Ch 12.3 help

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Ch 12.3 help

Post  periwinkle on Tue Nov 18, 2008 12:05 am

Question 1
A container holds 10.89 grams Oxygen molecules at 350.3 degrees K. What is the internal (translational kinetic) energy of the gas (see sheet 15,23 and consider that the Oxygen molecule has 2 atoms)? Indicate with a positive (negative) sign whether the internal energy is less (greater) than the average kinetic energy of the molecules.

Answer: -1485.970
How to solve:

Question 2
What is the average (translational) kinetic energy of the atoms contained in 1.895 moles of helium at 29.52 degrees C (see sheet 15,24) Indicate with a negative (positive) sign wheather the universal Boltzmann constant k is (is not) the ratio of the universal gas constant R and Avogadro's universal number Na?

Answer: -6.26526
How to solve:

Question 3
Calculate the root-mean-square (rms) velocity of Oxygen molecules at 65.95 degrees C (see sheet 24). Indicate with a negative (positive) sign whether all molecules have this velocity (this value is an average for the molecules present).

Answer: 513.9102
How to solve:

Question 4
What is the ratio of the average kinetic energy of Helium atoms over the average kinetic energy of Neon atoms both at 347.7 oK (see sheet 24) ? Indicate with a positive (negative) sign whether the corresponding ratio of the rms velocities is the same (not the same).

Answer: -1
How to solve:

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question 2

Post  guest17 on Thu Nov 20, 2008 4:35 pm

average kinetic energy = (3/2)kT

k= 1.38e^-23
T= 29.52+273.15 = 302.67

avg KE = (3/2)*1.38e^-23*302.67
= 6.265269e^-21

ans: -6.265269e^-21

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question 4

Post  guest17 on Thu Nov 20, 2008 7:21 pm

temperature is proportional to the average kinetic energy so:

347.7/347.7 = 1

ans is -1

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question 1

Post  guest17 on Thu Nov 20, 2008 7:42 pm

formula U=(3/2)nRT

first convert the grams given to moles

grams of substance/moles of substance =molar mass of substance/1 mole

10.89g/x=16*2/1

n= 10.89/32= .3403125

U= (3/2)*.3403125*8.31*350.3

U= -1485.971

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question 3

Post  guest 17 on Thu Nov 20, 2008 8:39 pm

v_rms= sqrt((2*KE_avg)/m)

KE_avg = (3/2)KT

T= 65.95 +273.15= 339.1

m= (16*2)*1.66e^-27

v_rms= sqrt((2*(3/2)*1.38e^-23*339.1)/(32*1.66e^-27))
=514

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Re: Ch 12.3 help

Post  question on Fri Nov 21, 2008 1:51 am

for #2 how do you write the answer on maple TA?

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Re: Ch 12.3 help

Post  for ques on Fri Nov 21, 2008 2:04 am

aren't we suppose to do something with the moles that was given and where did 1.38e^-23 come from?

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Re: Ch 12.3 help

Post  guestach on Fri Nov 21, 2008 7:29 pm

yo yo yo for number two i did the exact same formula but they keeps saying im wrong so whats with that... anyone have similar problem?
I did
(3/2)kT

Where k is boltzmanns constant, T is temp in kelvins and (3/2) is a number. Help me out please... and ill post up a really funny joke or something...

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Re: Ch 12.3 help

Post  Guest01 on Sat Nov 22, 2008 6:03 am

I did #2 as per above and got it correct. How did you proceed with the question? Try first finding (3/2) and multiplying that answer across.

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Re: Ch 12.3 help

Post  MAGguest on Sat Nov 22, 2008 8:15 pm

You might be getting #2 wrong because of how you are entering the answer into Maple TA. For example, instead of putting in "-6.265269e^-21" from the post above, try writing out the whole number... ".000000000000000000006265269" along with the negative. The formula "(3/2)*k*T" is correct.

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Re: Ch 12.3 help

Post  hello on Sat Nov 22, 2008 11:09 pm

i keep getting # 2 wrong. i did (3/2)KT ..my answer was -6.43456 but it did not work.

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Re: Ch 12.3 help

Post  Guest01 on Sun Nov 23, 2008 6:32 am

My answer was to four decimal places and i used E-XX instead of all those zeroes

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Re: Ch 12.3 help

Post  Guest_ on Sun Nov 23, 2008 11:13 pm

instead of "answer"e^-21 i put in answer*10^-21. hope that helps

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Re: Ch 12.3 help

Post  Guest01 on Mon Nov 24, 2008 12:01 am

If you use E then don't use the carrot (^). That may be messing up the answer.

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Re: Ch 12.3 help

Post  Guest 87 on Mon Nov 24, 2008 11:51 am

Can anyone please, explain why in question 4 the answer is negative

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Re: Ch 12.3 help

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