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Posted

How doI calculate this on a calculator?

 

ln2^10^22

 

this is the natural log of 2 raised to the power of 10 to the 22.

 

Ta

Posted

well, on a pocket calculator the short answer is, you don't. it will return '0' if it doesn't give you an error.

 

on a computer you may be able to find a program that will give you an answer.

Posted
well, on a pocket calculator the short answer is, you don't. it will return '0' if it doesn't give you an error.

 

on a computer you may be able to find a program that will give you an answer.

 

 

I see. I ask because in my book Chemistry in Context it states the following calculation fo entropy:-

 

S = k lnW where W is the number above and k is Boltzman's constant and the answer given is 0.096J/K

Posted

well then where dit the first calculation come from?

 

boltzman constant is 1.4*10^-23 J/K. a much more manageable number

 

this gives S = ln(2) * 1.4*10^-23 which is 9.7*10-24 J/K

Posted
well then where dit the first calculation come from?

 

boltzman constant is 1.4*10^-23 J/K. a much more manageable number

 

this gives S = ln(2) * 1.4*10^-23 which is 9.7*10-24 J/K

 

Which is several ordersof magnitude smaller then 0.096J/K Could the 10^22 come into it somewhere.

 

The whole calculation is S = 1.4 x 10^-23J/K x ln2^10^22 = 0.096J/K

Posted
that would give 2.135*10^-21 J/K

 

i misread the opening OP. use of brackets would make this a lot more readable.

 

What would give 2.135*10^-21 J/K ?

Posted (edited)
The whole calculation is S = 1.4 x 10^-23J/K x ln2^10^22

 

that would. assuming the last term is ln(2^10^22). wait, n/m my calculator is messing up on it. gave me three different values.

 

okay, one quick hand calculation(done in crayon on my wall :D) shows its 0.097J/K This is why machines will never trump humans, the inability to do entropy calculations with a quick scribble on the wall :P.

Edited by insane_alien
multiple post merged
Posted
that would. assuming the last term is ln(2^10^22). wait, n/m my calculator is messing up on it. gave me three different values.

 

okay, one quick hand calculation(done in crayon on my wall :D) shows its 0.097J/K This is why machines will never trump humans, the inability to do entropy calculations with a quick scribble on the wall :P.

 

So ow do you do it. :)

 

Baring in mind that ln2^10^22 is the natural log of 2 raised to the power of 10 to the 22.

Posted

well, there are certain little tricks you can use when dealing with logarithms that really cuts down on the size of 2^10^22.

 

such as ln(x^y) =y*ln(x) where we can simplify to (10^22)*ln(2)

 

then multiply by 1.4*10^-23 so we get 1.4*10^(-1) * ln(2) which equals 0.14*ln(2) plug that into a calculator and you have your answer.

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