For Realms Of Royalty Academy SS1 Student
CHARLES’ LAW
The effect of temperature changes on the volume of a given mass of a gas at a constant pressure is described by Charles. Charles’ law states that the volume of a given mass of gas is directly proportional to its temperature in Kelvin, provided that pressure remains constant.
The volume of the gas decreases as the temperature decreases, and increases as the temperature increases.
Mathematically, the law can be expressed as:
V∝T∴V=kT
or VT=k
Where V = volume
T = Kelvin Temperature
K = mathematical constant
A Representation of Charles’s law
For a direct relationship, when the temperature increases, the volume will also increase at the same rate and vice versa, at constant pressure .The diagram above shows that when V is decreasing, T is also decreasing and when V is increasing, T is also increasing thus, making the quotient constant.
Charles’s law can be represented graphically has shown below.
If we divide the varying gas volumes by the corresponding temperature in Kelvin, the result would always be a constant. This relationship can also be expressed in another form.
V1T1=V2T2∴V2=T2V1T1
Where V1 is the volume at temperature T1
V2 is the volume at temperature T2
ABSOLUTE ZERO
This is the temperature at which the volume of a gas is theoretically zero..At this temperature there is no motiom of any form and all gases have been liquefied or solidified. The value of the temperature is -2730C.
TEMPERATURE CONVERSION
To convert from Celsius scale to Kelvin scale, add 273 i.e. T = 0C + 273. This is because O0C = 273K.
To convert from Kelvin scale to Celsius scale, subtract 273. i.e 0C = T − 273.
Where T = Temperature in Kevin
0C = Temperature in Celsius.
Examples:
- Convert the following Celsius temperature to Kelvin temperature.
(a) 1000C (b) 00C (c) -570C
Solution
Recall: T = 0C + 273
(a) 1000C = (100 + 273) = 373k
0C =(0 + 273) = (0 + 273) = 373k
(b) −570c = (−57 + 273)k = (273 − 57) = 216k - Convert the following Kelvin temperatures to Celsius temperature.
(a) 298k (b) 405k (b) 285k (d) 0k
Solution
Recall 00c = k – 273
298k = (298 – 273)0C = 250C
405k = (405 – 273)0C = 120C
0k = (0 – 273)0C = − 2730C
Worked examples on Charles’s law - A gas occupies a volume of 20.0dm3 at 373k. Its volume at 746k at that pressure will be?
Here pressure is constant. Charles’s law will apply.
V1 = 20.0dm3
T1 = 273k
T2 = 746
Recall Charles’s law V1T1=V2T2V2=T2V1T1V2=20×746273=40.0dm3
EVALUATION:
State Charles’s law
Express the two laws mathematically
Draw two graphs to illustrate Charles’ law.
