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States Of Matter

Question
CBSEENCH11005900

State and explain Charle’s law.

Solution

Charle's law states:  Pressure remaining constant the volume of a given mass of gas increases or decreases by 1 over 273 space of space its volume at 0 degree straight C for each one-degree rise or fall in temperature. 
Mathematically,
 If V0 is the volume at 0°C, then the volume at various other temperatures can be written as:
     Volume space at space 10 degree straight C space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times 100 over denominator 273 end fraction
Volume space at space 40 degree straight C space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times 40 over denominator 273 end fraction
In space general comma
Volume space at space straight t degree straight C comma space space straight V subscript straight t space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times space straight t over denominator 273 end fraction
equals space straight V subscript 0 space open parentheses 1 plus straight t over 273 close parentheses
But if the gas is cooled to –273°C, then its volume becomes zero.
Volume space at space minus 273 degree straight C space equals space straight V subscript 0 open parentheses 1 minus 273 over 273 close parentheses space equals space 0
This implies that a gas at -273°C will have zero or no volume i.e. it will cease to exist. In actual practice, all gases liquefy before this temperature is reached. Also –273°C should be the lowest possible temperature because any further cooling would lead to a volume of less than zero or negative volume which is meaningless. Therefore, this temperature (–273° C) was termed as absolute zero of temperature.
Another statement of Charle’s law:
According to Charle’s law, volume of a given mass of a gas at different temperatures is related to its volume (V0) at 0°C as follows:
Volume at Volume space at space straight t subscript 1 superscript degree straight C comma
straight V subscript straight t subscript 1 end subscript space equals space straight V subscript 0 space open square brackets 1 plus straight t subscript 1 over 273 close square brackets space equals space straight V subscript 0 space open square brackets fraction numerator 273 space plus space straight t subscript 1 over denominator 273 end fraction close square brackets
straight V subscript straight t subscript 1 end subscript space equals space straight V subscript 0 straight T subscript 1 over 273 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Volume space at space straight t subscript 2 superscript degree straight C comma space space space space straight V subscript straight t subscript 2 end subscript space equals space straight V subscript 0 space open square brackets 1 plus straight t subscript 2 over 273 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight V subscript 0 open square brackets fraction numerator 273 plus straight t subscript 2 over denominator 273 end fraction close square brackets
straight V subscript straight t subscript 2 end subscript space equals space straight V subscript 0 space straight T subscript 2 over 273 space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Dividing (1) by (2), we get,
     
 straight V subscript straight t subscript 1 end subscript over straight V subscript straight t subscript 2 end subscript space equals space fraction numerator straight V subscript 0 straight T subscript 1 over denominator 273 end fraction cross times fraction numerator 273 over denominator straight V subscript 0 straight T subscript 2 end fraction space equals space straight T subscript 1 over straight T subscript 2
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[Pressure and mass of the gas constant]
Hence Charle’s law may also be stated as Pressure remaining constant, the volume of a given mass of a gas is directly proportional to absolute temperature i.e. straight V over straight T space equals space constant. space
Thus, if V1 is the initial volume of the gas at temperature T1 (in degree kelvin) and V2 is the final volume of the gas at temperature T2 (in degree kelvin), keeping pressure constant, then
straight V subscript 1 over straight T subscript 1 space equals space straight V subscript 2 over straight T subscript 2