Find the ratio 𝑓 of 𝑣 sub p over 𝑓 of 𝑣 sub rms for hydrogen gas at a temperature of 77.0 kelvin.
Use a molar mass of 2.02 grams per mole for hydrogen gas.
Okay, so in this question we’re trying to find a ratio 𝑓 of 𝑣 sub p over 𝑓 of 𝑣 sub rms.
And we’re trying to do this for hydrogen gas, which is at a temperature of 77.0 kelvin and which has a molar mass of 2.02 grams per mole.
So, what do 𝑓 of 𝑣 sub p and 𝑓 of 𝑣 sub rms actually represent?
Well, first of all, 𝑓 of 𝑣, where 𝑣 is a general velocity, represents the Maxwell–Boltzmann distribution.
And this looks something like the expression here.
Basically, it’s a distribution of the velocities of particles in an ideal gas at a given temperature.
So naturally, it depends on the masses of the particles in the gas, the temperature of the gas itself, and it gives a velocity distribution of the particles.
Now 𝑣 sub p and 𝑣 sub rms are special specific velocities.
Let’s look at 𝑣 sub p.
𝑣 sub p refers to the most probable or modal velocity.
It is basically the most commonly occurring or modal velocity of the Maxwell–Boltzmann distribution.
And 𝑣 sub rms is the root mean square velocity.
Luckily for us, there are already standard results that exist which have calculated the most probable velocity and the root mean square velocity of the Maxwell–Boltzmann distribution.
The results are as follows: 𝑣 sub p is equal to the square root of two multiplied by 𝑅, the molar gas constant, multiplied by the temperature of the gas 𝑇 divided by the molar mass 𝑀.
And 𝑣 sub rms, the root mean square velocity, is the square root of three 𝑅𝑇 over 𝑀.
Now we’re trying to find the ratio of 𝑓 of 𝑣 sub p divided by 𝑓 of 𝑣 sub rms.
So we can substitute in the values of 𝑣 sub p and 𝑣 sub rms into the Maxwell–Boltzmann distribution, starting with 𝑓 of 𝑣 sub p.
In the Maxwell–Boltzmann distribution, wherever we see a 𝑣, we replace it with 𝑣 sub p, and subsequently the expression for 𝑣 sub p.
And that’s exactly what we’ve done here.
We’ve taken this 𝑣 squared and replaced it with 𝑣 sub p squared and this 𝑣 squared and replaced it with 𝑣 sub p squared once again.
Simplifying a bit gives us the following equation.
Now let’s quickly look at this exponent here.
The exponent itself is negative lowercase 𝑚 over two 𝑘𝑇 multiplied by two 𝑅𝑇 over capital 𝑀.
Here, the lower case 𝑚 represents the mass of a hydrogen molecule, whereas the capital 𝑀 represents the molar mass of hydrogen.
However, we can simplify this whole expression here.
To do this, we need to remember what the ideal gas equation is.
The ideal gas equation tells us that the pressure exerted by a gas multiplied by the volume occupied by that gas is equal to the number of moles of gas multiplied by the molar gas constant multiplied by the temperature of the gas.
But there is another way to write the ideal gas equation.
And that is the following: the left-hand side stays the same.
We’ve got a product of 𝑝 and 𝑣.
On the right-hand side, we’ve once again got temperature, but this time we’ve got a capital 𝑁 and a 𝑘.
The capital 𝑁 represents the number of particles in that gas, not the number of moles of gas that we have, the number of particles.
And 𝑘 is the Boltzmann constant.
Now these two equations here are exactly the same; they’re equivalent.
And since we’ve got 𝑝, 𝑣 and 𝑇 on both sides of the equation in both cases, we can safely say that lowercase 𝑛 multiplied by 𝑅 must be the same as upper case 𝑁 multiplied by 𝑘.
We can rearrange this to solve for 𝑅 to give us 𝑅 is equal to uppercase 𝑁 over lowercase 𝑛 multiplied by 𝑘.
Now this fraction here, uppercase 𝑁 over lowercase 𝑛, simplifies a little bit, because that fraction is the number of particles that we have all together divided by the number of moles of particles that we have.
