How much work is done in moving a mole of electrons around a circuit

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To calculate the work done in moving a mole of electrons around a circuit with a 6V voltage, the formula used is W = QV, where Q is the total charge. The charge of one electron is -1.6 x 10^-19 coulombs, and one mole of electrons contains approximately 6.022 x 10^23 electrons. Therefore, the total charge for one mole is calculated by multiplying the charge of one electron by the number of electrons in a mole. The resulting work done is W = 6V * (1.6 x 10^-19 C * 6.022 x 10^23), which equals approximately 5.78 x 10^5 joules. This calculation clarifies the relationship between voltage, charge, and work in electrical circuits.
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i know this is basic suff, but forget it, can u please tell me how to do this question



how much work is done in moving a mole of electrons around a circut using a 6v voltage( 1 mole of electrons is 6.023 x 10 power 23 electrons


thanks

answer is w = qv = 5.78 x 10power5j
 
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Do you know the charge on an electron? To find the total charge, q, multiply the charge of one electron by the number of electrons...
 
yes the charge of an eleectron is -1.6 x10power-19

sorry don't understand,

v =w/q

w=vq

v is potential difference in volts
w is work done by field in joules
q is charge being ppushed in coulombs


so w= 6 x q( which i thought was the mole of electrons, which is 6.023 x 10power23 right?)
 
If you know q is charge, as measured in coulombs, why would you think it was 6.022 x 10^23, which is the number of electrons?

If you had one electron, the work done would be, W = QV = (1.6 x 10^-19)*6..
 
k, got confused there, i hate this topic, i like the space one!

i see now:) so


w = 6* ( 1.6 x 10^-19 x number of electrons (6.023 x 10^23)
 
Yes, that's correct :smile:.
 
thanks, i have been lazey thesee holidays:P

seen these forums once before and when i formated my computer i lost the address, just found it again, i quess this is British forum(the mother land!)
 
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