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OCR B advancing physics synoptic paper

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I think that for some more rough working you can assume that 3/2kT is roughly equal to kT as you are only looking for the order of magnitude.

anyone help with the range of values for the boltzmann factor? I need some explanation! Please?!

Reply 61

sas88

the energy of a particle is only a rough estimation, and is very rough, the book says its basically the principle that we are learning opposed to accuracy, both are acceptable and on marck schemes in additional guidance it normally gives the alternative answer as being okay, e.g if used KT in mark scheme it says 3/2 KT gives blah blah

3/2 is because there are three dimensions in which a particle can travel, but it gets more complicated and can sometimes be as much as like 6/2KT when u take into account like rotational spin on molecules made of more than one atom etc..

So basically either is fine! - (in reality gas is around 3/2KT and matter is a multiple of KT)

Reply 62

Vipul Solanki

sas88

in response to that, just wondered if ne1 wanted to derive KE from the gas laws? i.e how is 3/2KT derrived?

Cheeerssss

sas

PV = nRT
PV = 1/3 Nmc^2

nRT = 1/3 Nmc^2

times both side by 3/2

3/2nRT = 1/2Nmc^2 (Therefore kinectic energy = 3/2 nRT (for number of moles)

You know that R = Avagadro Constant (Na) x k
and

N = n x Avagadagro Constant (Na)

1/2mc^2 = (3/2)nRT / N (1)

you know that N/n = Na

therefore n/N = 1/Na (sub this in equation (1)

you get (3/2)RT / Na (2)

you know that R/Na = k (sub this in (2) )

therefore 3/2 kT

hence kinectic energy = 3/2 kT (1 molecule and 3/2 NkT (for number of molecules)

:biggrin:

PS Small n = number of moles
and N = number of molecules

Reply 63

Sephrenia

datr

I have another five exams after FP2 also...

ick...only 1 after that thank god...doing the maths AEA...

and Motoroller - I guess youre not on Edexcel maths then? and youre on a par with my boyfriend - he had 18 exams..hes done most of them now...hes had 4 exams int he past 2 days...

Reply 64

sas88

Nice one cheers great help

Reply 65

edp

How do i calculate the maximum theoretical value for the efficiency using the equation (line 34)

Reply 66

Slinc

Depends what values you get given.
If you used 293K and 373K then:
Max theoretical efficiency = 1 - (293/373) = 1 - 0.79 = 0.21 = 21%.

right cool cheers

Boltzmann factor anyone? Sorry I know this is annoying, but there isn't much online or in the textbook so I need some help! What are the ranges for -E/kT? Or is it for e^(-E/kT)?

Reply 69

Will11

i dnt know abar ranges for -e/kT, but e^(-E/kT) is the factor by which the number of molecules at the next energy has. e.g. if e^(-E/kT) = 0.5, and the groud energy level has 20 molecules, the next energy level will have 10, the next 5, etc... dnt no if tht helps but its all i know about it. sorry!

Reply 70

ed84c

Boltzmann is a ratio, so 0 and 1. If thats what you were asking? Boltzmann is the exponential you gave.

Reply 71

motoroller

no, for the factor E/kT

Reply 72

motoroller

OK, leaving for the exam now. Good luck!

Reply 73

Sephrenia

How did everyone find it?

Reply 74

motoroller

How was it? I thought that was a hard paper! It was so long!

Reply 75

mrsmann

Yeah but this has 4 pages of discussion on the pre-release whereas my thread is brand new specific for the exam! http://www.thestudentroom.co.uk/showthread.php?t=413995

Reply 76

motoroller

That space question with the aluminium string and the EMF was hard! I couldn't work out the answer!

Reply 77

cornishbrooksy

Hey,
thought the paper was ok, nothing un-expected, just a really odd set-out and alot of explain questions!!but also a good number of show that,think it went ok tho!