In other words, that fraction simplified to Avogadro’s number, the number of particles in one mole of gas and Avogadro’s number is written as capital 𝑁 subscript 𝐴.
So that’s the relationship between the molar gas constant 𝑅 and the Boltzmann constant 𝑘.
Now we can also find a relationship between the mass of one particle of gas and the molar mass of that gas because the molar mass of that gas is simply the mass of one mole of that gas.
And in one mole of gas, there is 6.02 times 10 to the power of 23 particles, where 6.02 times 10 to the power of 23 is Avogadro’s number.
So this is the mass of one particle, the lower case 𝑛.
This whole expression on the left-hand side is the mass of 6.02 times 10 to the power of 23 particles or Avogadro’s number worth of particles.
And that is the same as the molar mass capital 𝑀.
So we’ve got another relationship here.
We can substitute these two relationships into these two parentheses here.
To do this, first let’s neat things up a bit.
Now we’ve got the two expressions that we’ve just found, so let’s sub them in.
We’ll keep the first set of parentheses the same and we’ll substitute this 𝑅 with 𝑁 sub 𝐴 𝑘 and this capital 𝑀 with 𝑁 sub 𝐴 lowercase 𝑚, which means that we can see lots of things cancel.
Of course firstly, the twos cancel, then the lowercase 𝑚s cancel, the 𝑇s cancel, and the 𝑘s cancel as well.
And of course, the 𝑁 sub 𝐴s cancel.
In other words, this expression is just equal to negative one.
And so we can replace these parentheses with negative one.
And that makes life a lot easier for us.
So we found 𝑓 of 𝑣 sub p.
Let’s go about finding 𝑓 of 𝑣 sub rms.
Once again, we’ve plugged in values, this time of 𝑣 sub rms, into 𝑣 squared here and 𝑣 squared here in the Maxwell–Boltzmann distribution.
Now we can see that a similar thing will happen with these two parentheses as what we’ve done here on the right-hand side.
The only difference is that this time the three will not cancel with the two, but once again everything else will cancel.
And so we’re left with negative three over two.
And that also simplifies things a lot for us, because now we’re trying to find the ratio of 𝑓 of 𝑣 sub p over 𝑓 of 𝑣 sub rms.
So, we just need to divide one by the other.
And let’s get rid of these two equal signs and turn it into one.
Now we’ve got a fraction on both sides of the equation.
So let’s start cancelling.
Well we can see this entire parenthesis cancels with this one here.
Similarly, the four 𝜋s have to go.
The 𝑅𝑇 over 𝑀 also goes.
And so what’s left is this two here and this 𝑒 to the power of negative one and this three here and this 𝑒 to the power of negative three by two.
In other words, we’re left with two 𝑒 to the power of minus one over three 𝑒 to the power of minus three by two.
But we can simplify this even further.
𝑒 to the power of negative one over 𝑒 to the power of negative three by two is the same as 𝑒 to the power of negative one minus negative three by two.
And this results in the value of positive half as the exponent.
So the ratio of 𝑓 of 𝑣 sub p over 𝑓 of 𝑣 sub rms turns out to be two-thirds 𝑒 to the power of a half.
And this is just a number, we can evaluate this on our calculators.
It ends up being 1.099 dot dot dot so on and so forth.
Now we can notice something.
We haven’t actually used any of the values given to us in the question.
We haven’t used the fact that the temperature is 77.0 kelvin or the molar mass of hydrogen as 2.02 grams per mole.
So what gives?
Well basically this ratio is constant for all gases, all ideal gases at least.
So it doesn’t matter that we’ve got hydrogen gas.
We could have plugged in values anywhere earlier on in our calculations and worked with messy numbers instead, but this way is much simpler, because we cancelled a lot of stuff down, which meant that we got a nice little expression out at the end.
And the fact that this expression is independent of the molar mass or the temperature show us that this should be true for all ideal gases at all temperatures.
Now even though we haven’t really used any values from the question in our calculation, we should still give our answer to three significant figures because the values we’ve been given are to three significant figures.
And so our final answer is that the ratio of 𝑓 of 𝑣 sub p over 𝑓 of 𝑣 sub rms is equal to 1.10.