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Find the ratio ๐ of ๐ฃ sub p over ๐ of ๐ฃ sub rms for hydrogen gas at a temperature of 77.0 kelvin.
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Use a molar mass of 2.02 grams per mole for hydrogen gas.
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Okay, so in this question weโre trying to find a ratio ๐ of ๐ฃ sub p over ๐ of ๐ฃ sub rms.
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And weโre 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.
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So, what do ๐ of ๐ฃ sub p and ๐ of ๐ฃ sub rms actually represent?
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Well, first of all, ๐ of ๐ฃ, where ๐ฃ is a general velocity, represents the MaxwellโBoltzmann distribution.
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And this looks something like the expression here.
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Basically, itโs a distribution of the velocities of particles in an ideal gas at a given temperature.
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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.
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Now ๐ฃ sub p and ๐ฃ sub rms are special specific velocities.
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Letโs look at ๐ฃ sub p.
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๐ฃ sub p refers to the most probable or modal velocity.
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It is basically the most commonly occurring or modal velocity of the MaxwellโBoltzmann distribution.
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And ๐ฃ sub rms is the root mean square velocity.
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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 MaxwellโBoltzmann distribution.
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The results are as follows: ๐ฃ sub p is equal to the square root of two multiplied by ๐
, the molar gas constant, multiplied by the temperature of the gas ๐ divided by the molar mass ๐.
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And ๐ฃ sub rms, the root mean square velocity, is the square root of three ๐
๐ over ๐.
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Now weโre trying to find the ratio of ๐ of ๐ฃ sub p divided by ๐ of ๐ฃ sub rms.
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So we can substitute in the values of ๐ฃ sub p and ๐ฃ sub rms into the MaxwellโBoltzmann distribution, starting with ๐ of ๐ฃ sub p.
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In the MaxwellโBoltzmann distribution, wherever we see a ๐ฃ, we replace it with ๐ฃ sub p, and subsequently the expression for ๐ฃ sub p.
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And thatโs exactly what weโve done here.
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Weโve taken this ๐ฃ squared and replaced it with ๐ฃ sub p squared and this ๐ฃ squared and replaced it with ๐ฃ sub p squared once again.
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Simplifying a bit gives us the following equation.
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Now letโs quickly look at this exponent here.
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The exponent itself is negative lowercase ๐ over two ๐๐ multiplied by two ๐
๐ over capital ๐.
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Here, the lower case ๐ represents the mass of a hydrogen molecule, whereas the capital ๐ represents the molar mass of hydrogen.
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However, we can simplify this whole expression here.
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To do this, we need to remember what the ideal gas equation is.
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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.
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But there is another way to write the ideal gas equation.
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And that is the following: the left-hand side stays the same.
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Weโve got a product of ๐ and ๐ฃ.
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On the right-hand side, weโve once again got temperature, but this time weโve got a capital ๐ and a ๐.
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The capital ๐ represents the number of particles in that gas, not the number of moles of gas that we have, the number of particles.
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And ๐ is the Boltzmann constant.
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Now these two equations here are exactly the same; theyโre equivalent.
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And since weโve got ๐, ๐ฃ and ๐ on both sides of the equation in both cases, we can safely say that lowercase ๐ multiplied by ๐
must be the same as upper case ๐ multiplied by ๐.
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We can rearrange this to solve for ๐
to give us ๐
is equal to uppercase ๐ over lowercase ๐ multiplied by ๐.
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Now this fraction here, uppercase ๐ over lowercase ๐, 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.
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In other words, that fraction simplified to Avogadroโs number, the number of particles in one mole of gas and Avogadroโs number is written as capital ๐ subscript ๐ด.
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So thatโs the relationship between the molar gas constant ๐
and the Boltzmann constant ๐.
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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.
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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 Avogadroโs number.
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So this is the mass of one particle, the lower case ๐.
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This whole expression on the left-hand side is the mass of 6.02 times 10 to the power of 23 particles or Avogadroโs number worth of particles.
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And that is the same as the molar mass capital ๐.
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So weโve got another relationship here.
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We can substitute these two relationships into these two parentheses here.
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To do this, first letโs neat things up a bit.
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Now weโve got the two expressions that weโve just found, so letโs sub them in.
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Weโll keep the first set of parentheses the same and weโll substitute this ๐
with ๐ sub ๐ด ๐ and this capital ๐ with ๐ sub ๐ด lowercase ๐, which means that we can see lots of things cancel.
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Of course firstly, the twos cancel, then the lowercase ๐s cancel, the ๐s cancel, and the ๐s cancel as well.
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And of course, the ๐ sub ๐ดs cancel.
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In other words, this expression is just equal to negative one.
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And so we can replace these parentheses with negative one.
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And that makes life a lot easier for us.
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So we found ๐ of ๐ฃ sub p.
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Letโs go about finding ๐ of ๐ฃ sub rms.
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Once again, weโve plugged in values, this time of ๐ฃ sub rms, into ๐ฃ squared here and ๐ฃ squared here in the MaxwellโBoltzmann distribution.
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Now we can see that a similar thing will happen with these two parentheses as what weโve done here on the right-hand side.
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The only difference is that this time the three will not cancel with the two, but once again everything else will cancel.
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And so weโre left with negative three over two.
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And that also simplifies things a lot for us, because now weโre trying to find the ratio of ๐ of ๐ฃ sub p over ๐ of ๐ฃ sub rms.
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So, we just need to divide one by the other.
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And letโs get rid of these two equal signs and turn it into one.
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Now weโve got a fraction on both sides of the equation.
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So letโs start cancelling.
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Well we can see this entire parenthesis cancels with this one here.
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Similarly, the four ๐s have to go.
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The ๐
๐ over ๐ also goes.
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And so whatโs left is this two here and this ๐ to the power of negative one and this three here and this ๐ to the power of negative three by two.
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In other words, weโre left with two ๐ to the power of minus one over three ๐ to the power of minus three by two.
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But we can simplify this even further.
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๐ to the power of negative one over ๐ to the power of negative three by two is the same as ๐ to the power of negative one minus negative three by two.
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And this results in the value of positive half as the exponent.
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So the ratio of ๐ of ๐ฃ sub p over ๐ of ๐ฃ sub rms turns out to be two-thirds ๐ to the power of a half.
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And this is just a number, we can evaluate this on our calculators.
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It ends up being 1.099 dot dot dot so on and so forth.
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Now we can notice something.
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We havenโt actually used any of the values given to us in the question.
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We havenโt used the fact that the temperature is 77.0 kelvin or the molar mass of hydrogen as 2.02 grams per mole.
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So what gives?
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Well basically this ratio is constant for all gases, all ideal gases at least.
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So it doesnโt matter that weโve got hydrogen gas.
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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.
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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.
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Now even though we havenโt really used any values from the question in our calculation, we should still give our answer to three significant figures because the values weโve been given are to three significant figures.
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And so our final answer is that the ratio of ๐ of ๐ฃ sub p over ๐ of ๐ฃ sub rms is equal to 1.10.