# help! a2 thermal physicsWatch

#1
hi i need help with this question as i keep getting part b wrong

1 mole of hydrogen has a mass of 2.0 x 10(to the power of -3) kg.
At 0 centigrade and one atmosphere pressure, 1.0 x 10(to the power of 5)

it occupies a volume of 0.022 cubic metre

Calculate

a)r.m.s speed of a hydrogen molecule.
i got 1.8x10(to the power of 3) m/s

b) the kinetic energy of one mole of the gas.

rep+ up for grabs

Last edited by Remarqable M; 8 years ago
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8 years ago
#2
Guess you've tried and got 3240 J/mol right? What's the answer book reckon?
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8 years ago
#3
Guess you've tried and got 3240 J/mol right? What's the answer book reckon?
im thinking that too, but if not, the only alternative i could think of using would be "mean translational kinetic energy" which may be completely wrong....imo.

3/2 x (boltzmann constant) x (273) = 5.7 x 10^-21 J

i have no idea tbh.
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8 years ago
#4
(Original post by Remarqable M)
hi i need help with this question as i keep getting part b wrong

1 mole of hydrogen has a mass of 2.0 x 10(to the power of -3) kg.
At 0 centigrade and one atmosphere pressure, 1.0 x 10(to the power of 5)

it occupies a volume of 0.022 cubic metre

Calculate

a)r.m.s speed of a hydrogen molecule.
i got 1.8x10(to the power of 3) m/s

b) the kinetic energy of one mole of the gas.

rep+ up for grabs

Hey!
use formula: pressure= 1.3 (density) <c>
yes its straight forward: use pv=nrt where temperature = 273.15 in kevlin and sub in.
we know k.e = 1/2mc^2 = 3/2kt
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#5
(Original post by ibysaiyan)
Hey!
use formula: pressure= 1.3 (density) <c>
yes its straight forward: use pv=nrt where temperature = 273.15 in kevlin and sub in.
we know k.e = 1/2mc^2 = 3/2kt
thanks bro! i've tried that and got the righ answer!
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quote
8 years ago
#6
im thinking that too, but if not, the only alternative i could think of using would be "mean translational kinetic energy" which may be completely wrong....imo.

3/2 x (boltzmann constant) x (273) = 5.7 x 10^-21 J

i have no idea tbh.
Mean translational KE is given by the formula I suggested earlier... If you use 3/2 x (boltzmann constant) x (273) then you get the KE per molecule, gotta multiply by avagadro's number (as there is one mole of hydrogen molecules) to get the total KE

Would then get 3402 J/mol - pretty much same ballpark as the other formula. Gotta be one of these right?! Also ibysaiyan reckons either of these are the right formula, so if the answer book says otherwise im pretty stumped!
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8 years ago
#7
(Original post by ibysaiyan)
Hey!
use formula: pressure= 1.3 (density) <c>
yes its straight forward: use pv=nrt where temperature = 273.15 in kevlin and sub in.
we know k.e = 1/2mc^2 = 3/2kt
What the hell are you doing with 'c' there? You could lose marks if you use c instead of v, if nothing else because it is confusing. You have equated the thermal energy to half the rest energy.
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8 years ago
#8
What the hell are you doing with 'c' there? You could lose marks if you use c instead of v, if nothing else because it is confusing. You have equated the thermal energy to half the rest energy.
oh yea not the speed of light in vacuum , sorry about that.
Edit: strangely on the work of scheme there is a 'c' .
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8 years ago
#9
(Original post by Remarqable M)
thanks bro! i've tried that and got the righ answer!
hehe np bud anytime
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