An experiment you're designing needs a gas with γ = 1.50. You ...

Question

An experiment you're designing needs a gas with γ = 1.50. You recall from your physics class that no individual gas has this value, but it occurs to you that you could produce a gas with γ = 1.50 by mixing together a monatomic gas and a diatomic gas. What fraction of the molecules need to be monatomic?

Accepted Answer

Hello, let's go through with this practice problem. You are provided with two gas samples. A poly atomic where the ratio of the specific heat at constant pressure or CP to the specific heat at constant volume or CV is equal to four thirds and a mo atomic gas where the ratio of CP to CV is five thirds. You are asked to mix the two gasses to yield a gas with AC P to CV ratio of seven fifths. Calculate the percentage of poly atomic molecules in the mixture option. A 33%. B 78% C 22% D 67% and E 50%. In order to solve this problem, we're gonna need to do some energy analysis to figure out how the two gasses are going to mix together. Recall that the change in thermal energy for a gas is equal to the number of particles multiplied by the specific heat at constant volume. CV, multiplied by the temperature change. Let's expand this formula out a bit by using the CV value for both the mono atomic gas and the poly atomic gas. Let's start with the mono atomic gas. Now, for a mo atomic gas, the value for CV is always going to be three halves. So the energy for the mo atomic gas, which I'll call E sub one is equal to N, sub one multiplied by three halves and is also multiplied by the ideal gas constant R and multiplied by the temperature for the poly atomic gas. This is a little trickier because unlike for a mo atomic gas, there isn't only one value that we can expect a poly atomic gas to have or the specific heat at constant volume. Whatever that value is, is based on the number of degrees of freedom for the molecule, which we're not told in the problem. So this is the first kind of challenge of the problem to figure out what the value for CV is for the poly atomic gaps. But we still can figure it out. But we, by using the fact that we are given the ratio of CP to CV, we're told that it's four thirds for the poly atomic gas. So let's use that in com in combination with the fact that CP, the specific heat at constant pressure is equal to CV, the specific heat at constant volume plus the ideal gas constant. So according to the ratio, we're given four thirds, this is equal to the CP value for the polytonic gas. So CV plus R divided by CV, which can also be written if we divide out as one plus R divided by CV, let's continue, four thirds minus one is equal to one third. So one third is equal to the ideal gas constant divided by CV. And we're trying to solve this for CV. So we're just going to multiply both sides of the equation by CV. And then multiply both sides by three. So we find that CV is equal to three R. So it looks like this is our CV value for the poly atomic gas. What's important to realize is that this three can also be written in other ways and other ratios that are equal to three. Let's see what I mean. So going off of our earlier equation for the thermal energy, the energy of the poly atomic gas or E sub P is equal to the number of particles in the polytonic gas or N sub P multiplied by the ratio which right now looks like it's three multiplied by R multiplied by the temperature. So what this tells us is that if we put together the two energy equations, we found then the total thermal energy, the net thermal energy for the mixture is gonna be equal to the sum of them. So the energy for the mo atomic gas plus the energy for the poly atomic gas, this is N sub one multiplied by three halves RT plus N sub P multiplied by three RT. And we can simplify this by factoring out the R the T. And if, instead of writing three, we write six divided by two as the ratio per CV. Then we can factor out the one half as well. So we can rewrite our equation for the thermal energy as one half multiplied by RT, then multiplied by in parentheses three and one plus six and P. So this is our equation for the thermal energy of the two gasses put together, we can compare this to the general formula or the thermal energy of the total mixture, which is gonna be the total number of particles N one plus N two multiplied by the CV value for the combined gas multiplied by the temperature. And set all of this equal to the equation. We just found using the other method. Three N plus one plus six NP. Notice that the teas cancel out and we're left with an equation saying that N one plus N two multiplied by CV is equal to one half, R multiplied by three and one plus six NP. Now let's take a second to figure out what CV value we're actually looking for in the final gas because as the problem tells us, the final gas should have AC P to CV value of seven fifths. That should be the ratio. So let's figure out what CV should be using. The same method we figured out earlier where seven fifths is gonna be equal to CP or CV plus R divided by CV, which can be rewritten as one plus R divided by CV, subtracting one from both sides of the equation. This becomes 2/5 is equal to R divided by CV. And then solving for CV, this is equal to five halves multiplied by R. So now let's substitute that into our combined energy equation. So that's N one plus N two, multiplied by five pals of R equals one half of R multiplied by three and one plus three or six NP. Now we'll just do some more simplifying. So the RS will cancel out now that we've got an R on both sides, this half will also cancel out. So on the left side of the equation we're left with and then distributing the 55 and one plus five and two R MP is equal to three and one plus six NP. Then we'll combine the like variables. So five N, one minus three and one equals two and one equals six NP minus five NP equals a single NP. So two N one equals NP. So if we're looking for the ratio of the number of mo atomic particles to the number of poly atomic particles, we can see there are twice as many mo atomic particles. So the ratio of N one to NP is equal to 2 to 1. So what that tells us is that for every three particles, two of them are monotonic and one of them is poly atomic. So because the problem is asking for the percentage of poly atomic particles, we're gonna take one and divide by three which gives us 0.33 or 33%. So 33% of the gas particles in this mixture are poly atomic, which corresponds to option A. So option A is our answer and that is it for this problem. I hope this video helped you out. And if it did, please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

Key Concepts

Heat Capacity Ratio (γ)

Mixing Gases

Mole Fraction

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