1.

In terms of Charles. law explain why -273^(@)C is the lowest possible temperature.

Answer»

Solution :Eq. of Charle.s : `V_(t)=V_(0)((273.15+t)/(273.15))`….(Eq.-i)
Put `t=-273/15^(@)C` in above equation.
`THEREFORE V_(t)=V_(0)((273.15-273.5)/(273.15))`
`therefore V_(t)=V_(0)((0)/(273.15))`
`therefore V_(t)=0=V_(-273.15)`
Therefore the VOLUME BECOME zero at the temperature `- 273.15^(@)C`. Actually all the GASES liquifies before reaching such low absolute `(-273.15^(@)C)` temperature and this is the least probable temperature.
At this `- 273.15^(@)C` temperature the volume of the gases is zero and it is called absolute zero.


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