Question

How many moles of electrons pass through a circuit containing a \(100 \mathrm{V}\) battery and a \(2 \Omega\) resistor over a period of 10 seconds? (Note: \(\mathrm{F}=9.65 \times 10^{4} \frac{\mathrm{C}}{\mathrm{mol} e^{-}}\)) a. \(5.18 \times 10^{-3}\) moles b. 500 moles c. \(5.18 \times 10^{3}\) moles d. \(5.2 \times 10^{6}\) moles

Short answer

The number of moles of electrons is \(5.18 \times 10^{-3}\).

Step-by-step solution

Calculate the Current

Use Ohm's Law to find the current. Ohm's Law is given by \[ I = \frac{V}{R} \]Where:- \(V\) is the voltage (100 V)- \(R\) is the resistance (2 Ω)So, \[ I = \frac{100 \text{ V}}{2 \text{ Ω}} = 50 \text{ A} \]

Find the Total Charge

The current is the rate of flow of charge. To find the total charge that flows in 10 seconds, use \[ Q = I \times t \]Where:- \(Q\) is the charge in coulombs- \(I\) is the current (50 A)- \(t\) is the time in seconds (10 s)So, \[ Q = 50 \text{ A} \times 10 \text{ s} = 500 \text{ C} \]

Calculate Moles of Electrons

Use Faraday's constant to convert charge to moles of electrons. Faraday's constant \(F\) is given by \[ F = 9.65 \times 10^{4} \frac{\text{C}}{\text{mol} \ e^{-}} \]Thus, the moles of electrons is \[ \text{Moles of electrons} = \frac{Q}{F} = \frac{500 \text{ C}}{9.65 \times 10^{4} \frac{\text{C}}{\text{mol} \ e^{-}}} = 5.18 \times 10^{-3} \text{ moles} \]

Key concepts

Ohm's Law

Ohm's Law is a fundamental concept in electrical engineering and physics. It states that the current (\(I\)) passing through a conductor between two points is directly proportional to the voltage (\(V\)) across the two points and inversely proportional to the resistance (\(R\)) of the conductor. The law is mathematically expressed as: