A monatomic gas is adiabatically compressed to ⅛ of its initial volume. Does each of the following quantities change? If so, does it increase or decrease, and by what factor? If not, why not?

b. The mean free path.

Question

A monatomic gas is adiabatically compressed to ⅛ of its initial volume. Does each of the following quantities change? If so, does it increase or decrease, and by what factor? If not, why not?

b. The mean free path.

Accepted Answer

Hey, everyone in this problem, we're told that a neon sample undergoes adiabatic expansion from V naught to two V knot where V naught is the initial volume. And we're asked whether the mean free path will increase or decrease. And by what factor? So we have four answer choices here. Option A, it decreases by a factor of two. Option B, it decreases by a factor of one half. Option C, it does not change because it's an adiabatic process or option D, it increases by a factor of two. All right. So we wanna look at the mean free path and let's recall that the mean free is given by lambda is equal to B divided by four by root two R squared A. So we have the volume B in the numerator. Everything else is in the denominator, the R is going to be the radius of the molecule and N is the number of molecules in that gas. All right. So we wanna find the relationship between lambda knot and lambda F the initial and final mean free paths. So let's write those two out. We have that lamb the knot, the initial, let me do this in black instead. So we have lamp a knot, that initial mean free path is going to be equal to the volume V knot divided by or by root two are not squared and not in the final volume or sorry. The final mean free path lambda F is going to be equal to the final volume VF divided by four pi root too RF squared and F. So we wanna figure out how these two are related. OK. So the first thing we know is that the volume increases from V 0 to 2 V. No, that tells us that the final volume is to be not OK. So our final volume is to be, and so lamb to F is gonna be equal to, to be not all divided by or high route two RF squared. And all right, so we figured out how to relate the volume to the initial case. What about this value of RF and NF? Well, let's think about this R is the radius of the molecule we're dealing with neon when it undergoes audio about expansion, it does not change from being neon. OK? It is still neon gas that we're dealing with. So the radius of the molecule doesn't change what that means is that R not, is going to be equal to R, they're the exact same. Now, we can do the exact same thing for N the number of molecules. Again, we're dealing with neon even through adiabatic expansion, the number of molecules that we have in our sample is not going to change. And so the value of N knot is going to be equal to the value of NF OK. So what that tells us is that our final mean free path lambda F is gonna be equal to two V knot divided by four pi route two are not squared and no, OK. Because again, R and N are the same in the initial and final cases. And now we've written our mean free path for the final case in terms of all of these initial variable. But what you'll notice is that the lamb F, as we've written it here is just equal to two multiplied by lambda knot. It's two V knot divided by four pi squared of two are not squared divided by N knot. And that's exactly what we had for lamb and nut except multiplied by two. All right. So what this means is that the final mean free path is two times the initial mean free path, which tells us that the mean free path is going to increase by a factor of two which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

A monatomic gas is adiabatically compressed to ⅛ of its initial volume. Does each of the following quantities change? If so, does it increase or decrease, and by what factor? If not, why not? b. The mean free path.

Verified Solution

Key Concepts

Mean Free Path

Adiabatic Process

Gas Density