If the physical conditions remain constant, the voltage across a resistor is directly proportional to the current through it.
V ∝ I
V = kI
k = R = resistance
V = IR
The physical conditions are temperature, pressure and humidity etc.
E.g.1
When a filament lamp is switched on, the most notable physical factor – the temperature – goes up. Is Ohm’s law still valid in these circumstances for a light bulb?
Yes, only after the temperature of the filament becomes steady at some point.
E.g.2
The voltage across a resistor is 8V and the current through it is 2A. Find the resistance.
V = IR
8 = 2R
R = 4 Ω.
Change the strength of the power source, a battery in this case, and measure the current through the ammeter and the voltage across the resistor. Then plot a graph of V against I.
If the graph is a straight line that goes through the origin, it shows Ohm’s Law is correct.
The substances that obey Ohm’s Law are Ohmic conductors.
E.g. metals
The substances that do not obey Ohm’s Law are Non-Ohmic conductors.
E.g. semiconductor diodes, filament lamps
The I/V graph for Ohmic conductors is a straight line; for semiconductors and filament lamp, they are curvy.
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Calculating Kc from Known Equilibrium Amounts
Example: Calculate the value of the equilibrium constant, K_{c} , for the system shown, if 0.1908 moles of CO_{2}, 0.0908 moles of H_{2}, 0.0092 moles of CO, and 0.0092 moles of H_{2}O vapor were present in a 2.00 L reaction vessel were present at equilibrium.
[CO_{2}] = 0.1908 mol CO_{2}/2.00 L = 0.0954 M
[H_{2}] = 0.0454 M
[CO] = 0.0046 M
[H_{2}O] = 0.0046 M
Calculating K from Initial Amounts and One Known Equilibrium Amount
Example: Initially, a mixture of 0.100 M NO, 0.050 M H_{2}, 0.100 M H_{2} O was allowed to reach equilibrium (initially there was no N_{2} ). At equilibrium the concentration of NO was found to be 0.062 M. Determine the value of the equilibrium constant, K_{c} , for the reaction:
NO | H_{2} | N_{2} | H_{2}O | |
Initial Concentration (M) | 0.100 | 0.0500 | 0 | 0.100 |
Change in Concentration (M) |
– 2 x |
– 2 x |
+ x |
+ 2 x |
Equilibrium Concentration (M) | 0.062 |
The change in concentration of the NO was (0.062 M – 0.100M) = – 0.038 M. Thus -2 x = – 0.038 and x = 0.019. Note: the negative sign indicates a decreasing concentration, not a negative concentration. The changes in the other species must agree with the stoichiometry dictated by the balance equation. The hydrogen will also change by – 0.038 M, while the nitrogen will increase by + 0.019 M and the water will increase by + 0.038 M. From these changes we can complete the chart to find the equilibrium concentrations for each species.
NO | H_{2} | N_{2} | H_{2}O | |
Initial Concentration (M) | 0.100 | 0.0500 | 0 | 0.100 |
Change in Concentration (M) | – 0.038 | – 0.038 | + 0.019 | + 0.038 |
Equilibrium Concentration (M) | 0.062 | 0.012 | 0.019 | 0.138 |
Calculating Kc from Known Initial Amounts and the Known Change in Amount of One of the Species
Example: A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25^{o}C and allowed to reach equilibrium. It was found that the pressure of the nitrogen dioxide decreased by 0.952 atm. Estimate the value of K_{p} for this system:
N_{2}O_{4} | NO_{2} | |
Initial Pressure (atm)(kpa) | 3.00 | 2.00 |
Change in Pressure (atm)(kpa) | + 0.476 | – 0.952 |
Equilibrium Pressure (atm)(kpa) | 3.476 | 1.048 |
Physical Sciences Grade 10-11-12.
Physical Sciences Grade 10-11-12.
Physical Sciences Grade 10-11-12.
Research . Design . Education
Physical Sciences Grade 10-11-12.
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