times have been given to two significant figures. So our answer should also be in two significant figures. And so our final answer is that the gas’s final temperature is 150 Kelvin.

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