Chapter 24, Problem 1P

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro’s number of electrons from an initial point where the electric potential is 9.00 V to a point where the electric potential is −5.00 V? (The potential in each case is measured relative to a common reference point.)

To determine

The work done by the battery, generator, or some other source in moving Avogadro’s number of electron from an initial point to final point.

Answer to Problem 1P

The work done by the battery, generator, or some other source in moving Avogadro’s number of electron from an initial point to final point is $1.35\text{MJ}$.

Explanation of Solution

Given information: Electric potential at initial point is $9.00\text{V}$ and the electric potential at final point is $-5.00\text{V}$, number of electron is equal to Avogadro’s number.

The expression for the work done is given as,

$W=\Delta V\times Q$ (1)

The value of potential difference is calculated as,

$\Delta V=V_f-V_i$

Substitute $-5.00\text{V}$ for $V_f$ and $9.00\text{V}$ for $V_i$ in the above equation.

$\begin{array}{c}\Delta V=V_f-V_i\\ =-5.00\text{V}-9.00\text{V}\\ =-14.00\text{V}\end{array}$

The expression for the total charge is given as,

$Q=ne$

Here,

$Q$ is total charge.

$n$ is number of electrons.

$e$ is the change of 1 electron.

Substitute $6.022\times {10}^{23}$ for $n$ and $-1.6\times {10}^{-19}\text{C}$ in the above equation.

$\begin{array}{c}Q=6.022\times {10}^{23}\times -1.6\times {10}^{-19}\text{C}\\ \text{=}-\text{96352}\text{C}\end{array}$

Substitute $-96352\text{C}$ for $Q$ and $-14.00\text{V}$ for $\Delta V$ in the above equation.

$\begin{array}{c}W=-96352\text{C}\times \left(-14.00\text{V}\right)\\ =1348928\text{J}\\ =1.35\text{MJ}\end{array}$

Conclusion:

Therefore, the work done by the battery, generator, or some other source in moving Avogadro’s number of electron from an initial point to final point is $1.35\text{MJ}$.