Question

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance of 1.1nm. How much work would be required to increase the separation of the two ions to an infinite distance?

Answer 1

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$, where K is the Coulomb´s constant ($9x10^{9} N\frac{m^{2} }{C^{2} }$), $q_{1}$= Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$.

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$

Answer 2

The work that would be required to increase the separation of the two ions to an infinite distance is $\rm 2.1 \times 10^-^1^9\;J$.

What is work?

Work is defined as the equivalency of the application of force over a distance.

Work = Fs

Force can be calculated by the formula

$F = k \dfrac{q^1\times q^2}{ r^2}$

The total word done from the initial state to the final state is

$W = W_f - W_i = k \dfrac{q^N^a^+\times q^C^l^-}{ r^2f} -k \dfrac{q^N^a^+\times q(^C^l^-)^r^i}{ r^2i} \\\\\\= kq_(_N_a_+ _) q_(_C_l_-_) \dfrac{1}{r_f} - \dfrac{1}{r_i}$

$r_i = 1.1 \;nm = 1.1 \times 10^-^9 m \;and \;r_f = inifnite \;distance\\\\\\ W = (9\times 10^-^9) (1.6 \times 10 ^-^1^9) (-1.6 \times 10^1^9)[ \dfrac{1}{\alpha} -\dfrac{1}{1.1 \times 10^-^9} ] = 2.1 \times 10^-^1^9$

Thus, the work done is $\rm 2.1 \times 10^-^1^9\;J$

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