GREPhysics.NET
GR | #
<< Previous | Next >>
Login | Register
   
  GR0177 #48
Problem
GREPhysics.NET Official Solution    Alternate Solutions
This problem is still being typed.
Thermodynamics}RMS Speed


In case one forgets the RMS speed, one does not need to go through the formalism of deriving it with the Maxwell-Boltzmann distribution. Instead, one can approximate its dependence on mass and temperature by 3/2 kT = 1/2 m v^2. One thus has v \propto \sqrt{kT/m}. For the ratio of velocities, one has \frac{v_N}{v_O}=\frac{m_O}{m_N}. Plug in the given molecular masses for Oxygen and Nitrogen to get choice (C).

Incidentally, the trick for memorizing the diatomic gasses is Have No Fear Of Ice Cold Beer (Hydrogen, Nitrogen, Florine, Oxygen, Iodine, Chlorine, Bromine).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
asdfasdfasdf
2016-09-06 17:39:31		Forgetting anything about v_{RMS} I assumed that since it\'s at constant temperature, the oxygen and nitrogen should have the same kinetic energy:\r\n\\frac{1}{2}mv^2_O = \\frac{1}{2}mv^2_N\r\n\r\n\\Rightarrow 32v^2_O = 28v^2_N\r\n\r\n\\Rightarrow \\frac{v^2_O}{v^2_N} = \\frac{8}{7}\r\n\r\n\\Rightarrow \\frac{v_O}{v_N} = \\sqrt{\\frac{8}{7}}\r\n\r\n Is it just coincidental that I got the right answer for this? Or does it matter that we\'re dealing with the ratio of the v_{RMS}? Or that they\'re both diatomic?
asdfasdfasdf
2016-09-06 17:43:40		 Hm, I don\'t know latex, but when I previewed my post, it looked good, I\'m not sure what happened. Here\'s my non-latex answer:\r\n\r\n(1/2)(m)(v_O)^2 = (1/2)(m)(v_N)^2\r\n\r\n==> (32)(v_O)^2 = (28)(v_N)^2\r\n\r\n==> (v_N)^2 / (v_O)^2 = (32)/(28) = (8)/(7)\r\n\r\n==> (v_N) / (v_O) = sqrt(8/7)
Alternate Solution - Unverified
uhurulol
2014-10-20 17:39:18		This can be solved very simply by just knowing the V_{rms} formula,

V_{rms} = \sqrt{\frac{3RT}{M_m}

Manipulate this algebraically to find the ratio by plugging in 28 and 32 to the respective M_m slots (remember that we're finding a ratio, so the Mass units don't really matter so long as they are consistent throughout the equation,

\frac{\sqrt{\frac{3RT}{28}}}{\sqrt{\frac{3RT}{32}}} = \sqrt{\frac{\frac{3RT}{28}}{\frac{3RT}{32}}} = \sqrt{\frac{3RT * 32}{3RT * 28} = \sqrt{\frac{8}{7}}

As in choice (C). Hope this simple solution helps.		Alternate Solution - Unverified
Izaac
2012-08-30 01:06:02		 My way:

- They talk about a ratio. Log of a ratio is actually a difference: eliminate (E).

- They talk about a ROOT mean square speed, so as a hint let's consider (B) and (C) (besides those are almost the same, as if ETS was trying to trick you out of confusion).

- 5 seconds to get that 7/8 is actually \frac{m_{N_2}}{m_{O_2}}. We're looking for \frac{v_{N_2}}{v_{O_2}}. But speed proportional to mass sounds weird, so let's take the ratio inverse. That's (C).

As with all answers of the kind, it's not waterproof, but not so rarely it leads you to the right answer.		Alternate Solution - Unverified
Comments
asdfasdfasdf
2016-09-06 17:39:31		Forgetting anything about v_{RMS} I assumed that since it\'s at constant temperature, the oxygen and nitrogen should have the same kinetic energy:\r\n\\frac{1}{2}mv^2_O = \\frac{1}{2}mv^2_N\r\n\r\n\\Rightarrow 32v^2_O = 28v^2_N\r\n\r\n\\Rightarrow \\frac{v^2_O}{v^2_N} = \\frac{8}{7}\r\n\r\n\\Rightarrow \\frac{v_O}{v_N} = \\sqrt{\\frac{8}{7}}\r\n\r\n Is it just coincidental that I got the right answer for this? Or does it matter that we\'re dealing with the ratio of the v_{RMS}? Or that they\'re both diatomic?
asdfasdfasdf
2016-09-06 17:43:40		 Hm, I don\'t know latex, but when I previewed my post, it looked good, I\'m not sure what happened. Here\'s my non-latex answer:\r\n\r\n(1/2)(m)(v_O)^2 = (1/2)(m)(v_N)^2\r\n\r\n==> (32)(v_O)^2 = (28)(v_N)^2\r\n\r\n==> (v_N)^2 / (v_O)^2 = (32)/(28) = (8)/(7)\r\n\r\n==> (v_N) / (v_O) = sqrt(8/7)
Alternate Solution - Unverified
uhurulol
2014-10-20 17:39:18		This can be solved very simply by just knowing the V_{rms} formula,

V_{rms} = \sqrt{\frac{3RT}{M_m}

Manipulate this algebraically to find the ratio by plugging in 28 and 32 to the respective M_m slots (remember that we're finding a ratio, so the Mass units don't really matter so long as they are consistent throughout the equation,

\frac{\sqrt{\frac{3RT}{28}}}{\sqrt{\frac{3RT}{32}}} = \sqrt{\frac{\frac{3RT}{28}}{\frac{3RT}{32}}} = \sqrt{\frac{3RT * 32}{3RT * 28} = \sqrt{\frac{8}{7}}

As in choice (C). Hope this simple solution helps.		Alternate Solution - Unverified
Izaac
2012-08-30 01:06:02		 My way:

- They talk about a ratio. Log of a ratio is actually a difference: eliminate (E).

