Ch 12.3 help
Page 1 of 1 • Share
Ch 12.3 help
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 rootmeansquare (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:
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 rootmeansquare (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:
periwinkle Posts : 22
Join date : 20080917
question 2
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
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
guest17 Guest
question 4
temperature is proportional to the average kinetic energy so:
347.7/347.7 = 1
ans is 1
347.7/347.7 = 1
ans is 1
guest17 Guest
question 1
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
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
guest17 Guest
question 3
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
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
guest 17 Guest
Re: Ch 12.3 help
aren't we suppose to do something with the moles that was given and where did 1.38e^23 come from?
for ques Guest
Re: Ch 12.3 help
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...
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...
guestach Guest
Re: Ch 12.3 help
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.
Guest01 Posts : 133
Join date : 20080919
Re: Ch 12.3 help
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.
MAGguest Guest
Re: Ch 12.3 help
i keep getting # 2 wrong. i did (3/2)KT ..my answer was 6.43456 but it did not work.
hello Guest
Re: Ch 12.3 help
My answer was to four decimal places and i used EXX instead of all those zeroes
Guest01 Posts : 133
Join date : 20080919
Re: Ch 12.3 help
If you use E then don't use the carrot (^). That may be messing up the answer.
Guest01 Posts : 133
Join date : 20080919
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

