Big number!

[Gareth56]

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

[insane_alien]

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.

[Gareth56]

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

[insane_alien]

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

[Gareth56]

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

[insane_alien]

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

 

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

[Gareth56]

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 ?

[insane_alien]

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 ) 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 .

Edited July 21, 2008 by insane_alien
multiple post merged

[Gareth56]

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 ) 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 .

 

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.

[insane_alien]

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.

[Gareth56]

Ah those pesky old logs I was told about all those years ago

 

Many thanks