- They talk about a ROOT mean square speed, so as a hint let's consider (B) and (C) (besides those are almost the same, as if ETS was trying to trick you out of confusion).

- 5 seconds to get that 7/8 is actually \frac{m_{N_2}}{m_{O_2}}. We're looking for \frac{v_{N_2}}{v_{O_2}}. But speed proportional to mass sounds weird, so let's take the ratio inverse. That's (C).

As with all answers of the kind, it's not waterproof, but not so rarely it leads you to the right answer.		Alternate Solution - Unverified
mpdude8
2012-04-20 21:23:52		Look at things like momentum, kinetic energy, etc., mass and speed are always on the same side of the equation (i.e., inversely related). So, the more massive thing should have the slower speed, and this ratio should be greater than 1.

The fact that it's the root mean square told me there's probably square roots involved, but I had it narrowed down to two choices anyway.		NEC
Albert
2009-10-28 05:21:26		Neat trick Yosun! Have No Fear Of Ice Cold Beer, loved it :)NEC
theFireman
2009-09-11 21:50:00		Also simply graham's law of effusion:

v_rms_A/v_rms_B =Sqrt(mass_B/mass_A)		NEC
wangjj0120
2008-11-06 20:33:25		Wooops.
i am sorry.
I mean, Kinetic energy includs translation, rotation and vibration kinetic energy. Total energy is equal to kinetic energy and potential energy. and temperature only involves translation energy.
NEC
lattes
2008-10-15 12:47:56		instead of \frac{v_{N}}{v_{O}}=\frac{m_{O}}{m_{N} it should be \frac{v_{N}}{v_{O}}=\sqrt{\frac{m_{O}}{m_{N}}.Typo Alert!
Ning Bao
2008-02-01 06:50:39		Smaller things go faster, but not that much fast ->C.

I love non-math physics.
a19grey2
2008-11-05 22:47:51		 Wow, good catch. I was stupid enough to solve the problem before looking at the answers. The best advice I can give everyone is to always look at the answers before trying to do the problem.

This lets you check limits and units before wasting time. Also, it lets you know how accurate you will need to be in your calculations/approximations to narrow down the choices.
f4hy
2009-11-06 23:24:58		 How does this help decide between 1.069 vs 1.3. Both to me are not that much faster.
NEC
Mexicana
2007-09-28 17:56:58		Just a small correction. The kinetic energy of O_2 and N_2 cannot really be approximated as \frac{3}{2}kT as Yosun is stating... In fact for a diatomic molecule the number of degrees of freedom is 5 and so, the energy of both O_2 and N_2 is \frac{5}{2}kT. However, this doesnt make a difference in the answer as we are asked for a ratio of rms velocities and both molecules are diatomic. This point would make a difference in the answer if for example we were dealing with a mixture of diatomic with monotic gases.
Richard
2007-10-25 21:45:17		 Thanks for pointing that out...I was about to.

Diatomic molecules have three translational degrees of freedom and two rotational degrees of freedom:
5/2kT
hayim
2008-11-05 07:20:06		 note that at higher temperatures a diatomic molecule actually has 7 degrees of freedom contributing to its energy, the additional 2 being vibrational kinetic and vibrational potential of the internuclear bond.

i guess an 'ideal' diatomic gas ignores these 2 since they are not significant at low temperatures (ie the molecule more approximates a rigid dumbell), but at high temps these will come into play and so Cv=\frac{7}{2}kT
wangjj0120
2008-11-06 20:22:45		 Mexicana:
your comment is not exactly correct. 5KT/2 is internal energy, which is equal to kinetic energy(3KT/2) plus potential energy(2KT/2 for rotation energy here). but temperature, (or kinetic temperature) only involves translational degree of freedom, that is kinetic energy here.
wangjj0120
2008-11-06 20:34:09		 Wooops.
i am sorry.
I mean, Kinetic energy includs translation, rotation and vibration kinetic energy. Total energy is equal to kinetic energy and potential energy. and temperature only involves translation energy.
jmason86
2009-07-14 22:12:39		 Word to Wang.. I was about to comment the same.
NEC

Post A Comment!
You are replying to:
This can be solved very simply by just knowing the V_{rms} formula, V_{rms} = \sqrt{\frac{3RT}{M_m} Manipulate this algebraically to find the ratio by plugging in 28 and 32 to the respective M_m slots (remember that we're finding a ratio, so the Mass units don't really matter so long as they are consistent throughout the equation, \frac{\sqrt{\frac{3RT}{28}}}{\sqrt{\frac{3RT}{32}}} = \sqrt{\frac{\frac{3RT}{28}}{\frac{3RT}{32}}} = \sqrt{\frac{3RT * 32}{3RT * 28} = \sqrt{\frac{8}{7}} As in choice (C). Hope this simple solution helps.

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

© 2005-2007 by Yosun Chang, All rights reserved.

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.